Eureka Math

Engage NY Eureka Math 6th Grade Module 2 Lesson 1 Answer Key

Eureka Math Grade 6 Module 2 Lesson 1 Example Answer Key

Example 1.
\(\frac{3}{4}\) gallon of batter is poured equally into 2 bowls. How many gallons of batter are in each bowl?
Answer:
→ Since the whole is being put into 2 equal parts or bowls, this is the partitive model of division. What division expression can we write to match this story?
→ \(\frac{3}{4}\) ÷ 2
→ We can also think of this problem as asking, “3 fourths is 2 groups of what?” Let’s take a look at how to solve this using an area model. What is the whole? How much batter is being shared?
→ Three-fourths gallon
→ The story tells us that the batter is poured equally into 2 bowls. How can we show this in the model?
→ We can draw another line, horizontally, to partition the model into 2 equal units.
→ Now our model shows 3 fourths being partitioned into 2 equal parts. Could we also say that our model shows \(\frac{1}{2}\) of 3 fourths?
→ Yes, it’s just like in the Opening Exercise where dividing by 2 and multiplying by are the same.
→ Let’s label our model to show that. What is half of 3 fourths?
Half of 3 fourths is 3 eighths.
→ Yes, and what is 3 fourths divided by 2?
→ It’s also 3 eighths!
→ We said that we could also think of this problem as, “3 fourths is 2 groups of what?” Is it true that 3 fourths is 2 groups of 3 eighths? Use a repeated addition or multiplication sentence to support your response.
→ Yes, it’s true. 3 eighths plus 3 eighths is 6 eighths, which is equal to 3 fourths. →. It’s true.
2 × \(\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\).

\(\frac{3}{4}\) ÷ 2 = \(\frac{1}{2}\) of \(\frac{3}{4}\) = \(\frac{3}{8}\)
\(\frac{3}{8}\) gallon of batter in each bowl.

Example 2.
\(\frac{3}{4}\) gallon of batter is poured equally into 2 bowls. How many gallons of batter are in each bowl?
Answer:
→ Again, this is partitive division problem since we’re told that there are 6 parts, or that the lasagna is being shared equally among 6 friends. Write a division expression to represent this story problem.
→ \(\frac{3}{4}\) ÷ 6
→ Remember when solving this problem, we can also think, “3 fourths is 6 groups of what?” Let’s draw another area model to solve. Think, what is our whole? Draw a model to show how much lasagna is being shared.
→ Now, partition your model again to show how it can be equally shared by 6 friends.
→ Remind me; dividing by 6 is the same as multiplying by what?
→ Multiplying by \(\frac{1}{6}\)
→ Look at our model. Explain to your neighbor how it shows division by 6 and also shows multiplication by \(\frac{1}{6}\).
→ Now write a multiplication expression that is equal to 3 fourths divided by 6.
→ \(\frac{1}{6}\) of \(\frac{3}{4}\) or \(\frac{1}{6}\) × \(\frac{3}{4}\)

\(\frac{3}{4}\) ÷ 6 = \(\frac{1}{6}\) of \(\frac{3}{4}\) = \(\frac{3}{24}\) = \(\frac{1}{8}\)

→ According to our model, what fraction of the pan will each friend get?
→ 3 twenty-fourths of the pan
→ Express 3 twenty-fourths in its simplest form.
1 eighth

Example 3.
A rope of length \(\frac{2}{5}\) m is cut into 4 equal cords. What is the length of each cord?
Answer:
→ Again, this is a partitive division problem since we’re told that there are 4 parts or that the rope is cut equally into 4 cords. Write a division expression to represent this story problem.
\(\frac{2}{5}\) ÷ 4
→ For this example, let’s draw a tape diagram and a number line.
→ The length of rope is cut into 4 equal cords. How can we show that in our models?
Partition the 2 shaded units into 4 equal parts.

→ We originally drew fifths and shaded 2 of them to represent the length of the rope, but now what fractional unit does our model show?
Tenths
→ Answer the question. What is the length of each cord?
Each cord is \(\frac{1}{10}\) meter long.
\(\frac{2}{5} \div 4=\frac{1}{4} \text { of } \frac{2}{5}=\frac{2}{20}=\frac{1}{10}\)
Each chord is \(\frac{1}{10}\) m.
\(\frac{2}{5}\) ÷ 4 = 4 tenths ÷ 4 = 1 tenth = \(\frac{1}{10}\)
→ Take a look at the division sentence I’ve written using unit language. How do our models support this thought? How does the use of unit language support your understanding of this division problem?
→ We can also construct a number line to support our solution. When the number line is drawn beneath the tape diagram, we can see the similarities between the two models.
Encourage students to study the models and discuss the similarities between dividing by 4 and multiplying by \(\frac{1}{4}\). Students should articulate that both operations yield the same result.

Eureka Math Grade 6 Module 2 Lesson 1 Exercise Answer Key

Exercises 1.
Fill in the blanks to complete the equation. Then, find the quotient and draw a model to support your solution.

Answer:

\(\frac{1}{2} \div 3=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)

Exercises 2.

Answer:

\(\frac{1}{3} \div 4=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)
\(\frac{1}{3}\) ÷ 4 = 4 twefiths ÷ 4 = 1 twelfth = \(\frac{1}{12}\)

Find the value of each of the following.

Exercises 3.
\(\frac{1}{4}\) ÷ 5
Answer:
\(\frac{1}{4} \div 5=\frac{1}{5} \text { of } \frac{1}{4}=\frac{1}{20}\)

Exercises 4.
\(\frac{3}{4}\) ÷ 5
Answer:
\(\frac{3}{4} \div 5=\frac{1}{5} \text { of } \frac{3}{4}=\frac{3}{20}\)

Exercises 5.
\(\frac{1}{5}\) ÷ 4
Answer:
\(\frac{1}{5} \div 4=\frac{1}{4} \text { of } \frac{1}{5}=\frac{1}{20}\)

Solve. Draw a model to support your solution.

Exercises 6.
\(\frac{3}{5}\)pt. of juice is poured equally into 6 glasses. How much juice is in each glass?
Answer:

Eureka Math Grade 6 Module 2 Lesson 1 Problem Set Answer Key

Find the value of each of the following in its simplest form.

Question 1.
a. \(\frac{1}{3}\) ÷ 4
Answer:
\(\frac{1}{12}\)

b. \(\frac{2}{5}\) ÷ 4
Answer:
\(\frac{1}{10}\)

c. \(\frac{4}{7}\) ÷ 4
Answer:
\(\frac{1}{7}\)

Question 2.
a. \(\frac{2}{5}\) ÷ 3
Answer:
\(\frac{2}{15}\)

b. \(\frac{5}{6}\) ÷ 5
Answer:
\(\frac{1}{6}\)

c. \(\frac{5}{8}\) ÷ 10
Answer:
\(\frac{1}{16}\)

Question 3.
a. \(\frac{6}{7}\) ÷ 3
Answer:
\(\frac{2}{7}\)

b. \(\frac{10}{8}\) ÷ 5
Answer:
\(\frac{1}{4}\)

c. \(\frac{20}{6}\) ÷ 2
Answer:
\(\frac{5}{3}\)

Question 4.
4 loads of stone weigh \(\frac{2}{3}\) ton. Find the weight of 1 load of stone.
Answer:
\(\frac{2}{3}\) ÷ 4 = \(\frac{1}{6}\)
Each load of stone weighs \(\frac{1}{6}\) ton.

Question 5.
What is the width of a rectangle with an area of in and a length of 10 inches?
Answer:
\(\frac{5}{8}\) ÷ 10 = \(\frac{1}{16}\)
The width of the rectangle is \(\frac{1}{16}\) in.

Question 6.
Lenox ironed \(\frac{1}{4}\) of the shirts over the weekend. She plans to split the remainder of the work equally over the next 5 evenings.
a. What fraction of the shirts will Lenox iron each day after school?
Answer:
\(\frac{3}{4}\) ÷ 5 = \(\frac{3}{20}\)
Lenox will iron \(\frac{3}{20}\) of the shirts each day after school.

b If Lenox has 40 shirts how many shirts wlll she need to iron on Thursday and Friday?
Answer:
\(\frac{3}{20}\)(40) = 6
Lenox will need to iron 6 shirts on Thursday and 6 shirts on Friday.
Therefore, Lenox will need to iron 12 shirts on Thursday and Friday.

Question 7.
Bo paid bills with \(\frac{1}{2}\) of his paycheck and put \(\frac{1}{5}\) of the remainder in savings. The rest of his paycheck he divided equally among the college accounts of his 3 children.
a. What fraction of his paycheck went into each child’s account?
Answer:
\(\frac{1}{2}\left(\frac{1}{5}\right)=\frac{1}{10}\). This means \(\frac{1}{10}\) of Bo’s paycheck goes into savings. \(\frac{1}{2}+\frac{1}{10}=\frac{5}{10}+\frac{1}{10}=\frac{6}{10}=\frac{3}{5}\).
This means that \(\frac{3}{5}\) of Bo’s paycheck goes to bills and savings, which leaves \(\frac{2}{5}\) of his paycheck for college accounts.
\(\frac{2}{5}\) ÷ 3 = \(\frac{2}{15}\)
Therefore, Bo put \(\frac{2}{15}\) of his paycheck in each child’s account.

b. If Bo deposited $400 in each child’s account, how much money was in Bo’s original paycheck?
Answer:
$400 is \(\frac{2}{15}\) group of what size?
2 unIts = $400
1 unit= $400 ÷ 2 = $200
15 units = 15 × $200 = $3, 000
Bo was originally paid $3, 000.

Eureka Math Grade 6 Module 2 Lesson 1 Exit Ticket Answer Key

Write an equivalent multiplication expression. Then, find the quotient In its simplest form. Use a model to support your response.

Question 1.
\(\frac{1}{4}\) ÷ 2
Answer:

\(\frac{1}{4} \div 2=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\)

Question 2.
\(\frac{2}{3}\) ÷ 6
Answer:

\(\frac{2}{3} \div 6=\frac{1}{6} \times \frac{2}{3}=\frac{2}{18}=\frac{1}{9}\)

Engage NY Eureka Math Grade 6 Module 5 Lesson 2 Answer Key

Eureka Math Grade 6 Module 5 Lesson 2 Exploratory Challenge Answer Key

Exploratory Challenge:

Question a.
Use the shapes labeled with an X to predict the formula needed to calculate the area of a right triangle. Explain your prediction.
Formula for the area of right triangles:
Area of the given triangle: ________
Answer:
Formula for the area of right triangles:
A = \(\frac{1}{2}\) × base × height or A = \(\frac{\text { base } \times \text { height }}{2}\)

Area of the given triangle:
A = \(\frac{1}{2}\) × 3 in. × 2 in. = 3 in²

Question b.
Use the shapes labeled with a Y to determine if the formula you discovered in part (a) is correct.

Does your area formula for triangle Y match the formula you got for triangle X?
Answer:
Answers will vary; however, the area formulas should be the same if students discovered the correct area
formula.

If so, do you believe you have the correct formula needed to calculate the area of a right triangle? Why or why not?
Answer:
Answers will vary.

If not, which formula do you think is correct? Why?
Answer:
Answers will vary.

Area of the given triangle:

A = \(\frac{1}{2}\) × 3 in. × 3 in. = 4.5 im ²

Eureka Math Grade 6 Module 5 Lesson 2 Exercise Answer Key

Exercises:

Calculate the area of each right triangle below. Each figure is not drawn to scale.

Question 1.

A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (8 ft.) (15 ft.)
= 60 ft²

Question 2.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (11.4 cm) (17.7 cm)
= 100.89 cm²

Question 3.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (6 in.) (8 in.)
= 24 in²

Question 4.

Answer:

Question 5.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (32.7 km) (21.4 km)
= 349.89 km²

Question 6.
Mr. Jones told his students they each need half of a piece of paper. Calvin cut his piece of paper horizontally, and Matthew cut his piece of paper diagonally. Which student has the larger area on his half piece of paper? Explain.

Answer:
After cutting the paper, both Calvin and Matthew have the same area. Calvin cut his into two rectangles that are each half the area of the original piece of paper. Matthew cut his paper into two equivalent right triangles that are also half the area of the original piece of paper.

Question 7.
Ben requested that the rectangular stage be split into two equal sections for the upcoming school play. The only instruction he gave was that he needed the area of each section to be half of the original size. If Ben wants the stage to be split into two right triangles, did he provide enough information? Why or why not?
Answer:
Ben did not provide enough information because the stage may be split horizontally or vertically through the middle of the rectangle. This would result in two equal pieces, but they would not be right triangles.

Question 8.
If the area of a right triangle is 6. 22 sq. in. and its base is 3. 11 in., write an equation that relates the area to the height, h, and the base. Solve the equation to determine the height.
Answer:
6.22 = \(\frac{1}{2}\) (3. 11)h
6.22 = (1.555)h
6.22 ÷ 1.555 = (1. 555)h ÷ 1.555
4 = h
Therefore, the height of the right triangle is 4 in.

Eureka Math Grade 6 Module 5 Lesson 2 Problem Set Answer Key

Calculate the area of each right triangle below. Note that the figures are not drawn to scale.

Question 1.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (31.2 cm) (9.1 cm) = 141.96 cm²

Question 2.

Answer:

Question 3.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (2.4 in.) (3.2 in.)
= 3.84 in²

Question 4.

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (11 mm) (60 mm)
= 330 mm²

Question 5.

Answer:

Question 6.
Elania has two congruent rugs at her house. She cut one vertically down the middle, and she cut diagonally through the other one.

After making the cuts, which rug (labeled A, B, C, or D) has the larger area? Explain.
Answer:
All of the rugs are the same size after making the cuts. The vertical line goes down the center of the rectangle, making two congruent parts. The diagonal line also splits the rectangle into two congruent parts because the area of a right triangle is exactly half the area of the rectangle.

Question 7.
Give the dimensions of a right triangle and a parallelogram with the same area. Explain how you know.
Answer:
Answers will vary.

Question 8.
If the area of a right triangle is \(\frac{9}{16}\) sq. ft. and the height is \(\frac{3}{4}\) ft., write an equation that relates the area to the base, b, and the height. Solve the equation to determine the base.
Answer:

Therefore, the base of the right triangle is 1\(\frac{1}{2}\) ft.

Eureka Math Grade 6 Module 5 Lesson 2 Exit Ticket Answer Key

Question 1.
Calculate the area of the right triangle. Each figure is not drawn to scale.
Answer:

A = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) (8 in.)(6 in.) = 24 in²

Question 2.
Dan and Joe are responsible for cutting the grass on the local high school soccer field. Joe cuts a diagonal line through the field, as shown in the diagram below, and says that each person is responsible for cutting the grass on one side of the line. Dan says that this is not fair because he will have to cut more grass than Joe. Is Dan correct? Why or why not?

Answer:
Dan is not correct. The diagonal line Joe cut in the grass would split the field into two right triangles. The area of each triangle is exactly half the area of the entire field because the area formula for a right triangle is
A = \(\frac{1}{2}\) × base × height.

Engage NY Eureka Math Grade 6 Module 5 Lesson 1 Answer Key

Eureka Math Grade 6 Module 5 Lesson 1 Exercise Answer Key

Opening Exercise:

Question 1.
Name each shape.

a.
Answer:

b.
Answer:

c.
Answer:

d.
Answer:

e.
Answer:

Exercises:

Question 1.
Find the area of each parallelogram below. Note that the figures are not drawn to scale.

a.
Answer:
A = bh
= (6 cm)(4 cm)
= 24 cm²

b.
A = bh
= (25 m)(8 m)
= 200 m²

c.
A = bh
= (12 ft.)(7 ft.)
84 ft²

Question 2.
Draw and label the height of each parallelogram. Use the correct mathematical tool to measure (in inches) the base and height, and calculate the area of each parallelogram.

a.
Answer:

A = bh = (0.5 in.) (2 in.) = 1 in²

b.
Answer:

A = bh = (1.5 in.) (2 in.) = 3 in²

c.
Answer:

A = bh = (1 in.) (1 in.) = 1 in²

Question 3.
If the area of a parallelogram is \(\frac{35}{42}\) cm² and the height is \(\frac{1}{7}\) cm, write an equation that relates the height, base, and area of the parallelogram. Solve the equation.
Answer:

Therefore, the base is 5\(\frac{5}{6}\) cm.

Eureka Math Grade 6 Module 5 Lesson 1 Problem Set Answer Key

Draw and label the height of each parallelogram.

Question 1.

Answer:

Question 2.

Answer:

Calculate the area of each parallelogram. The figures are not drawn to scale.

Question 3.

Answer:
A = bh
= (13 cm) (6 cm)
= 78 cm²

Question 4.

Answer:
A = bh
= (1.2 ft.) (12.8 ft)
= 15.36 ft²

Question 5.

Answer:

Question 6.

Answer:

Question 7.
Brittany and Sid were both asked to draw the height of a parallelogram. Their answers are below.

Are both Brittany and Sid correct? If not, who is correct? Explain your answer
Answer:
Both Brittany and Sid are correct because both of their heights represent a line segment that is perpendicular to the base and whose endpoint is on the opposite side of the parallelogram.

Question 8.
Do the rectangle and parallelogram below have the same area? Explain why or why not.

Answer:
Yes, the rectangle and parallelogram have the same area because if we cut off the right triangle on the left side of the parallelogram, we can move it over to the right side and make the parallelogram into a rectangle. After transforming the parallelogram into a rectangle, both rectangles would have the same dimensions; therefore, their areas would be the same.

Question 9.
A parallelogram has an area of 20.3 cm² and a base of 2. 5 cm. Write an equation that relates the area to the base and height, h. Solve the equation to determine the height of the parallelogram.
Answer:
20.3 = (2.5)(h)
20.3 ÷ 2.5 = (2. 5)(h) ÷ 2.5
8.12 = h
Therefore, the height of the parallelogram is 8. 12 cm.

Eureka Math Grade 6 Module 5 Lesson 1 Exit Ticket Answer Key

Calculate the area of each parallelogram. Note that the figures are not drawn to scale.

Question 1.

Answer:
A = bh = (20 ft.) (10 ft.) = 200 ft²

Question 2.

Answer:
A = bh = (5 cm) (35 cm) = 175 cm²

Question 3.

Answer:
A = bh = (7 m) (2 m) = 14 m²

Eureka Math Grade 6 Module 5 Lesson 1 Multiplication of Fractions Answer Key

Multiplication of Fractions I – Round 1

Directions: Determine the product of the fractions and simplify.

Question 1.
\(\frac{1}{2}\) × \(\frac{3}{4}\)
Answer:
\(\frac{3}{8}\)

Question 2.
\(\frac{5}{6}\) × \(\frac{5}{7}\)
Answer:
\(\frac{25}{42}\)

Question 3.
\(\frac{3}{4}\) × \(\frac{7}{8}\)
Answer:
\(\frac{21}{32}\)

Question 4.
\(\frac{4}{5}\) × \(\frac{8}{9}\)
Answer:
\(\frac{32}{45}\)

Question 5.
\(\frac{1}{4}\) × \(\frac{3}{7}\)
Answer:
\(\frac{3}{28}\)

Question 6.
\(\frac{5}{7}\) × \(\frac{4}{9}\)
Answer:
\(\frac{20}{63}\)

Question 7.
\(\frac{3}{5}\) × \(\frac{1}{8}\)
Answer:
\(\frac{3}{40}\)

Question 8.
\(\frac{2}{9}\) × \(\frac{7}{9}\)
Answer:
\(\frac{14}{81}\)

Question 9.
\(\frac{1}{3}\) × \(\frac{2}{5}\)
Answer:
\(\frac{2}{15}\)

Question 10.
\(\frac{3}{7}\) × \(\frac{5}{8}\)
Answer:
\(\frac{15}{56}\)

Question 11.
\(\frac{2}{3}\) × \(\frac{9}{10}\)
Answer:
\(\frac{18}{30}\) = \(\frac{3}{5}\)

Question 12.
\(\frac{3}{5}\) × \(\frac{1}{6}\)
Answer:
\(\frac{3}{30}\) = \(\frac{1}{10}\)

Question 13.
\(\frac{2}{7}\) × \(\frac{3}{4}\)
Answer:
\(\frac{6}{28}\) = \(\frac{3}{14}\)

Question 14.
\(\frac{5}{8}\) × \(\frac{3}{10}\)
Answer:
\(\frac{15}{80}\)
= \(\frac{3}{16}\)

Question 15.
\(\frac{4}{5}\) × \(\frac{7}{8}\)
Answer:
\(\frac{28}{40}\) = \(\frac{7}{10}\)

Question 16.
\(\frac{8}{9}\) × \(\frac{3}{4}\)
Answer:
\(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 17.
\(\frac{3}{4}\) × \(\frac{4}{7}\)
Answer:
\(\frac{12}{28}\) = \(\frac{3}{7}\)

Question 18.
\(\frac{1}{4}\) × \(\frac{8}{9}\)
Answer:
\(\frac{8}{36}\) = \(\frac{2}{9}\)

Question 19.
\(\frac{3}{5}\) × \(\frac{10}{11}\)
Answer:
\(\frac{30}{55}\) = \(\frac{6}{11}\)

Question 20.
\(\frac{8}{13}\) × \(\frac{7}{24}\)
Answer:
\(\frac{56}{312}\) = \(\frac{7}{39}\)

Question 21.
2\(\frac{1}{2}\) × 3\(\frac{3}{4}\)
Answer: \(\frac{75}{8}\) = 9\(\frac{3}{8}\)

Question 22.
1\(\frac{4}{5}\) × 6\(\frac{1}{3}\)
Answer:
\(\frac{171}{15}\) = 11\(\frac{2}{5}\)

Question 23.
8\(\frac{2}{7}\) × 4\(\frac{5}{6}\)
Answer:
\(\frac{1682}{42}\) = 40\(\frac{1}{21}\)

Question 24.
5\(\frac{2}{5}\) × 2\(\frac{1}{8}\)
Answer:
\(\frac{459}{40}\) = 11\(\frac{19}{40}\)

Question 25.
4\(\frac{6}{7}\) × 1\(\frac{1}{4}\)
Answer:
\(\frac{170}{28}\) = 6 \(\frac{1}{14}\)

Question 26.
2\(\frac{2}{3}\) × 4\(\frac{2}{5}\)
Answer:
\(\frac{176}{15}\) = 11 \(\frac{11}{15}\)

Question 27.
6\(\frac{9}{10}\) × 7\(\frac{1}{3}\)
Answer:
\(\frac{1518}{30}\) = 50 \(\frac{3}{5}\)

Question 28.
1\(\frac{3}{8}\) × 4\(\frac{2}{5}\)
Answer:
\(\frac{242}{40}\) = 6 \(\frac{1}{20}\)

Question 29.
3\(\frac{5}{6}\) × 2\(\frac{4}{15}\)
Answer:
\(\frac{782}{90}\) = 8 \(\frac{31}{45}\)

Question 30.
4\(\frac{1}{3}\) × 5
Answer:
\(\frac{65}{3}\) = 21 \(\frac{2}{3}\)

Multiplication of Fractions I – Round 2

Directions: Determine the product of the fractions and simplify.

Question 1.
\(\frac{5}{6}\) × \(\frac{1}{4}\)
Answer:
\(\frac{5}{24}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{5}{7}\)
Answer:
\(\frac{10}{21}\)

Question 3.
\(\frac{1}{3}\) × \(\frac{2}{5}\)
Answer:
\(\frac{2}{15}\)

Question 4.
\(\frac{5}{7}\) × \(\frac{5}{8}\)
Answer:
\(\frac{25}{56}\)

Question 5.
\(\frac{3}{8}\) × \(\frac{7}{9}\)
Answer:
\(\frac{21}{72}\) = \(\frac{7}{24}\)

Question 6.
\(\frac{3}{4}\) × \(\frac{5}{6}\)
Answer:
\(\frac{15}{24}\) = \(\frac{5}{8}\)

Question 7.
\(\frac{2}{7}\) × \(\frac{3}{8}\)
Answer:
\(\frac{6}{56}\) = \(\frac{3}{28}\)

Question 8.
\(\frac{1}{4}\) × \(\frac{3}{4}\)
Answer:
\(\frac{3}{16}\)

Question 9.
\(\frac{5}{8}\) × \(\frac{3}{10}\)
Answer:
\(\frac{15}{80}\) = \(\frac{3}{16}\)

Question 10.
\(\frac{6}{11}\) × \(\frac{1}{2}\)
Answer:
\(\frac{6}{22}\) = \(\frac{3}{11}\)

Question 11.
\(\frac{6}{7}\) × \(\frac{5}{8}\)
Answer:
\(\frac{30}{56}\) = \(\frac{15}{28}\)

Question 12.
\(\frac{1}{6}\) × \(\frac{9}{10}\)
Answer:
\(\frac{9}{60}\) = \(\frac{3}{20}\)

Question 13.
\(\frac{3}{4}\) × \(\frac{8}{9}\)
Answer:
\(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 14.
\(\frac{5}{6}\) × \(\frac{2}{3}\)
Answer:
\(\frac{10}{18}\) = \(\frac{5}{9}\)

Question 15.
\(\frac{1}{4}\) × \(\frac{8}{11}\)
Answer:
\(\frac{8}{44}\) = \(\frac{2}{11}\)

Question 16.
\(\frac{3}{7}\) × \(\frac{2}{9}\)
Answer:
\(\frac{6}{63}\) = \(\frac{2}{21}\)

Question 17.
\(\frac{4}{5}\) × \(\frac{10}{13}\)
Answer:
\(\frac{40}{65}\) = \(\frac{8}{13}\)

Question 18.
\(\frac{2}{9}\) × \(\frac{3}{8}\)
Answer:
\(\frac{6}{72}\) = \(\frac{1}{12}\)

Question 19.
\(\frac{1}{8}\) × \(\frac{4}{5}\)
Answer:
\(\frac{4}{40}\) = \(\frac{1}{10}\)

Question 20.
\(\frac{3}{7}\) × \(\frac{2}{15}\)
Answer:
\(\frac{6}{105}\) = \(\frac{2}{35}\)

Question 21.
1\(\frac{1}{2}\) × 4\(\frac{3}{4}\)
Answer:
\(\frac{57}{8}\) = 7\(\frac{1}{8}\)

Question 22.
2\(\frac{5}{6}\) × 3\(\frac{3}{8}\)
Answer:
\(\frac{459}{48}\) = 9\(\frac{9}{16}\)

Question 23.
1\(\frac{7}{8}\) × 5\(\frac{1}{5}\)
Answer:
\(\frac{390}{40}\) = 9\(\frac{3}{4}\)

Question 24.
6\(\frac{2}{3}\) × 2\(\frac{3}{8}\)
Answer:
\(\frac{380}{24}\) = 15\(\frac{5}{6}\)

Question 25.
7\(\frac{1}{2}\) × 3\(\frac{6}{7}\)
Answer:
\(\frac{405}{14}\) = 28\(\frac{13}{14}\)

Question 26.
3 × 4\(\frac{1}{3}\)
Answer:
\(\frac{39}{3}\) = 13

Question 27.
2\(\frac{3}{5}\) × 5\(\frac{1}{6}\)
Answer:
\(\frac{403}{30}\) = 13\(\frac{13}{30}\)

Question 28.
4\(\frac{2}{5}\) × 7
Answer:
\(\frac{154}{5}\) = 30\(\frac{4}{5}\)

Question 29.
1\(\frac{4}{7}\) × 2\(\frac{1}{2}\)
Answer:
\(\frac{55}{14}\) = 3\(\frac{13}{14}\)

Question 30.
3\(\frac{5}{6}\) × \(\frac{3}{10}\)
Answer:
\(\frac{69}{60}\) = 1\(\frac{3}{20}\)

Engage NY Eureka Math Grade 6 Module 6 Lesson 22 Answer Key

Eureka Math Grade 6 Module 6 Lesson 22 Exploratory Challenge Answer Key

A statistical study involves the following four-step investigative process:

Step 1: Pose a question that can be answered by data.
Step 2: Collect appropriate data.
Step 3: Summarize the data with graphs and numerical summaries.
Step 4: Answer the question posed in Step 1 using the numerical summaries and graphs.

Now it is your turn to be a researcher and to present your own statistical study. In Lesson 17, you posed a statistical question, proposed a plan to collect data to answer the question, and collected the data. In Lesson 21, you created a poster or an outline of a presentation that included the following: the statistical question, the plan you used to collect the data, graphs and numerical summaries of the data, and an answer to the statistical question based on your data. Use the following table to organize your presentation.

Answer:

Engage NY Eureka Math Grade 6 Module 6 Lesson 21 Answer Key

Eureka Math Grade 6 Module 6 Lesson 21 Exploratory Challenge Answer Key

Exploratory Challenge: Annual Rainfall in the State of New York

Question 1.
The National Climate Data Center collects data throughout the United States that can be used to summarize the climate of a region. You can obtain climate data for a state, a city, a county, or a region. If you were interested in researching the climate in your area, what data would you collect? Explain why you think these data would be important in a statistical study of the climate in your area.
Answer:
Answers will vary. Anticipate that student responses will include things like temperature, amount of rainfall, number of sunny days per year, number of tornadoes per year, etc.

For this lesson, you will use yearly rainfall data for the state of New York that were compiled the National Climate Data Center. The following data are the number of inches of rain (averaged over various locations in the state) for the years from 1983 to 2012 (30 years).

Use the four steps to carry out a statistical study using these data.

Step 1: Pose a question that can be answered by data.

What is a statistical question that you think can be answered with these data? Write your question in the template provided for this lesson.

Step 2: Collect appropriate data.

The data have already been collected for this lesson. How do you think these data were collected? Recall that the data are the number of inches of rain (averaged over various locations in the state) for the years from 1983 to 2012 (30 years). Write a summary of how you think the data were collected in the template for this lesson.

Step 3: Summarize the data with graphs and numerical summaries.

A good first step might be to summarize the data with a dot plot. What other graph might you construct? Construct a dot plot or another appropriate graph in the template for this lesson.

What numerical summaries will you calculate? What measure of center will you use to describe a typical value for these data? What measure of variability will you calculate and use to summarize the variability of the data? Calculate the numerical summaries, and write them in the template for this lesson.

Step 4: Answer your statistical question using the numerical summaries and graphs.

Write a summary that answers the question you posed in the template for this lesson.

Eureka Math Grade 6 Module 6 Lesson 21 Template Answer Key

Template for Lesson 21

Step 1: Pose a question that can be answered by data.

What is a statistical question that you think can be answered with these data? Write your question in the template provided for this lesson.

It is important that students are reminded of the two most important parts of the definition of a statistical question. A statistical question is (1) a question that is answered by data and (2) a question that anticipates the data will vary. As students examine the data, point out to them that there is variability.

Although students may use different wording in their statistical questions, it is anticipated that most students will form a question that essentially asks, “What is the typical annual rainfall in New York?”

Step 2: How do you think the data were collected?

The data have already been collected for this lesson. How do you think these data were collected? Recall that the data are the number of inches of rain (averaged over various locations in the state) for the years from 1983 to 2012 (30 years). Write a summary of how you think the data were collected in the template for this lesson.

The data are provided in this lesson. This step is more challenging as they carry out their own statistical study because they need to explain the plan they developed to collect their data. For this lesson, allow students to speculate on how the National Climate Data Center might have collected these data.

As the data represent the annual rainfall for the state of New York, the center had to collect rainfall totals from several reporting weather centers around the state. They calculated an average of those levels for each day of the year. At the end of the year, the National Climate Data Center added those daily results together. Students might be asked how a rainfall level is measured at a weather center. A rain gauge might be a good visual to share with students.

Step 3: Construct graphs and calculate numerical summaries of the data.

1. A good first step might be to summarize the data with a dot plot. What other graph might you construct? Construct a dot plot or another appropriate graph in the template for this lesson.

2. What numerical summaries will you calculate? What measure of center will you use to describe a typical value for these data? What measure of variability will you calculate and use to summarize the variability of the data? Calculate the numerical summaries, and write them in the template for this lesson.

This step represents most of the work students are expected to do in this lesson. As a first step, encourage students who are not sure how to start summarizing the data to construct a dot plot. A blank grid is provided at the end of the Teacher Notes that can be duplicated for students who may need some structure in making a dot plot. This grid could also be used if any student decides to develop a box plot or a histogram of the data distribution.

Students examine the dot plot and make decisions about the data distribution. For example, is the distribution
approximately symmetric, or is the distribution skewed? This dot plot shows a data distribution that is approximately symmetric.

Based on the decision that the distribution is approximately symmetric, students should proceed to calculate the mean as a measure of center and the MAD as a measure of variability. Some students might also choose to investigate this distribution with a box plot to answer the question about the symmetry.

The box plot does not indicate a perfectly symmetrical distribution; however, it is approximately symmetric. The mean and the median of this data distribution are nearly equal to each other. Based on this decision, students should calculate the mean and the MAD.

The mean rainfall is 43 inches (to the nearest inch), and the mean absolute deviation (MAD) is 3.9 or 4 inches (to the nearest inch).

At the end of the Teacher Notes is a table that could be used for students who need structure in calculating the MAD. There are several steps in calculating the MAD, and some students may need help in organizing these steps.

Step 4: Answer your statistical question using the numerical summaries and graphs.

Write a summary that answers the question you posed in the template for this lesson.

This step asks students to write a short summary interpreting the graphs and numerical summaries. Students should connect this back to their statistical questions. Students would indicate that the typical rainfall for New York is about 43 inches per year. They would also indicate that a typical distance from the mean is about 4 inches.

Eureka Math Grade 6 Module 6 Lesson 21 Problem Set Answer Key

In Lesson 17, you posed a statistical question and created a plan to collect data to answer your question. You also constructed graphs and calculated numerical summaries of your data. Review the data collected and your summaries.

Based on directions from your teacher, create a poster or an outline for a presentation using your own data. On your poster, indicate your statistical question. Also, indicate a brief summary of how you collected your data based on the plan you proposed in Lesson 17. Include a graph that shows the shape of the data distribution, along with summary measures of center and variability. Finally, answer your statistical question based on the graphs and the numerical summaries.

Share the poster you will present in Lesson 22 with your teacher. If you are instructed to prepare an outline of the presentation, share your outline with your teacher.

Eureka Math Grade 6 Module 6 Lesson 21 Exit Ticket Answer Key

Question 1.
Based on the statistical question you are investigating for your project, summarize the four steps you are expected to complete as part of the presentation of your statistical study.
Answer:
Step 1:
State my statistical question. My question is based on collecting data that will vary.

Step 2:
Devise a plan to collect data. I prepared a question to ask the students in my class. (Allow students to explain the question they asked, the responses they received, and the method they used for recording answers.)

Step 3:
Summarize my data. I prepared a dot plot of the responses to the question. My dot plot indicated that the responses to my question were skewed to the left; therefore, I used the median of the data distribution to describe my center and the IQR to describe the variability. (Allow for a summary of the specific median or mean and a specific summary of the variability as the MAD or IQR.)

Step 4:
Based on my graphs and numerical summaries, I answered my question.

Engage NY Eureka Math Grade 6 Module 6 Lesson 20 Answer Key

Eureka Math Grade 6 Module 6 Lesson 20 Example Answer Key

Example 1: The Great Lakes Yellow Perch

Scientists collected data from many yellow perch because they were concerned about the survival of the yellow perch. What data do you think researchers might want to collect about these perch?
Answer:
Answers will vary. Anticipate that students also suggest collecting data on physical characteristics of the fish, such as weight, length, etc. Some students may say they would look for evidence of disease. Students might also suggest collecting data on age, but this might be difficult to determine. As this lesson progresses, students see that it is possible to estimate the age of a perch from its length.

Scientists captured yellow perch from a lake in this region. They recorded data on each fish and then returned each fish to the lake. Consider the following histogram of data on the length (in centimeters) for a sample of yellow perch.

Exercises 1 – 11:

Scientists were concerned about the survival of the yellow perch as they studied the histogram.

Exercise 1.
What statistical question could be answered based on this data distribution?
Answer:
Answers will vary. A possible statistical question would be, “What is a typical length of the Great Lakes yellow perch?”

Exercise 2.
Use the histogram to complete the following table.

Answer:

Exercise 3.
The length of each fish in the sample was measured and recorded before the fish was released back into the lake. How many yellow perch were measured in this sample?
Answer:
100 fish were measured in this sample.

Exercise 4.
Would you describe the distribution of the lengths of the fish in the sample as a skewed distribution or as an approximately symmetric distribution? Explain your answer.
Answer:
The data distribution is a skewed distribution with the tail to the left.

Exercise 5.
What percentage of fish in the sample were less than 10 centimeters in length?
Answer:
15 of the 100 fish measured in this sample had a length of less than 10 centimeters, so 15% of the fish are less than 10 centimeters.

Exercise 6.
If the smallest fish In this sample was 2 centimeters in length, what is your estimate of an interval of lengths that would contain the lengths of the shortest 25% of the fish? Explain how you determined your answer.
Answer:
25% of the fish ore represented in the first three intervals. If the smallest value in the first interval is known, then an estimate of the interval of the smallest 25% of the fish is 2 centimeters to 15 centimeters. determined this by considering the histogram bars at the low end and looking for intervals that would represent 25 fish.

Exercise 7.
If the length of the largest yellow perch was 29 centimeters, what is your estimate of an interval of lengths that would contain the lengths of the longest 25% of the fish?
Answer:
In a similar way, there are 25 fish in the interval 25 to 30 centimeters in length. If the longest fish were measured at 29 centimeters, then an estimate of an interval containing the upper 25% would be 25 to 29 centimeters.

Exercise 8.
Estimate the median length of the yellow perch in the sample. Explain how you determined your estimate.
Answer:
To estimate the median length, I would Identify o length for which approximately 50% of the fish would be above and approximately 50% of the fish would be below. Starting from the smallest lengths, an estimate of the median would be located within the 20 to 25 centimeters interval.

The same interval would be identified if started with the largest lengths. As the actual values of the lengths of the fish are not known, any estimate within that interval would be a good estimate. For example, an estimate of 23 centimeters would be a good estimate.

Exercise 9.
Based on the shape of this data distribution, do you think the mean length of a yellow perch would be greater than, less than, or the same as your estimate of the median? Explain your answer.
Answer:
Because the data distribution is skewed, the smaller lengths will pull an estimate of the mean to the left of the median. Therefore, an estimate of the mean would be less than the estimate of the median.

Exercise 10.
Recall that the mean length is the balance point of the distribution of lengths. Estimate the mean length for this sample of yellow perch.
Answer:
Answers will vary, but an estimate of a length in the 15 to 20 centimeters interval would show an understanding of this idea. For example, 17 or 18 centimeters would be a good estimate of the mean.

Exercise 11.
The length of a yellow perch is used to estimate the age of the fish. Yellow perch typically grow throughout their lives. Adult yellow perch have lengths between 10 and 30 centimeters. How many of the yellow perch in this sample would be considered adult yellow perch? What percentage of the fish in the sample are adult fish?
Answer:
85 fish are counted in the intervals that represent 10 to 30 centimeters. Therefore, 85% of the fish in this sample are estimated to be adult fish.

Example 2: What Would a Better Distribution Look Like?

Yellow perch are part of the food supply of larger fish and other wildlife in the Great Lakes region. Why do you think that the scientists worried when they saw the histogram of fish lengths given previously in Exercise 2.
Answer:
The small percentage of short fish means that there are fewer young yellow perch than older yellow perch.

Sketch a histogram representing a sample of 100 yellow perch lengths that you think would indicate the perch are not in danger of dying out.

Answer:
Answers will vary, but students’ histograms should have a larger number of fish with short lengths.

Exercises 12 – 17: Estimating the Variability in Yellow Perch Lengths

You estimated the median length of yellow perch from the first sample in Exercise 8. It is also useful to describe variability in the length of yellow perch. Why might this be important? Consider the following questions:

Exercise 12.
In several previous lessons, you described a data distribution using the five-number summary. Use the histogram and your answers to the questions in previous exercises to provide estimates of the values for the five-number summary for this sample:
Minimum (mm) value =
Ql value =
Median =
Q3 value =
Maximum (max) value =
Answer:
Minimum (min) value = 2 centimeters
Q1 value = 15 centimeters
Median = 23 centimeters
Q3 value = 25 centimeters
Maximum (max) value = 29 centimeters

Because students are estimating these values from a histogram, expect a variety of reasonable answers. Any values that are reasonably close to the values shown here should be considered acceptable.

Exercise 13.
Based on the five-number summary, what is an estimate of the value of the interquartile range (IQR) for this data distribution?
Answer:
Based on the above estimates, an estimate of the interquartile range (IQR) would be as follows:
25 centimeters – 15 centimeters = 10 centimeters.
Actual student answers will vary depending on their answers to Exercise 12.

Exercise 14.
Sketch a box plot representing the lengths of the yellow perch in this sample.

Answer:

Exercise 15.
Which measure of center, the median or the mean, Is closer to where the lengths of yellow perch tend to cluster?
Answer:
For a skewed distribution, the median is closer to where the lengths of yellow perch tend to cluster.

Exercise 16.
What value would you report as a typical length for the yellow perch in this sample?
Answer:
Encourage students to use the estimate of the median value in the previous questions as o typical value for the yellow perch.

Exercise 17.
The mean absolute deviation (or MAD) or the interquartile range (IQR) is used to describe the variability in a data distribution. Which measure of variability would you use for this sample of perch? Explain your answer.
Answer:
When the median is selected as the measure of center for a typical value (because the distribution is not approximately symmetric), then the interquartile range would be selected as the measure of variability. In this case, 10 centimeters, or the IQR determined in Exercise 13, would be the measure of the variability.

Eureka Math Grade 6 Module 6 Lesson 20 Problem Set Answer Key

Another sample of Great Lake yellow perch from a different lake was collected. A histogram of the lengths for the fish in this sample is shown below.

Question 1.
If the length of a yellow perch is an indicator of its age, how does this second sample differ from the sample you investigated in the exercises? Explain your answer.
Answer:
The second sample has more fish with lengths in the intervals corresponding to shorter lengths. Therefore, more of the fish are younger.

Question 2.
Does this histogram represent a data distribution that is skewed or that is nearly symmetrical?
Answer:
This distribution is also skewed. However, the tail of this distribution is to the right, or toward the longer lengths.

Question 3.
What measure of center would you use to describe a typical length of a yellow perch in this second sample? Explain your answer.
Answer:
Students should recommend the median of the data distribution is a description of a typical value of the length of the yellow perch because this distribution is skewed.

Question 4.
Assume the smallest perch caught was 2 centimeters in length, and the largest perch caught was 29 centimeters in length. Estimate the values in the five-number summary for this sample:
Answer:
Q1, Q3, and median values are not as clear cut in this distribution as in Exercise 4, so allow a wider range of acceptable answers.
Minimum (min) value = 2 centimeters
Q1 value = 4 centimeters (a value greater than 2 but within the interval of 0 to 5 centimeters)
Median value = 7 centimeters (o value within the interval of 5 to 10 centimeters)
Q3 value = 12 centimeters (a value within the interval of 10 to 15 centimeters)
Maximum (max) value = 29 centimeters.

Question 5.
Based on the shape of this data distribution, do you think the mean length of a yellow perch from this second sample would be greater than, less than, or the same as your estimate of the median? Explain your answer.
Answer:
An estimate of the mean would be greater than the median length because the values in the tail, or to the right of the median, pull the mean in that direction. Consider estimating the mean as the balance point of this distribution.

(If students have problems with estimating the balance point, consider providing them with a representation similar to the representation used to introduce a balance point in earlier lessons. Use a ruler with coins (or weights) taped to locations that would represent a skewed distribution. This representation may help students sense the point of balance.)

Question 6.
Estimate the mean value of this data distribution.
Answer:
An estimate of the mean would be a value slightly larger than the median value. For example, a mean of 10 or 11 centimeters would be a reasonable estimate of a balance point.

Question 7.
What is your estimate of a typical length of a yellow perch in this sample? Did you use the mean length from Problem 5 for this estimate? Explain why or why not.
Answer:
Since the median was selected as the appropriate estimate of a measure of center, a value of 7 centimeters (or whatever students used to estimate the median) would be an estimate of a typical value for a yellow perch from this sample.

Question 8.
Would you use the MAD or the IQR to describe variability in the length of Great Lakes yellow perch in this sample? Estimate the value of the measure of variability that you selected.
Answer:
Students should use the IQR to describe the variability because the data distribution is skewed, and the median was used as a measure of a typical value. An estimate of the IQR based on the above estimates would be as follows:
12 centimeters – 4 centimeters = 8 centimeters.

Eureka Math Grade 6 Module 6 Lesson 20 Exit Ticket Answer Key

Question 1.
Great Lake yellow perch continue to grow until they die. What does the histogram in Example 1 indicate about the ages of the perch in the sample?
Answer:
The histogram indicates that most of the perch are in the intervals corresponding to the longest lengths. Because length is related to age, the histogram indicates that there are more older fish.

Question 2.
What feature of the histogram in Example 1 indicates that the values of the mean and the median of the data distribution will not be equal?
Answer:
The histogram indicates that the shape of the data distribution is skewed. For skewed distributions, the mean and the median are not equal.

Question 3.
Adult yellow perch have lengths between 10 and 30 centimeters. Would a perch with a length equal to the median length be classified as an adult or a pre-adult fish? Explain your answer.
Answer:
A perch equal to the median length would be classified as an adult fish. The median is estimated to be between 20 and 25 centimeters in length. Adult fish are 10 centimeters or more in length.

Engage NY Eureka Math Grade 6 Module 6 Lesson 19 Answer Key

Eureka Math Grade 6 Module 6 Lesson 19 Exercise Answer Key

There was also a 15-kilometer race. The ages of the 55 participants in that race appear below.

Does the longer race appear to attract different runners in terms of age? Below are side-by-side box plots that may help answer that question. Side-by-side box plots are two or more box plots drawn using the same scale. What do you notice about the two box plots?

Both the range and the IQR are larger for the longer race, which means there is more variability in the ages for the long race than in the ages for the short race.

Exercises 1 – 6:

Exercise 1.
Based on the box plots, estimate the values in the five-number summary for the ages in the 15-kilometer race data set.
Answer:
Minimum – 16, Q1 = 30, Median = 37, Q3 = 48, ond Maximum = 81.

Exercise 2.
Do the two data sets have the same median? If not, which race had the higher median age?
Answer:
The two data sets do not have the same median; the 15 km race has a slightly higher median: 37 years of age compared to 35 years of age for the 5 km race.

Exercise 3.
Do the two data sets have the same IQR? If not, which distribution has the greater spread in the middle 50% of its distribution?
Answer:
The two data sets do not have the same IQR; the 15 km race has a slightly higher IQR: 18 years of age compared to 11 years of age for the 5 km race. The ages of runners in the middle 50% for the 15 km race are more spread out.

Exercise 4.
Which race had the smaller overall range of ages? What do you think the range of ages is for the 15-kilometer race?
Answer:
The 5 km race hod the smaller range of ages: 56 compared to 65 for the 15 km race.

Exercise 5.
Which race had the oldest runner? About how old was this runner?
Answer:
The 15 km race had the oldest runner at 81 years of age. The oldest runner for the 5 km race was only 74.

Exercise 6.
Now, consider just the youngest 25% of the runners in the 15-kilometer race. How old was the youngest runner in this group? How old was the oldest runner in this group? How does that compare with the 5-kilometer race?
Answer:
For the 15 km race, the youngest 25% of the runners had ages between the minimum and Q1. The youngest runner was 16 years old because this was the minimum for this distribution. The oldest runner in this group was 30 years old or younger because Q1 was 30. For the 5 km race, this youngest runner was 18 years old, and the oldest runner in the bottom 25% was 30 years old or younger (both distributions have the same Q1).

Exercises 7 – 12: Comparing Box Plots

In 2012, Major League Baseball had two leagues: an American League of 14 teams and a National League of 16 teams. Jesse wondered if American League teams have higher batting averages and on-base percentages. (Higher values are better.) Use the following box plots to investigate.

Exercise 7.
Was the highest American League team batting average very different from the highest National League team batting average? Approximately how large was the difference, and which league had the higher maximum value?
Answer:
The highest batting averages for both leagues appear to be around 0.274. (This value must be estimated from the box plot; so other answers that are dose to 0.274 are acceptable.)

Exercise 8.
Was the range of the American League team batting averages very different or only slightly different from the range of the National League team batting averages?
Answer:
They appear to be only slightly different, with the American League range being slightly higher. The American League minimum (0.234) is slightly lower than the National League minimum (0,236), and both leagues appear to have the same maximum.

Exercise 9.
Which league had the higher median team batting average? Given the scale of the graph and the range of the data sets, does the difference between the median values for the two leagues seem to be small or large? Explain why you think it is small or large.
Answer:
The American League has a higher median batting average of about 0.258, while the median batting average for the National League median is about 0.252. In answering the second part of this question, make sure that students look at the difference in the medians. Students might say that they think this is a small difference because 0.006 is a small number.

However, a better answer would consider the scale and the range of the data. The difference of 0.006 is roughly \(\frac{1}{6}\) of the range for the National League, which suggests that the difference may be large enough to be considered meaningful.

Exercise 10.
Based on the box plots below for on-base percentage, which three summary values (from the five-number summary) appear to be the same or virtually the same for both leagues?

Answer:
The Q1, median, and maximum appear to be roughly the same.

Exercise 11.
Which league’s data set appears to have less variability? Explain.
Answer:
The National League data set appears to have less variability, as it has a smaller IQR and smaller range.

Exercise 12.
Recall that Jesse wondered if American League teams have higher batting averages and on-base percentages. Based on the box plots given above, what would you tell Jesse?
Answer:
Answers will vary.

It does look like the American League teams tend to have higher team batting averages. The median and upper quartile for the American League team batting average box plot are greater than the median and upper quartile for the National League.

However, there is at least one team in the American League with a team batting average that is lower than every team in the National League, and there is at least one team in the National League that has a team batting average that is about the same as the American League team with the highest batting average.

For on-base percentage, the median and lower quartile and the maximum are about the same for both leagues. So it does not really look like it would be reasonable to say that the American League teams tend to have higher team on-base percentages than National League teams.

Eureka Math Grade 6 Module 6 Lesson 19 Problem Set Answer Key

Question 1.
College athletic programs are separated into divisions based on school size, available athletic scholarships, and other factors. A researcher wondered if members of swimming and diving programs in Division I (usually large schools that offer athletic scholarships) tend to be taller than the swimmers and divers in Division III programs (usually smaller schools that do not offer athletic scholarships).

To begin the investigation, the researcher creates side-by side box plots for the heights (in inches) of 41 male swimmers and divers at Mountain Vista University (a Division I program) and the heights (in inches) of 10 male swimmers and divers at Eaglecrest College (a Division III program).

a. Which data set has the smaller range?
Answer:
Eaglecrest College has a smaller range.

b. True or false: A swimmer who had a height equal to the median for the Mountain Vista University would be taller than the median height of swimmers and divers at Eaglecrest College.
Answer:
True

c. To be thorough, the researcher will examine many other colleges’ sports programs to further investigate the claim that members of swimming and diving programs in Division I are generally taller than the swimmers and divers in Division III. But given the graph above, in this initial stage of her research, do you think that the claim might be valid? Carefully support your answer using summary measures or graphical attributes.
Answer:
Based on just these two teams, it looks like the claim may be correct. A large portion of the Mountain Vista University distribution is higher than the maximum value of the Eaglecrest College distribution. The median value for the Mountain Vista University appears to be 4 inches higher than the median value of the Eogkcrest College distribution.

Question 2.
Data on the weights (in pounds) of 100 polar bears and 50 grizzly bears are summarized in the box plots shown below.

a. True or false: At least one of the polar bears weighed more than the heaviest grizzly bear. Explain how you know.
Answer:
True. The lower quartile and the median weight for the polar bears were greater than the maximum weight for the grizzly bears, so at least 75% of the polar bears weighed more than the heaviest grizzly bear.

b. True or false: Weight differs more from bear to bear for polar bears than for grizzly bears. Explain how you know.
Answer:
True. The weights of the grizzly bears are more compact than the weights of the polar bears. The grizzly bear weight distribution has a smaller range and IQR than the polar bear weight distribution.

c. Which type of bear tends to weigh more? Explain.
Answer:
Polar bears tend to have greater weights. The median weight for grizzly bears is much smaller than the median polar bear weight, and about half of the grizzly bear weight distribution is lower than the minimum weight for the polar bears.

Question 3.
Many movie studios rely heavily on viewer data to determine how a movie will be marketed and distributed. Recently, previews of a soon-to-be-released movie were shown to 300 people. Each person was asked to rate the movie on a scale of 0 to 10, with 10 representing “best movie I have ever seen” and O representing “worst movie I have ever seen.”
Below are some side-by-side box plots that summarize the ratings by gender and by age.

For 150 women and 150 men:

For 3 age groups:

a. Does it appear that the men and women rated the film in a similar manner or in a very different manner? Write a few sentences explaining your answer using comparative information about center and variability.
Answer:
It appears that the men and women rated the film in a very similar manner: The box plots show the same quartile values, the same medians, and the same maximums. The only difference is that the minimum rating from a man was slightly lower than the minimum rating from a woman.

b. It appears that the film tended to receive better ratings from the older members of the group. Write a few sentences using comparative measures of center and spread or aspects of the graphical displays to justify this claim.
Answer:
For the two oldest age groups, the Q1. median, Q3, and maximum values are all higher than the 18 – 24 counterparts. In fact, the Q1 value for each of these two older groups equals the median rating of the youngest group, and the median value for each of these two older groups equals the Q3 rating of the youngest group.

Additionally, while the two oldest groups have similar distributions, the minimum score of the oldest group was much higher than the minimum value of the 25 – 39 group. This means that none of the 40 – 64 respondents rated the movie with a score as low as a 4 (as was the case in the 25 – 39 age group).

Eureka Math Grade 6 Module 6 Lesson 19 Exit Ticket Answer Key

Hay is used to feed animals such as cows, horses, and goats. Almost \(\frac{1}{3}\) of the hay grown in the United States comes from just five states. Is this because these states have more acres planted in hay, or could it be because these states produce more hay per acre than other states? The following box plots show the distribution of hay produced (in tons) per acre planted in hay for three different regions: 22 eastern states, 14 midwestern states, and 12 western states.

Question 1.
Which of the three regions’ data sets has the least variability? Which has the greatest variability? To explain how you chose your answers, write a sentence or two that supports your choices by comparing relevant summary measures (such as median and lOR) or aspects of the graphical displays (such as shape and variability).
Answer:
The East data set has the least variability, as it has the smallest range and the smallest IQR. The West data set has the greatest variability, as it has the largest range and the largest IQR.

Question 2.
True or false: The western state with the smallest hay yield per acre has a higher hay yield per acre than at least half of the midwestern states. Explain how you know this is true or how you know this is false.
Answer:
This is true; the minimum value of the West data set is higher than the median value of the Midwest data set. Therefore, this minimum value for the West must be higher than at least half of the midwestern states’ values.

Question 3.
Which region typically has states with the largest hay yield per acre? To explain how you chose your answer, write a sentence or two that supports your choice by comparing relevant summary measures or aspects of the graphical displays.
Answer:
The West typically has states with the largest hay yield per acre. Over half of the western states have hay yields that are higher than any yield in either of the other two regions Also, some western yields are more than two times the largest eastern and midwestern yields.

Engage NY Eureka Math Grade 6 Module 6 Lesson 18 Answer Key

Eureka Math Grade 6 Module 6 Lesson 18 Example Answer Key

Example 1: Summary Information from Graphs

Here is a data set of the ages (in years) of 43 participants who ran in a 5-kilometer race.

Here are some summary statistics, a dot plot, and a histogram for the data:

Minimum = 18, Q1 = 30, Median = 35, Q3 = 41, Maximum = 74; Mean = 36.8, MAD = 8.1

Eureka Math Grade 6 Module 6 Lesson 18 Exercise Answer Key

Exercises 1 – 7:

Exercise 1.
Based on the histogram, would you describe the shape of the data distribution as approximately symmetric or as skewed? Would you have reached this same conclusion looking at the dot plot?
Answer:
Both graphs show a slightly skewed right data distribution.

Exercise 2.
If there had been 500 participants instead of just 43, would you use a dot plot or a histogram to display the data?
Answer:
Dot plots do not work as well for large data sets, so with 500 ages, I would probably use a histogram to display the data distribution.

Exercise 3.
What is something you can see in the dot plot that is not as easy to seem the histogram?
Answer:
When using the histogram, we cannot determine the exact minimum or maximum age – for example, we only know that the minimum age is between 15 and 20 years of age. Also, we can only approximate the median (we cannot figure out the exact median value from a histogram).

Since the dot plot provides us with a dot for each observation, we can see the actual data values. With the dot plot, we can see that the minimum is 18. The median is the 22nd observation (since there are 43 observations), and the 22nd dot counting from left to right is 35 (we cannot be that precise with the histogram). We can also see that the oldest runner (74) appears to be a more extreme departure from the rest of the data in the dot plot.

Exercise 4.
Do the dot plot and the histogram seem to be centered in about the same place?
Answer:
Yes. As both graphs are based on the same data, they should generally communicate the same information regarding center.

Exercise 5.
Do both the dot plot and the histogram convey information about the variability in the age distribution?
Answer:
Yes, as both graphs are based on the same data, they should generally communicate the same information regarding variability. However, as mentioned earlier, the oldest runner (74) appears to be o more extreme departure from the rest of the data ¡rl the dot plot. This does not show up as much in the histogram.

Exercise 6.
If you did not have the original data set and only had the dot plot and the histogram, would you be able to find the value of the median age from the dot plot?
Answer:
Yes. See the response to Exercise 3.

Exercise 7.
Explain why you would only be able to estimate the value of the median if you only had a histogram of the data.
Answer:
The median is the 22nd ordered observation in this data set since there are 43 observations. Counting from left to right, we know that the first 21 observations are in the first 4 intervals: 15 – 20(1 value), 20 – 25(3 values), 25 – 30 (4 values), and 30 – 3 5 (13 values). Cumulatively, we have encountered the lowest observations by the time we are finished with the 30 – 35 interval. So, the 22nd value must be in the next interval, which is 35 – 40 years of age. We just cannot determine the exact value from the histogram.

Exercises 8 – 12: Graphs and Numerical Summaries

Exercise 8.
Suppose that a newspaper article was written about race. The article included the histogram shown here and also said, “The race attracted many older runners this year. The median age was 45.” Based on the histogram, how can you tell that this is an incorrect statement?

Answer:
Several answers are possible.
From the histogram, it appears that less than half of the runners are 45 or older (or alternatively, more than half of the runners are younger than 45). The value of the median is in the 35 – 40 age interval.

Exercise 9.
One of the histograms below is another correctly drawn histogram for the runners’ ages. Select the correct histogram, and explain how you determined which graph is correct (and which one is incorrect) based on the summary measures and dot plot.

Answer:
One of the objectives of this question is to reinforce the idea that there is more than one way to draw a proper histogram for a distribution. This question is especially detail-oriented because students need to carefully reconcile components of the histogram with the data set given in Example 1.

The histogram on the right is the correct one because it is consistent with the dot pot/data. Most notably, the histogram on the right correctly shows there are 3 runners in the 10 – 22 age group, while the left histogram shows only 1 runner in the 12 – 2 4 age group (and there are actually 4 runners in that interval). Other intervals in the left histogram do not match the dot plot/data (e.g., the 48-60 group), so several answers are possible.

Exercise 10.
The histogram below represents the age distribution of the population of Kenya in 2010.

a. How do we know from the graph above that the first quartile (Q1) of this age distribution is between 5 and 10 years of age?
Answer:
Since a histogram should display information that is consistent with summary measures, we are seeking a data value such that 25% of the distribution is at or below that value. While the 0 – 5 age group represents approximately the lowest 17%. the next group (age 5 – 10) appears to account for approximately the next 15% of the distribution. This means that about 17% is below 5 and about 32% is below 10. So, the 25% mark would be somewhere between 5 and 10 years.

b. Someone believes that the median age in Kenya is about 30. Based on the histogram, is 30 years a good estimate of the median age for Kenya? Explain why it is or why it is not.
Answer:
The median does not appear to be 30 years of age. Specifically, the 50th percentile estimated by adding approximate percentages (and/or visually assessing the point at which the area seems split evenly) appears to be in the 15 – 20 age group.

Exercise 11.
The histogram below represents the age distribution of the population of the United States in 2010. Based on the histogram, which of the following ranges do you think includes the median age for the United States: 20 – 30, 30 – 40, or 40 – 50? Why?

Answer:
Using similar arguments as described in the response to Exercise 10 part (b), the median appears to be in the 30 – 40 age group, most likely in the 35 – 40 interval.

Exercise 12.
Consider the following three-dot plots. Note: The same scale is used in each dot plot.

a. Which dot plot has a median of 8? Explain why you selected this dot plot over the other two.
Answer:
Plot 3 – It is the only distribution visually centered near 8, and one can tell that the 22 ordered observation (the median in this case) is 8.

b. Which dot plot has a mean of 9.6? Explain why you selected this dot plot over the other two.
Answer:
Plot 1 – This dot plot is the only dot plot for which there could be a mean value as high as 9.6. It is the only dot plot with values of 11, 12, and 13, and there are several of these values.

c. Which dot plot has a median of 6 and a range of S? Explain why you selected this dot plot over the other two.
Answer:
Plot 2 – It is the only dot plot that appears to be centered at 6. It is also the only dot plot with o range of 5 (each of the other dot plots has a range of 8).

Eureka Math Grade 6 Module 6 Lesson 18 Problem Set Answer Key

Question 1.
The following histogram shows the amount of coal produced (by state) for the 20 largest coal-producing states in 2011. Many of these states produced less than 50 million tons of coal, but one state produced over 400 million tons (Wyoming). For the histogram, which one of the three sets of summary measures could match the graph? For each choice that you eliminate, give at least one reason for eliminating the choice.

a. Minimum = 1, Q1 = 12, Median = 36, Q3 = 57, Maximum = 410; Mean = 33, MAD = 2.76
b. Minimum = 2, Q1 = 13.5, Median = 27.5, Q3 = 44, Maximum = 439; Mean = 54.6, MAD = 52.36
c. Minimum = 10, Q1 = 37.5, Median = 62, Q3 = 105, Maximum = 439; Mean = 54.6, MAD = 52.36
Answer:
The correct answer is (b).
Choice (a) would not work because Q3 (the average of the 15th and 16th ordered observations) must be less than 50 since both the 15th and 16th ordered observations are less than 50. The mean is most likely greater than (not less than) the median given the skewed right nature of the distribution and the large outlier. The MAD value is most likely much larger than 2.76 given the presence of the outlier and its distance from the cluster of remaining observations.

Choice (c) would not work. Since there are 20 observations, the median (the average of the 10th and 11th ordered observations) must be less than 50, since both the 10th and 11th ordered observations are less than 50. Likewise, the Q3 (the average of the 15th and 16th ordered observations) must be less than 50, since both the 15th and 16th observations are less than 50. The mean is most likely greater than (not less than) the median given the skewed right nature of the distribution and the large outlier.

Question 2.
The heights (rounded to the nearest inch) of the 41 members of the 2012 – 2013 University of Texas Men’s Swimming and Diving Team are shown in the dot plot below.

a. Use the dot plot to determine the 5-number summary (minimum, lower quartile, median, upper quartile, and maximum) for the data set.
Answer:
Min = 66, Q1 = 71, Median = 73, Q3 = 75, and Max = 80

b. Based on this dot plot, make a histogram of the heights using the following intervals: 66 to < 68 inches, 68 to < 70 inches, and so on.
Answer:

Question 3.
Data on the weight (in pounds) of 143 wild bears are summarized in the histogram below.

Which one of the three-dot plots below could be a dot plot of the bear weight data? Explain how you determined which the corrected plot.

Answer:
Dot plot 1 is the correct choice. Dot plot 2 does not have the same shape as the data distribution shown in the histogram, and it is centered to the right of where the histogram is centered. Although dot plot 3 has the same shape as the distribution shown in the histogram, it is not centered in the same place as the histogram,

Eureka Math Grade 6 Module 6 Lesson 18 Exit Ticket Answer Key

Question 1.
Many states produce maple syrup, which requires tapping sap from a maple tree. However, some states produce more pints of maple syrup per tap than other states. The following dot plot shows the pints of maple syrup yielded per tap in each of the 10 maple syrup-producing states in 2012.

a. Minimum = 0.66, Q1 = 1.26, Median = 1.385, Q3 = 1.71, Maximum = 1.95, Range = 2.4; Mean = 1.95, MAD = 0.28
b. Minimum = 0.66, Q1 = 1.26, Median = 1.71, Q3 = 1.92, Maximum = 1.95, Range = 1.29; Mean = 1.43, MAD = 2.27
c. Minimum = 0.66, Q1 = 1.26, Median = 1.385, Q3 = 1.71, Maximum = 1.95, Range = 1.29; Mean = 1.43, MAD = 0.28
Answer:
The correct answer is (c).
Choice (a) would not work because the range is too large. For the data set, the difference between maximum and minimum is only 1.29 pints. Also, the mean would not be that close to (or the same os) the maximum value in this case.

Choice (b) would not work because a median value of 1.71 would be too high. By estimating the dot values, we see that the 5th and 6th ordered observations (the median for a data set of 10 items) are near 1.4. Also, the MAD is much too large because the range of the data set is only 1.29 pints.

Question 2.
Which one of the three histograms below could be a histogram for the data displayed in the dot plot in Problem 1? For each histogram that you eliminate, give at least one reason for eliminating it as a possibility.

Answer:
The correct answer is the histogram (a).

Histograms (b) and (c) ore for data sets with similar shape features to the correct graph (histogram (a)), but the range and distribution of values do not match the data set in the dot plot. For example, histogram (b) would not be valid, as it is based on 3 data values of 2 pints or more, and there are no values that large in the original dot plot.

Also, the smallest value in the histogram (b) is at least 1 pint, and the actual data set contains a value less than 1 pint. Histogram (c) is based on values that are smaller than many of those in the dot plot; in fact, all of the values in the histogram (c) are less than 1.02 pints, and nearly all of the 10 observations in the actual data set are greater than 1.02.

Engage NY Eureka Math Grade 6 Module 6 Lesson 17 Answer Key

Eureka Math Grade 6 Module 6 Lesson 17 Review of Statistical Questions Answer Key

Exploratory Challenge:

Review of Statistical Questions:

Statistical questions you investigated in this module included the following:

1. How many hours of sleep do sixth-graders typically get on a night when there is school the next day?
2. What is the typical number of books read over the course of 6 months by a sixth-grader?
3. What is the typical heart rate of a student in a sixth-grade class?
4. How many hours does a sixth-grader typically spend playing a sport or a game outdoors?
5. What are the head circumferences of adults interested in buying baseball hats?
6. How long is the battery life of a certain brand of batteries?
7. How many pets do students have?
8. How long does it take students to get to school?
9. What is a typical daily temperature in New York City?
10. What is the typical weight of a backpack for students at a certain school?
11. What is the typical number of french fries in a large order from a fast-food restaurant?
12. What is the typical number of minutes a student spends on homework each day?
13. What is the typical height of a vertical jump for a player in the NBA?

Question 1.
What do these questions have in common?
Answer:
All of these questions are statistical questions because they can be answered by collecting data, and there is variability in the data.

Question 2.
Why do several of these questions include the word typical?
Answer:
Answers will vary. Students may focus on the idea that typical suggests finding a single number that summarizes the data collected to answer the statistical question.

Eureka Math Grade 6 Module 6 Lesson 17 Review of Statistical Investigation Answer Key

A Review of a Statistical Investigation:

Recall from the very first lesson in this module that a statistical question is a question answered by data that you anticipate will vary. Let’s review the steps of a statistical investigation.

Step 1: Pose a question that can be answered by data.
Step 2: Collect appropriate data.
Step 3: Summarize the data with graphs and numerical summaries.
Step 4: Answer the question posed in Step 1 using the numerical summaries and graphs.

The first step is to pose a statistical question. Select one of the questions investigated in this module, and write it in the following Statistical Study Review Template.

The second step is to collect the data. In all of these investigations, you were given data. How do you think the data for the question you selected in Step 1 were collected? Write your answer in the summary below for Step 2.

The third step involves the various ways you summarize the data. List the various ways you summarized the data in the space for Step 3.

Question 1.
Statistical Study Review Template

Answer:
Step 1: Pose a statistical question.
How many hours of sleep do sixth-graders typically get on a night when there is school the next day? (Lesson 3)

Step 2: Collect the data.
Students from a sixth-grade class might have been asked to indicate how many hours they slept. The data would consist of the answers from all of the students.

Step 3: Summarize the data.
The first summary was to organize the data in a dot plot. This data set indicated a nearly symmetrical data distribution. Numerical summaries of the mean and the MAD or the median and IQR would provide o description of the typical number of hours of sleep for sixth-grade students and a measure of how much variability there was in the sleep times.

Step 4: Answer the question.
From the data, we can see that the typical sixth-grader gets about 8. 5 hours of sleep when there is school the next day because this is the mean of the data. The MAD is only 1 hour, so there is not o lot of variability in the data.

Eureka Math Grade 6 Module 6 Lesson 17 Developing Statistical Questions Answer Key

Developing Statistical Questions:

Now it is your turn to answer a statistical question based on the data you collect. Before you collect the data, explore possible statistical questions. For each question, indicate the data that you would collect and summarize to answer the question. Also, indicate how you plan to collect the data.

Think of questions that could be answered by data collected from members of your class or school or data that could be collected from recognized websites (such as the American Statistical Association and the Census at School project). Your teacher will need to approve both your question and your plan to collect data before data are collected.

As a class, explore possibilities for a statistical investigation. Record some of the ideas discussed by your class using the following table.

After discussing several of the possibilities for a statistical project, prepare a statistical question and a plan to collect the data. After your teacher approves your question and data collection plan, begin collecting the data.

Carefully organize your data as you begin developing the numerical and graphical summaries to answer your statistical question. In future lessons, you will be directed to begin creating a poster or an outline of a presentation that will be shared with your teacher and other members of your class.

Complete the following to present to your teacher:

1. The statistical question for my investigation is:

2. Here is how I propose to collect my data. (Include how you are going to collect your data and a clear description of what you plan to measure or count.)

Eureka Math Grade 6 Module 6 Lesson 17 Problem Set Answer Key

Your teacher will outline steps you are expected to complete In the next several days to develop this project. Keep In mind that the first step is to formulate a statistical question. With one of the statistical questions posed In this lesson or with a new one developed in this lesson, describe your question and pian to collect and summarize data. Complete the process as outlined by your teacher.

A formal Problem Set has not been added to this lesson. However, teachers are encouraged to design a Problem Set based on students’ progress during this lesson. The following options are possible ideas for designing a Problem Set. Teacher discretion in organizing the project is important.

Option 1:
Students who struggled with completing the four-step table developed around one of the questions used in this lesson should be encouraged to select a different question and complete the table for this second question. The first three steps provide students with a structure for connecting a question to a plan for collecting and then summarizing data.

These steps were illustrated in the previous lessons; however, in this lesson, students need to bring the steps together. Once students understand these steps for a given question, they are ready to formulate their own questions and data collection plans.

Option 2:
Students provide a question and a data collection plan to the teacher for review as outlined in the lesson. Direct students to complete the four-step table for their questions and plans. Using the table provided in this lesson is an excellent way for students to organize their progress, plus it provides a good record for the teacher to understand how students are thinking at the beginning of this project.

For students ready to begin this process, direct them to provide a summary of their statistical questions, a plan for collecting the data, and a summary of the data they anticipate to collect. Although these steps were discussed in the lesson, organizing this into a table similar to the one presented in the lesson provides a summary of student progress. Periodically ask students during the next several days for an update on their progress by providing a summary of the table used in this lesson.

Option 3:
For students going beyond the questions outlined in the lesson, they need to provide specific descriptions of what they plan to research and how they plan to collect the data.

For example, if a student proposed to explore research on honeybees, make sure the student clearly indicates the statistical question (or the question to be answered by data that are anticipated to vary), where she plans to obtain the data, and how she plans to summarize the data. A brief written report or summary of student progress might constitute a workable Problem Set option for students at this level.

Eureka Math Grade 6 Module 6 Lesson 17 Exit Ticket Answer Key

Question 1.
What is a statistical question?
Answer:
A statistical question is one that can be answered by collecting data and where there is variability in the data.

Question 2.
What are the four steps in a statistical investigation?
Answer:
Step 1: Pose a statistical question.
Step 2: Collect data.
Step 3: Summarize the data.
Step 4: Answer the statistical question posed.

Engage NY Eureka Math Grade 6 Module 6 Lesson 16 Answer Key

Eureka Math Grade 6 Module 6 Lesson 16 Exercise Answer Key

Exercise 1: Supreme Court Chief Justices

Exercise 1.
The Supreme Court is the highest court of law in the United States, and it makes decisions that affect the whole country. The chief justice is appointed to the court and is a justice the rest of his lite unless he resigns or becomes ill. Some people think that this means that the chief justice serves for a very long time. The first chief justice was appointed in 1789.

The table shows the years in office for each of the chief justices of the Supreme Court as of 2013:

Use the table to answer the following:

a. Which chief justice served the longest term, and which served the shortest term? How many years did each of these chief justices serve?
Answer:
John Marshall had the longest term, which was 34 years. He served from 1801 to 1835. John Rutledge served the shortest term, which was 1 year in 1795.

b. What is the median number of years these chief justices have served on the Supreme Court? Explain how you found the median and what it means In terms of the data.
Answer:
First, you have to put the data in order. There are 17 justices, so the median would fall at the 9th value (11 years) counting from the top or from the bottom. The median is 11. Approximately half of the justices served less than or equal to 11 years, and half served greater than or equal to 11 years.

c. Make a box plot of the years the justices served. Describe the shape of the distribution and how the median
and IQR relate to the box plot.
Answer:
The distribution seems to have more justices serving a small number of years (on the lower end). The range (max – mm) is 33 years, from 1 year to 34 years. The IQR is 18 – 6. 5 = 11.5, so about half of the chief justices had terms in the 11.5-year interval from 6.5 to 18 years.

d. Is the median halfway between the least and the most number of years served? Why or why not?
Answer:
The halfway point on the number line between the smallest number of years served, 1, and the greatest number of years served, 34, is 17.5, but because the data are clustered in the lower end of the distribution, the median, 11, is to the left of (smaller than) 17.5. The middle of the interval from the smallest to the largest data value has no connection to the median. The median depends on how the data are spread out over the interval.

Exercises 2 – 3: Downloading Songs

Exercise 2.
A broadband company timed how long it took to download 232 four-minute songs on a dial-up connection. The dot plot below shows their results.

a. What can you observe about the download times from the dot plot?
Answer:
The smallest time was a little bit less than 5 minutes, and the largest is a little bit more than 17 minutes. Most of the times seem to be between 8 to 13 minutes.

b. Is it easy to tell whether or not 12.5 minutes is in the top quarter of the download times?
Answer:
You cannot easily tell from the dot plot.

c. The box plot of the data is shown below. Now, answer parts (a) and (b) above using the box plot.
Answer:

Answer for part (a) based on the box plot: About half of the times are above 10.6 minutes. The distribution is roughly symmetric around the median. About half of the times are between 8.7 minutes and 12.2 minutes.
Answer for part (b) based on the box plot: 12.5 is above Q3, so it was in the top quarter of the data.

d. What are the advantages of using a box plot to summarize a large data set? What are the disadvantages?
Answer:
With lots of data, the dots in a dot plot overlap, and while you can see general patterns, it is hard to really get anything quantifiable. The box plot shows at least an approximate value for each of the five-number summary measures and gives a pretty good idea of how the data are spread out.

The disadvantage of box plots is that the specific values in the data set are not given.
Teacher note: It may be useful to have a brief class discussion of the advantages/disadvantages of dot plots and box plots. You can use the dot plot in part (a) and the box plot in part (b) to facilitate this discussion.

Exercise 3.
Molly presented the box plots below to argue that using a dial-up connection would be better than using a broadband connection. She argued that the dial-up connection seems to have less variability around the median even though the overall range seems to be about the same for the download times using broadband. What would you say?

Answer:
The scales are different for the two plots, and so are the units, so you cannot just look at the box plots. The time using broadband is centered near seconds to download the song while the median for dial-up is almost minutes for a song. This suggests that broadband is going to be faster than dial-up.

Teacher note:
This is an important point. Make sure that students understand the importance of using the same scale if box plots are being constructed to compare two data distributions.

Exercises 4 – 5: Rainfall

Exercise 4.
Data on the average rainfall for each of the twelve months of the year were used to construct the two-dot plots below.

a. How many data points are in each dot plot? What does each data point represent?
Answer:
There are 12 data points in the St. Louis dot plot. There are also 12 data points in the San Francisco dot plot. Each data point represents the average monthly precipitation in inches for one month.

b. Make a conjecture about which city has the most variability in the average monthly amount of precipitation and how this would be reflected in the lQRs for the data from both cities.
Answer:
San Francisco has the most variability in the average monthly amount of precipitation. It should have the largest IQR of the two cities.

c. Based on the dot plots, what are the approximate values of the interquartile ranges (IQRs) for the average monthly precipitations for each city? Use the lQRs to compare the cities.
Answer:
The answers that follow are based on estimates from the dot plot. Students might get slightly different values. For St. Louis, the IQR is 4.2 – 3.2 = 1; for San Francisco, the IQR is 3.9 – 0.2 = 3.7. About the middle half of the monthly precipitation amounts in St. Louis are within 1 inch of each other. In San Francisco, the middle half of the monthly precipitation amounts are within about 4 inches of each other.

d. In an earlier lesson, the average monthly temperatures were rounded to the nearest degree Fahrenheit. Would it make sense to round the amount of precipitation to the nearest inch? Why or why not?
Answer:
Answers will vary. Possible answers include: It would not make sense because the numbers are pretty close together, or yes, it would make sense because you would still get a good idea of how the precipitation varied.

If you rounded to the nearest inch, the IQR for San Francisco would be 4 because three of the values round to 0, and three of the values round to 5. The IQR for St. Louis would be 1 because most of the values round to 3 or 4. In both cases, that is pretty close to the IQR found in part (c).

Exercise 5.
Use the data from Exercise 4 to answer the following.

a. Make a box plot of the monthly precipitation amounts for each city using the same scale.

Answer:

b. Compare the percent of months that have above 2 inches of pi-ecipitation for the two cities. Explain your thinking.
Answer:
In St. Louis, the average amount of precipitation each month is always over 2 inches, while this happens, at most, for half of the months In San Francisco because the median amount of precipitation is just above 1 inch.

c. How does the top 25% of the average monthly precipitations compare for the two cities?
Answer:
The top 25% of the precipitation amounts in the two cities are spread over about the same interval (about 4 to 5 inches). St. Louis has a bit more spread; the top 25% in St. Louis are between 4.2 inches and 4.8 inches, while the top 25% in San Francisco are all very close to 4.5 inches.

d. Describe the intervals that contain the smallest 25% of the average monthly precipitation amounts for each city.
Answer:
In St. Louis, the smallest 25% of the monthly averages are between about 2.5 inches and 3.0 inches; in San Francisco, the smallest averages are much lower, ranging from O to 0.2 inches.

e. Think about the dot plots and the box plots. Which representation do you think helps you the most in understanding how the data vary?
Answer:
Answers will vary. Some sample answers are provided below.
The dot plot because we can see individual values.
The box plot because it just shows how the data are spread out in each of the four sections.

Eureka Math Grade 6 Module 6 Lesson 16 Problem Set Answer Key

Question 1.
The box plots below summarize the ages at the time of the award for leading actress and leading actor Academy Award winners.

a. Based on the box plots, do you think it is harder for an older woman to win an Academy Award for a best actress than it is for an older man to win the best actor award? Why or why not?
Answer:
Answers will vary. Students might take either side as long as they give an explanation for why they made the choice they did that is based on the box plots.

b. The oldest female to win an Academy Award was Jessica Tandy in 1990 for Driving Miss Daisy. The oldest actor was Henry Fonda for On Golden Pond in 1982. How old were they when they won the award? How can you tell? Were they a lot older than most of the other winners?
Answer:
Henry Fonda was 76, and Jessica Tandy was 80. I know this because those are the maximum values. You cannot tell if there were actors or actresses that were nearly as old.

c. The 2013 winning actor was Daniel Day-Lewis for Lincoln. He was 55 years old at that time. What can you say about the percent of male award winners who were older than Daniel Day-Lewis when they won their Oscars?
Answer:
He was in the upper quarter and one of the older actors. Fewer than 25% of the male award winners were older than Daniel Day-Lewis.

d. Use the information provided by the box plots to write a paragraph supporting or refuting the claim that fewer older actresses than actors win Academy Awards.
Answer:
Overall, the box plot for actresses starts about 10 years younger than actors and is centered around a lower age than the box plot for- actors. The median age for actresses who won the award is 33, and for actors it is 42. The upper quartile is also lower for actresses, 41, compared to 49 for actors.

The range for actresses’ ages is larger, 80 – 21 = 59, compared to the range for actors, 76 – 29 = 47. About \(\frac{3}{4}\) of the actresses who won the award were younger than the median age for the men.

Question 2.
The scores of sixth and seventh graders on a test about polygons and their characteristics are summarized in the box plots below.

a. In which grade did the students do the best? Explain how you can tell.
Answer:
Three-fourths of the seventh-grade students did better than half of the sixth graders. You can tell by comparing Q1 for Grade 7 to the median for Grade 6. Therefore, the seventh-grade students performed the best.

b. Why do you think two of the data values in Grade 7 are not part of the line segments?
Answer:
The highest and lowest scores were pretty far away from the other scores, so they were marked separately.

c. How do the median scores for the two grades compare? Is this surprising? Why or why not?
Answer:
The median score in Grade 7 was higher than the median in Grade 6. This makes sense because the seventh graders should know more than the sixth graders.

d. How do the lQRs compare for the two grades?
Answer:
The middle half of the Grade 7 scores were close together in a span of about 11 with the median around 66.
The middle half of the Grade 6 scores were spread over a larger span, about 17 points from about 50 to 67.

Question 3.
A formula for the IQR could be written as Q3 – Q1 = IQR. Suppose you knew the IQR and the Q1. How could you find the Q3?
Answer:
Q3 = IQR + Q1. Add the lower quartile to the IQR.

Question 4.
Consider the statement, “Historically, the average length of service as chief justice on the Supreme Court has been less than 15 years; however, since 1969 the average length of service has increased.” Use the data given in Exercise 1 to answer the following questions.

a. Do you agree or disagree with the statement? Explain your thinking.
Answer:
The mean number of years as chief justice overall is about 13. The mean number of years since 1969 is about 14.7. Even though the mean has increased, it does not seem like a big difference because there have only been three justices since then to cover o span of 44 years (and three times 13 is 39, so not enough to really show an increasing trend).

b. Would your answer change if you used the median number of years rather than the mean?
Answer:
The median overall was 11 years; the median since 1969 was 17 years, which is considerably larger. This seems to justify the statement.

Eureka Math Grade 6 Module 6 Lesson 16 Exit Ticket Answer Key

Data on the number of pets per family for students in a sixth-grade class are summarized in the box plot below:

Question 1.
Can you tell how many families have two pets? Explain why or why not.
Answer:
You cannot tell from the box plot. You only know that the lower quartile (Q]) is 2 pets. You do not know how many families are included in the data set.

Question 2.
Given the box plot above, which of the following statements are true? If the statement is false, modify it to make the statement true.

a. Every family has at least one pet.
Answer:
True

b. About one-fourth of the families have six or more pets.
Answer:
True

c. Most of the families have three pets.
Answer:
False, because you cannot determine the number of any specific data value. Revise to “You cannot determine the number of pets most families have.”

d. About half of the families have two or fewer pets.
Answer:
False. Revise to “About half of the families have three or fewer pets.”

e. About three-fourths of the families have two or more pets.
Answer:
True

Engage NY Eureka Math Grade 6 Module 6 Lesson 15 Answer Key

Eureka Math Grade 6 Module 6 Lesson 15 Example Answer Key

Example 1: Tootsie Pops

Ninety-four people were asked to grab as many Tootsie Pops as they could hold. Here is a box plot for these data. Are you surprised?
Answer:
Answers will vary. Students might indicate that they are surprised that some people were able to hold as many as 40 Tootsie Pops or that they are surprised at how much person-to-person variability there is. They may also comment on the fact that about half the people were able to hold between 18 and 23 Tootsie Pops.

Eureka Math Grade 6 Module 6 Lesson 15 Exercise Answer Key

Exercises 1 – 5:

Exercise 1.
What might explain the variability in the number of Tootsie Pops that the 94 people were able to hold?
Answer:
Answers will vary. Possible answers include the size of people’s hands, handspan, and whether a person is flexible in moving his fingers.

Exercise 2.
Use a box plot to estimate the values in the five-number summary.
Answer:
Min = 7, Q1 = 18, Median = 20, Q3 = 22, Max = 42.

Exercise 3.
Describe how the box plot can help you understand differences in the numbers of Tootsie Pops people could hold.
Answer:
The maximum of about 42 and minimum of about 7 indicate that there is o lot of variability in the number of Tootsie Pops that people can hold, with the numbers covering a range of about 35 Tootsie Pops. The “box” port of the box plot shows that about half of the peopk con hold within about 2 Tootsie Pops of the median, which was 20 Tootsie Pops.

Exercise 4.
Here is Jayne’s description of what she sees in the box plot. Do you agree or disagree with her description? Explain your reasoning.
“One person could hold as many as 42 Tootsie Pops. The number of Tootsie Pops people could hold was really different and spread about equally from 7 to 42. About one-half of the people could hold more than 20 Tootsie Pops.”
Answer:
You cannot tell that they are evenly spread—the “box” part of the box plot contains about half of the values for the number of Tootsie Pops. However, the box is only four units long. That means half of the people were bunched over those four numbers.

Exercise 5.
Here is a different box plot of the same data on the number of Tootsie Pops 94 people could hold.

a. Why do you suppose there are five values that are shown as separate points and are labeled?
Answer:
Maybe because they are far away from most of the other values. It shows that more than half ofthe data an from about 12 to 27 Tootsie Pops.

b. Does knowing these data values change anything about your responses to Exercises 1 to 4 above?
Answer:
Not really, except maybe to say that only a few of the people could hold more than 30 Tootsie Pops; the rest held fewer than that. And only a few people could hold 10 or fewer Tootsie Pops.

Exercises 6 – 10: Maximum Speeds

The maximum speeds of selected birds and land animals are given in the tables below.

Exercise 6.
As you look at the speeds, what strikes you as interesting?
Answer:
Answers will vary. Some students might suggest birds are really fast, especially the falcon. Others may notice that only two of the speeds have decimals. The speeds of specific animals might strike students as interesting.

Exercise 7.
Do birds or land animals seem to have the greatest variability in speeds? Explain your reasoning.
Answer:
It looks like the speeds of the birds vary a lot, as they go from 60 mph for some birds to 242 mph for others. The speeds of the land animals vary but not as much; they go from 9 mph to 75 mph.

Exercise 8.
Find the five-number summary for the speeds in each data set. What do the five-number summaries tell you about the distribution of speeds for each data set?
Answer:
Land animal five-number summary: Min = 9, Q1 = 32.5, Median = 43.97, Q3 = 50, Max = 75
Bird five-number summary: Min = 60, Q1 = 76, Median = 97.5, Q3 = 105.5, Max = 242
The summaries give me a sense of the range or span of the speeds (maximum-minimum speed) and how the speeds are grouped around the median.

Exercise 9.
Use the five-number summaries to make a box plot for each of the two data sets.

Answer:

Exercise 10.
Write several sentences describing the speeds of birds and land animals.
Answer:
Answers will vary. At least one bird flies really fast: the falcon at 242 mph. Three-fourths of the birds fly less than 106 mph, and the slowest bird flies 60 mph. The land animals’ running speeds are slower, ranging from 9 mph to 75 mph. The middle half of the speeds for land animals is between 32.5 mph and 50 mph.

Exercises 11 – 15: What is the Same, and What is Different?

Consider the following box plots, which show the number of correctly answered questions on a 20-question quiz for students in three different classes.

Exercise 11.
Describe the variability in the scores of each of the three classes.
Answer:
The range (max-min) is the same for all three classes, and so is the median, but the intervals that contain the middle half of the scores (the length of the box part of the box plot) are different. The third class has a small box, so the scores in the middle are close together.

In Class 2, the minimum and lower quartile are the same scores, and the maximum and upper quartile are also the same score, so lots of scores are piled at the ends of the range. The middle half of the scores in Class 1 are spread out more than Class 3 but not as much as Class 2.

Teacher Note: The box plot for Class 2 may be difficult for students to interpret at first. If they have trouble, consider discussing the example provided earlier in the teacher flotes that shows how it is possible that the minimum and the lower quartile might be equal and how the upper quartile and the maximum might be equal.

Exercise 12.
a. Estimate the interquartile range for each of the three sets of scores.
Answer:
Class 1 IQR = 10; Class 2 IQR = 15; Class 3 IQR = 6

b. What fraction of students would have scores In the interval that extends from the lower quartile to the upper quartile?
Answer:
About one-half.

c. What does the value of the IQR tell you about how the scores are distributed?
Answer:
For Class 1, half of the scores are spread over an interval of width 10, and for Class 3, half of the scores are bunched together over an interval of width 5. For Class 2, the middle half of the data is spread over an interval of width 15, and in fact, because the quartiles are equal to the minimum and the maximum for Class 2, all of the data values are included in this interval with data values bunched up at the minimum and the maximum.

Exercise 13.
Which class do you believe performed the best? Be sure to use information from the box plots to back up your answer.
Answer:
1. Answers will vary. A few sample answers are provided.
2. Class 3, as it has the smallest IQR. About half of the students scored close to the median score. Scores were more consistent for this class.
3. Approximately 25% of the students in Class 1 scored 18 or higher compared to 25% of the students in Class 3 who scored 15 or higher. Therefore, Class 1 performed the best.
4. In Class 2, several students must have scored near the top in order for the Q3 and maximum to be the same.
Therefore, Class 2 performed the best.

Exercise 14.
a. Find the IQR for the three data sets in the first two examples: maximum speed of birds, maximum speed of
land animals, and number of Tootsie Pops.
Answer:
Land animals: 50 – 32. 5for an IQR of 17.5
Birds: 105.5 – 76 for an IQR of 29.5
Tootsie Pops: 22 – 18 for an IQR of 4

b. Which data set had the highest percentage of data values between the lower quartile and the upper quartile? Explain your thinking.
Answer:
All of the data sets should have about half of the data values between the quartiles.

Exercise 15.
A teacher asked students to draw a box plot with a minimum value at 34 and a maximum value at 64 that had an interquartile range of 10. Jeremy said he could not draw just one because he did not know where to put the box on the number line. Do you agree with Jeremy? Why or why not?
Answer:
Jeremy is correct since a box with a width of 10 could be drawn anywhere between the minimum and maximum values.

Eureka Math Grade 6 Module 6 Lesson 15 Problem Set Answer Key

Question 1.
The box plot below summarizes the maximum speeds of certain kinds of fish.

a. Estimate the values in the five-number summary from the box plot.
Answer:
Answers will vary. Min = 35 mph; Q1 = 39 mph; Median = 42 mph; Q3 = 48 mph; Max = 68 mph

b. The fastest fish is the sailfish at 68 mph, followed by the marlin at 50 mph. What does this tell you about the speed of the fish speeds in the top quarter of the box plot?
Answer:
The Q3 is about 48, so all but one of the fish in the top quarter ore between 48 mph and 50 mph.

c. Use the five-number summary and the IQR to describe the speeds of the fish.
Answer:
The speeds of fish vary from 35 mph to 68 mph. The IQR is 9 mph; the middle half of the speeds are between 39 mph and 48 mph. Half of the speeds are less than 42 mph.

Question 2.
Suppose the interquartile range for the number of hours students spent playing video games during the school week was 10. What do you think about each of the following statements? Explain your reasoning.

a. About half of the students played video games for 10 hours during a school week.
Answer:
This may not be correct, as you know the width of the interval that contains the middle half of the times was 10, but you do not know where it starts or stops. You do not know the lower or upper quartile.

b. All of the students played at least 10 hours of video games during the school week.
Answer:
This may not be correct for the same reason as in part (a).

c. About half of the class could have played video games from 10 to 20 hours a week or from 15 to 25 hours.
Answer:
Either could be correct, as the only information you have is the IQR of 10, and the statement says ‘could be,” not “is.”

Question 3.
Suppose you know the following for a data set: The minimum value is 130, the lower quartile is 142, the IQR is 30, half of the data are less than 168, and the maximum value is 195.

a. Think of a context for which these numbers might make sense.
Answer:
Answers will vary. For example, one possibility is the number of calories in a serving of fruit.

b. Sketch a box plot.
Answer:

c. Are there more data values above or below the median? Explain your reasoning.
Answer:
The number of data values on either side of the median should be about the same: one-half of all of the data.

Question 4.
The speeds for the fastest dogs are given in the table below.

a. Find the five-number summary for this data set and use it to create a box plot of the speeds.
Answer:

Min = 20, Q1 = 28, Median = 30, Q3 = 36, Max = 45

b. Why is the median not in the center of the box?
Answer:
The median is not in the center of the box because about \(\frac{1}{4}\) of the speeds are between 30 and 36, and another \(\frac{1}{4}\) are closer together, between 28 and 30.

c. Write a few sentences telling your friend about the speeds of the fastest dogs.
Answer:
Half of the dogs run faster than 30 mph the fastest dog in the list is the greyhound with a speed of 45 mph. The slowest dog in the list is the Australian cattle dog with a speed of 20 mph. The middle 50% of the speeds are between 28 mph and 36 mph.

Eureka Math Grade 6 Module 6 Lesson 15 Exit Ticket Answer Key

Question 1.
Given the following information, create a box plot, and find the IQR.
For a large group of dogs, the shortest dog was 6 inches, and the tallest was 32 inches. One-half of the dogs were taller than 18 inches. One-fourth of the dogs were shorter than 15 inches. The upper quartile of the dog heights was 23 inches.

Answer:

IQR = 23 – 15 = 8
The IQR is 8.

Engage NY Eureka Math Grade 6 Module 6 Lesson 14 Answer Key

Eureka Math Grade 6 Module 6 Lesson 14 Example Answer Key

Example 1: Time to Get to School

Consider the statistical question, “What is the typical amount of time it takes for a person in your class to get to school?” The amount of time it takes to get to school in the morning varies for the students in your class. Take a minute to answer the following questions. Your class will use this information to create a dot plot.

Write your name and an estimate of the number of minutes it took you to get to school today on a sticky note.
Answer:
Answers will vary

What were some of the things you had to think about when you made your estimate?
Answer:
Answers will vary. Some examples include: Does it count when you have to wait in the car for your sister? I usually walk, but today I got a ride. Does it matter that we had to go a different way because the road was closed? The bus was late.

Exercises 1 – 4:
Here is a dot plot of the estimates of the times it took students in Mr. S’s class to get to school one morning.
Mr. S’s Class

Mr. S’s Class

Exercise 1.
Put a line on the dot plot that you think separates the times Into two groups – one group representing the longer times and the other group representing the shorter times.
Answer:
Answers may vary. Some might put the dividing line between 15 and 20.

Exercise 2.
Put another line on the dot plot that separates out the times for students who live really close to the school. Add another line that separates out the times for students who take a very long time to get to school.
Answer:
Responses will be different. Some might put a line at 30 and a line at 10.

Exercise 3.
Your dot plot should now be divided into four sections. Record the number of data values in each of the four sections.
Answer:
Answers will vary. Depending on the divisions, 7 or 8 in the lower one, 9 in the next, 5 in the next, and 5 in the upper section.

Exercise 4.
Share your marked-up dot plot with some of your classmates. Compare how each of you divided the dot plot into four sections.
Answer:
Different responses; students should recognize that the divisions might be close but that some are different.

Exercises 5 – 7: Time to Get to School

The times (in minutes) for the students in Mr. S’s class have been put in order from smallest to largest and are shown below.

5 5 5 5 7 8 8 10 10 12 12 12 12 15 15 15 15 22 22 25 25 25 30 30 35 45 60

Exercise 5.
What is the value of the median time to get to school for students in Mr. S’s class?
Answer:
There are 27 times in the data set, so the median is the 14th value in the ordered list. The median is 15.

Exercise 6.
What is the value of the lower quartile? The upper quartile?
Answer:
The lower quartile is the 7th value in the ordered list, and the upper quartile is the 21st value in the ordered list. The lower quartile is 8, and the upper quartile is 25.

Exercise 7.
The lines on the dot plot below indicate the location of the median, the lower quartile, and the upper quartile. These lines divide the data set into four parts. About what fraction of the data values are in each part?
Mr. S’s Class

Answer:
There are about \(\frac{3}{4}\) of the data values in each part.

Example 2: Making a Box Plot

A box plot is a graph made using the following five numbers: the smallest value in the data set, the lower quartile, the median, the upper quartile, and the largest value in the data set.

To make a box plot:
• Find the median of all of the data.
• Find Q1, the median of the bottom half of the data, and Q3, the median of the top half of the data.
• Draw a number line, and then draw a box that goes from Q1 to Q3.
• Draw a vertical line in the box at the value of the median.
• Draw a line segment connecting the minimum value to the box and a line segment that connects the maximum value to the box.

You will end up with a graph that looks something like this:

Now, use the given number line to make a box plot of the data below.
20, 21, 25, 31, 35, 38, 40, 42, 44

Min=
Q1=
Median =
Q3=
Max =
Answer:

The five-number summary is as follows:
Min = 20
Q1 = 23
Median = 35
Q3 = 41
Max = 44

Exercises 8 – 11: A Human Box Plot

Consider again the sticky note that you used to write down the number of minutes it takes you to get to school. If possible, you and your classmates will form a human box plot of the number of minutes it takes students in your class to get to school.

Exercise 8.
Find the median of the group. Does someone represent the median? If not, who is the closest to the median?
Answer:
Answers will vary depending on the data.

Exercise 9.
Find the maximum and minimum of the group. Who are they?
Answer:
Answers will vary depending on the data.

Exercise 10.
Find Q1 and Q3 of the group. Does anyone represent Q1 or Q3? If not, who is the closest to Q1? Who is the closest to Q3?
Answer:
Answers will vary depending on the data.

Exercise 11.
Sketch the box plot for this data set.
Answer:
Answers will vary depending on the data.

Eureka Math Grade 6 Module 6 Lesson 14 Problem Set Answer Key

Question 1.
Dot plots for the amount of time it took students in Mr. S’s and Ms. J’s classes to get to school are below.

a. Make a box plot of the times for each class.
Answer:

Mr. S’s five-number summary: 5, 10, 15, 25, 60
Ms. J’s five-number summary: 5, 16, 20, 28, 40

b. What is one thing you can see in the dot plot that you cannot see in the box plot? What is something that is easier to see In the box plot than in the dot plot?
Answer:
The dot plot shows individual times, which you cannot see in the box plot. The box plot shows the location of the median and of the lower and upper quartiles.

Question 2.
The dot plot below shows the vertical jump of some NBA players. A vertical jump is how high a player can jump from a standstill. Draw a box plot of the heights for the vertical jumps of the NBA players above the dot plot.

Answer:

Question 3.
The mean daily temperatures in degrees Fahrenheit for the month of February for a certain city are as follows:
4, 11, 14, 15, 17, 20, 30, 23, 20, 35, 35, 31, 34, 23, 15, 19, 39, 22, 15, 15, 19, 39, 22, 23, 29, 26, 29, 29

a. Make a box plot of the temperatures.
Answer:

Five number summary: 4, 16, 22.5, 29.5, 39

b. Make a prediction about the part of the United States you think the city might be located in. Explain your reasoning.
Answer:
Answers will vary. The city was probably somewhere in the northern states, either in the Midwest or Northeast, maybe Montana or Wyoming, because the temperatures are typically pretty cold in those regions.

c. Describe the temperature data distribution. Include a description of the center and spread.
Answer:
The IQR is 29. 5°F – 16°F, or 13.5°F. Half of the temperatures were near the middle between 16°F and 29.5°F. The median is 22.5°F. A quarter of the temperatures are less than 16°F but greater than or equal to 4°F. A quarter of the temperatures are greater than 29.5°F and less than or equal to 3 9°F.

Question 4.
The box plot below summarizes data from a survey of households about the number of dogs they have. Identify each of the following statements is true or false. Explain your reasoning in each case.

a. The maximum number of dogs per house is 8.
Ans;
True, because the line segment at the top goes to 8.

b. At least of the houses have 2 or more dogs.
Answer:
True, because 2 is the median.

c. All of the houses have dogs.
Answer:
False, because the lower line segment starts at 0, so at least one household does not have a dog as a pet.

d. Half of the houses surveyed have between 2 and 4 dogs.
Answer:
False, because only about 25% of the houses would have between 2 and 4 dogs.

e. Most of the houses surveyed have no dogs.
Answer:
False, because at least \(\frac{3}{4}\) of those surveyed had 1 or more dogs.

Eureka Math Grade 6 Module 6 Lesson 14 Exit Ticket Answer Key

Question 1.
Sulee explained how to make a box plot to her sister as follows:

“First, you find the smallest and largest values and put a mark halfway between them, and then put a mark halfway between that mark and each end. So, if 10 is the smallest value and 30 is the largest value, you would put a mark at 20. Then, another mark belongs halfway between 20 and 10, which would be at 15.

And then one more mark belongs halfway between 20 and 30, which would be at 25. Now, you put a box around the three middle marks, and draw lines from the box to the smallest and largest values.” Here is her box plot. What would you say to Sulee?

Answer:
Sulee is wrong. This is not the correct way to create a box plot. Sulee did not find the median or the quartiles using the data values; she just divided up the length between the smallest and largest numbers into four equal sections. For a box plot, the sections will not always have the same length, but there will be the same number of observations in the sections.

Teacher note: This Exit Ticket problem addresses a very common student misconception about box plots. Make sure students understand how the median and quartiles are used to create the four sections of the box plot.

Engage NY Eureka Math Grade 6 Module 6 Lesson 13 Answer Key

Eureka Math Grade 6 Module 6 Lesson 13 Exercise Answer Key

Exercises 1 – 4: More French Fries

Exercise 1.
In Lesson 12, you thought about the claim made by a chain restaurant that the typical number of french fries in a large bag was 82. Then, you looked at data on the number of fries in a bag from three of the restaurants.

a. How do you think the data were collected, and what problems might have come up in collecting the data?
Answer:
Answers will vary. They probably went to the restaurants and ordered a bunch of large bags of french fries. Sometimes the fries are broken, so they might have to figure out what to do with those – either count them as a whole, discord them, or put them together to make whole fries.

b. What scenario(s) would give counts that might not be representative of typical bags?
Answer:
Answers will vary. Different workers might put different amounts in a bog, so if you bought the bags at lunch, you might have different numbers than if you did it in the evening. The restaurants might weigh the bags to see that the weight was constant despite the size of the fries, so you could have some weight of fries even though you had different counts for the bags.

Exercise 2.
The medians of the top half and the medians of the bottom half of the data for each of the three restaurants are as follows: Restaurant A – 87. 5 and 77; Restaurant B – 83 and 76; Restaurant C – 84 and 78. The difference between the medians of the two halves are called the interquartile range, or IQR.

a. What is the IQR for each of the three restaurants?
Answer:
The IQR for Restaurant A is 87.5 – 77 = 10.5; Restaurant B is 83 – 76 = 7; Restaurant C is 84 – 78 = 6.

b. Which of the restaurants had the smallest IQR, and what does that tell you?
Answer:
Restaurant Chad the smallest IQR. This indicates that the spread around the median number of fries is smaller than for either of the other two restaurants. About half of the data are within a range of 6 fries and near the median, so the median is a pretty good description of what is typical.

c. The median of the bottom half of the data is called the lower quartile (denoted by Q1), and the median of the top half of the data ¡s called the upper quartile (denoted by Q3). About what fraction of the data would be between the lower and upper quartiles? Explain your thinking.
Answer:
About \(\frac{1}{2}\), or 50%, of the counts would be between the quartiles because about \(\frac{1}{4}\) of the counts are between the median and \(\frac{1}{4}\) of the lower quartile, and of the counts are between the median and the upper quartile.

Exercise 3.
Why do you think that the median of the top half of the data is called the upper quartile and the median of the bottom half of the data is called the lower quartile?
Answer:
Answers will vary. Students might say that quartile is related to quarter, and the lower quartile, the median, and the upper quartile divide the data into four sections with about one fourth, or a quarter, of the data values in each section.

Exercise 4.

a. Mark the quartiles for each restaurant on the graphs below.

Answer:

b. Does the IQR help you decide which of the three restaurants seems most likely to really have 82 fries in a typical large bag? Explain your thinking.
Answer:
The IQR does help decide which restaurant is most likely to have 82 fries in a typical large bag because the IQR explains the variability of the data. Because Restaurant Chas the smallest IQR, the middle half of the counts of the number of fries in a bag is really close to the median. In addition, Restaurant C also has the smallest range.

Exercise 5: When Should You Use the IQR?

Exercise 5.
When should you use the IQR? The data for the 2012 salaries for the Lakers basketball team are given in the two plots below. (See Problem 5 in the Problem Set from Lesson 12.)

a. The data are given in hundreds of thousands of dollars. What would a salary of 40 hundred thousand dollars be?
Answer:
The salary would be $4, 000,000.

b. The vertical lines on the top plot show the mean and the mean plus and minus the MAD. The bottom plot shows the median and the IQR. Which interval is a better picture of the typical salaries? Explain your thinking.
Answer:
The median and the IQR seem to represent the typical salaries better than the mean plus or minus the MAD. The mean salary is above all but five of the salaries. Both the mean and the MAD are affected by the three unusually large salaries in the data set.

Exercise 6: On Your Own with IQRs

Exercise 6.
Create three different examples where you might collect data and where that data might have an IQR of 20. Define a median in the context of each example. Be specific about how the data might have been collected and the units involved. Be ready to describe what the median and IQR mean in each context.
Answer:
It may be difficult for students to get started on the exercise. if students struggle, consider a class discussion of one of the examples that follow. Student answers will vary.

Some examples that could be included in this exercise are as follows: number of books read by students during a school year (some students read a lot of books, while other students may not read as many), number of movies viewed at a theater during the lost year by students in a class, number of text messages students receive during a specific day (for example, on Monday), number of commercials on TV during a specific time period that are about buying a car, number of different states students have visited, number of healthy trees on certain blocks of a city, and number of students in each classroom of a school during a specific time period.

Remind students that the goal is to have them think of data that if collected might have an IQR of approximately 20, meaning that the middle half of the data would spread out over an interval of length 20. These ideas also allow students to start thinking of the process of actually collecting data that are needed later.

Eureka Math Grade 6 Module 6 Lesson 13 Example Answer Key

Example 1: Finding the IQR

Read through the following steps. If something does not make sense to you, make a note, and raise it during class discussion. Consider the data: 1, 1,3,4,6,6,7,8, 10, 11, 11, 12, 15, 15, 17, 17, 17

Creating an IQR:

a. Put the data in order from smallest to largest.
Answer:
The data are already ordered.
1, 1,3,4,6,6,7,8, 10,11, 11, 12, 15, 15, 17,17, 17.

b. Find the minimum and maximum.
Answer:
The minimum data point is 1, and the maximum is 17.

c. Find the median.
Answer:
There are 17 data points, so the ninth one from the smallest or the largest is the median.

d. Find the lower quartile and upper quartile.
Answer:
The lower quartile (Q1) is halfway between the 4th and 5th data points (the average of 4 and 6), or 5, and the upper quartile (Q3) is halfway between the 13th and the 14th data points (the average of 15 and 15), or 15.

e. Calculate the IQR by finding the difference between Q3 and Q1.
Answer:
IQR = 15 – 5 = 10.

Eureka Math Grade 6 Module 6 Lesson 13 Problem Set Answer Key

Question 1.
The average monthly high temperatures (in degrees Fahrenheit) for St. Louis and San Francisco are given in the table below.

a. How do you think the data might have been collected?
Answer:
Someone at a park or the airport or someplace probably records the temperature every hour of every day and then takes oil of the highest daily temperatures and finds the mean.

b. Do you think it would be possible for \(\frac{1}{4}\) of the temperatures in the month of July for St. Louis to be 95°F or above? Why or why not?
Answer:
Yes, it is possible. The mean temperature in St. Louis for July is 89° F. There are 31 days in July, s0 \(\frac{1}{4}\) of the days would be about 8 days. Student answers should have 8 or more temperatures that are above 95 °F, and then the other values could be anything that would result in an overall mean of 89 °F.

For example, if the temperature was 95°F for 5 days, 100°Ffor 3 days, and 86°F for all of the rest of the days, there would be 8 days with temperatures of 95 °F or higher, and the mean for the month would be 89°F.

c. Make a prediction about how the values of the IQR for the temperatures for each city compare. Explain your thinking.
Answer:
San Francisco probably has the smaller IQR because those temperatures do not seem to vary as much as the St. Louis temperatures.

d. Find the IQR for the average monthly high temperature for each city. How do the results compare to what you predicted?
Answer:
The IQRs for San Francisco and St. Louis are 6.5°F and 33°F, respectively. This result matches my prediction in part (c).

Question 2.
The plot below shows the years in which each of 100 pennies were made.

a. What does the stack of 17 dots at 2012 representing 17 pennies tell you about the age of these pennies in 2014?
Answer:
17 pennies were made in 2012, and they would be 2 years old in 2014.

b. Here is some information about the sample of 100 pennies. The mean year they were made is 1994; the first
year any of the pennies were made was 1958; the newest pennies were made in 2012; Q1 is 1984, the median is 1994, and Q3 is 2006; the MAD is 11.5 years. Use the information to indicate the years in which the middle half of the pennies were made.
Answer:
In this case, the IQR is 22 years, so the middle half of the pennies was mode over an interval 22 years.

Question 3.

In each of parts (a) – (c), create a data set with at least 6 values such that it has the following properties:

a. A small IQR and a big range (maximum-minimum)
Answer:
Answers will vary. One example is (0, 100, 50, 50, 50, 50, 50) where the range is 100 and the IQR is 0.

b. An IQR equal to the range
Answer:
Answers will vary. One example is (10, 10, 10, 15, 20, 20, 20).

c. The lower quartile is the same as the median.
Answer:
Answers will vary. One example is (1, 1, 1, 1, 1, 5, 6.7).

Question 4.
Rank the following three data sets by the value of the IQR.

Answer:
Data set 1 has the smallest IQR at about 14 data; set 2 has the next smallest IQR at about 22 and data set 3 has the largest IQR at about 41.

Question 5.
Here are the number of fries in each of the bags from Restaurant A:
80, 72, 77, 80, 90, 85, 93, 79, 84, 73, 87, 67, 80, 86, 92, 88, 86, 88, 66, 77

a. Suppose one bag of fries had been overlooked and that bag had only 50 fries. If that value is added to the data set, would the IQR change? Explain your reasoning.
Answer:
The IQR would be larger, 12.5, because the median number of fries would be at 80 now instead of 82, which would make the lower quartile at 75 instead of 77.

b. Will adding another data value always change the IQR? Give an example to support your answer.
Answer:
No. It depends on how many values you have in the data set and what value is added. For example, if the set of data is (2,2,2,6,9,9,9), the IQR is 9 – 2 = 7. If you add another 6, the IQR would stay at 7.

Eureka Math Grade 6 Module 6 Lesson 13 Exit Ticket Answer Key

Question 1.
On the dot plot below, insert the following words in approximately the correct position. Maximum Minimum IQR Median Lower Quartile (Q1) Upper Quartile (Q3)

Answer:

Question 2.
Estimate the IQR for the data set shown in the dot plot.
Answer:
The IQR is approximately 22.

Engage NY Eureka Math Grade 6 Module 6 Lesson 12 Answer Key

Eureka Math Grade 6 Module 6 Lesson 12 Example Answer Key

Suppose a chain restaurant (Restaurant A) advertises that a typical number of french fries in a large bag is 82. The dot plot shows the number of french fries in a sample of twenty large bags from Restaurant A.

Sometimes it is useful to know what point separates a data distribution into two equal parts, where one part represents the upper half of the data values and the other part represents the lower half of the data values. This point is called the median. When the data are arranged in order from smallest to largest, the same number of values will be above the median point as below the median.

Exercises 1 – 3:

Exercise 1.
You just bought a large bag of fries from the restaurant. Do you think you have exactly 82 french fries? Why or why not?
Answer:
The number of fries in a bag seems to wary greatly from bag to bag. No bag had exactly 82 fries, so mine probably will not. The bags that were in the sample had from 66 to 93 french fries.

Exercise 2.
How many bags were in the sample?
Answer:
20 bags were port of the sample.

Exercise 3.
Which of the following statement(s) would seem to be true for the given data? Explain your reasoning.
a. Half of the bags had more than 82 fries in them.
b. Half of the bags had fewer than 82 fries in them.
c. More than half of the bags had more than 82 fries in them.
d. More than half of the bags had fewer than 82 fries in them.
e. If you got a random bag of fries, you could get as many as 93 fries.
Answer:
Statements (a) and (b) are true because there are 10 bags above 82 fries and 10 bags below 82 fries. Also, statement (e) is true because that happened once, so ¡t could probably happen again.

Example 2:
Examine the dot plot below.

a. How many data values are represented on the dot plot above?
Answer:
There are 28 data values on the dot plot.

b. How many data values should be located above the median? How many below the median? Explain.
Answer:
There should be 14 data values above the median and 14 data values below the median because the median represents the middle value in o sorted data set.

c. For this data set, 14 values are 80 or smaller, and 14 values are 85 or larger, so the median should be between 80 and 85. When the median falls between two values in a data set, we use the average of the two middle values. For this example, the two middle values are 80 and 85. What is the median of the data presented on the dot plot?
Answer:
The median of the dot plot is 82.5.

d. What does this information tell us about the data?
Answer:
The median tells us half of the students in the class scored below an 82. 5 on the science test, and the other half of the students scored above 82.5 on the science test.

Example 3:

Use the information from the dot plot In Example 2.

a. What percentage of students scored higher than the median? Lower than the median?
Answer:
50% of the students scored higher than the median, and 50% of the students scored lower than the median.

b. Suppose the teacher made a mistake, and the student who scored a 65 actually scored a 71. Would the median change? Why or why not?
Answer:
The median would not change because there would still be 14 scores below 82 5 and 14 scores above 82.5.

c. Suppose the student who scored a 65 actually scored an 89. Would the median change? Why or why not?
Answer:
The median would change because now there would be 13 scores below 82. 5 and 15 scores above 82.5, so 82.5 would not be the median.

Example 4:

A grocery store usually has three checkout lines open on Saturday afternoons. One Saturday afternoon, the store manager decided to count how many customers were waiting to check out at 10 different times. She calculated the median of her ten data values to be 8 customers.

a. Why might the median be an important number for the store manager to consider?
Answer:
Answers will vary. For example, students might point out that this means that half the time there were mœe than 8 customers waiting to check out. if there are only 3 checkout lines open, there would be a lot of people waiting to check out. She might want to consider having more checkout lines open on Saturday afternoons.

b. Give another example of when the median of a data set might provide useful information. Explain your thinking.
Answer:
Answers will vary.
Possible responses: When the data are about how much time students spend doing homework, it would be interesting to know the amount of time that more than half of the students spend on homework. If you are looking at the number of points earned in a competition, it would be good to know what number separates the top half of the competitors from the bottom half.

Exercises 4 – 5: A Skewed Distribution

Exercise 4.
The owner of the chain decided to check the number of french fries at another restaurant in the chain. Here are the data for Restaurant B: 82, 83, 83, 79, 85. 82, 78, 76, 76, 75, 78, 74, 70, 60, 82, 82, 83, 83, 83

a. How many bags of fries were counted?
Answer:
19 bags of fries were counted.

b. Sallee claims the median is 75 because she sees that 75 is the middle number in the data set listed on the previous page. She thinks half of the bags had fewer than 75 fries because there are 9 data values that come before 75 in the list, and there are 9 data values that come after 75 in the list. Do you think she would change her mind if the data were plotted in a dot plot? Why or why not?
Answer:
Yes. You cannot find the median unless the data are organized from least to greatest. Plotting the number of fries in each bag on a dot plot would order the data correctly. You would probably get a different halfway point because the data above are not ordered from least to greatest.

c. Jake said the median was 83. What would you say to Jake?
Answer:
83 is the most common number of fries in the bags (5 bags had 83 fries), but It is not in the middle of the data.

d. Betse argued that the median was halfway between 60 and 85, or 72.5. Do you think she is right? Why or why not?
Answer:
She is wrong because the median is not calculated from the distance between the largest and smallest value in the data set. This is not the same as finding a point that separates the ordered data into two ports with the same number of values in each part.

e. Chris thought the median was 82. Do you agree? Why or why not?
Answer:
Chris is correct because ¡f you order the numbers, the middle number will be the 10th number in the ordered list, with at most 9 bags that have more than 82 fries and at most 9 bogs that have fewer than 82 fries.

Exercise 5.
Calculate the mean, and compare it to the median. What do you observe about the two values? If the mean and median are both measures of center, why do you think one of them is smaller than the other?
Answer:
The mean is 78.6, and the median is 82. The bag with only 60 fries decreased the value of the mean.

Exercises 6 – 8: Finding Medians from Frequency Tables

Exercise 6.
A third restaurant (Restaurant C) tallied the number of fries for a sample of bags of french fries and found the results below.

a. How many bags of fries did they count?
Answer:
They counted 26 bags of fries.

b. What is the median number of fries for the sample of bags from this restaurant? Describe how you found your answer.
Answer:
79.5; I took half of 26, which is 13, and then counted 13 tallies from 86 to reach 80. I also counted 13 tallies from 75 to reach 79. The point halfway between 79 and 80 is the median.

Exercise 7.
Robere wanted to look more closely at the data for bags of fries that contained a smaller number of fries and bags that contained a larger number of fries. He decided to divide the data into two parts. He first found the median of the whole data set and then divided the data set into the bottom half (the values in the ordered list that are before the median) and the top half (the values in the ordered list that are after the median).

a. List the 13 values in the bottom half. Find the median of these 13 values.
Answer:
75 75 76 77 77 78 78 78 79 79 79 79 79
The median of the lower half is 78.

b. List the 13 values of the top half. Find the median of these 13 values.
Answer:
80 80 80 80 81 82 84 84 84 85 85 85 86
The median of the top half is 84.

Exercise 8.
Which of the three restaurants seems most likely to really have 82 fries in a typical bag? Explain your thinking.
Answer:
Answers will vary. The data sets for Restaurants A and B both have a median of 82. Look for answers that consider how much the data values vary around 82. Restaurant B seems to have the most bogs closest to a count of 82. The data set for Restaurant C has a median of 79.5, but the data values are not very spread out, and most are close to 82, so some students might make a case for Restaurant C.

Eureka Math Grade 6 Module 6 Lesson 12 Problem Set Answer Key

Question 1.
The amount of precipitation in each of the western states in the United States is given in the table as well as the dot plot.

a. How do the amounts vary across the states?
Answer:
Answers will vary. The spread is pretty large: 54.2 inches. Nevada has the lowest precipitation at 9.5 inches per year. Hawaii, Alaska, and Washington have more rain than most of the states. Hawaii has the most precipitation with 63.7 inches, followed by Alaska at 58.3 inches.

b. Find the median. What does the median tell you about the amount of precipitation?
Answer:
The median is 15.9 inches. Half of the western states have more than 15.9 inches of precipitation per year, and half have less.

c. Do you think the mean or median would be a better description of the typical amount of precipitation? Explain your thinking.
Answer:
The mean at 24.8 inches reflects the extreme values, while the median seems more typical at 15.9 inches.

Question 2.
Identify the following as true or false. If a statement is false, give an example showing why.

a. The median is always equal to one of the values in the data set.
Answer:
False. If the middle two values in the ordered data set are 1 and 5, the median is 3, and 3 is not in the set.

b. The median is halfway between the least and greatest values in the data set.
Answer:
False. For example, looking at the number of french fries per bog for Restaurant A in Example 1, the median is 82, which is not halfway between 66 and 93 (79.5).

c. At most, half of the values in a data set have values less than the median.
Answer:
True

d. In a data set with 25 different values, if you change the two smallest values in the data set to smaller values, the median will not be changed.
Answer:
True

e. If you add 10 to every value in a data set, the median will not change.
Answer:
False. The median will increase by 10 as well. If the data set is 1, 2, 3, 4, 5, the median is 3. For the data set 11, 12, 13, 14,15, the median is 13.

Question 3.
Make up a data set such that the following is true:

a. The data set has 11 different values, and the median is 5.
Answer:
Answers will vary. If the numbers are whole numbers, the set would be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

b. The data set has 10 values, and the median is 25.
Answer:
Answers will vary. One answer is to have ten values that are all 25’s.

c. The data set has 7 values, and the median is the same as the least value.
Answer:
Answers will vary. One answer is to have 1, 1, 1, 1, 2, 3, 4.

Question 4.
The dot plot shows the number of landline phones that a sample of people have in their homes.

a. How many people were in the sample?
Answer:
There are 25 people in the sample.

b. Why do you think three people have no landline phones in their homes?
Answer:
Possible answers: Some people might only have cell phones or some people may not be able to afford a phone or may not want a phone.

c. Find the median number of phones for the people in the sample.
Answer:
The median number of phones per home is 2.

Question 5.
The salaries of the Los Angeles Lakers for the 2012 – 2013 basketball season are given below. The salaries in the table are ordered from largest to smallest.

a. Just looking at the data, what do you notice about the salaries?
Answer:
Possible answer: A few of the salaries for the big stars like Kobe Bryant are really big, while others are very small in comparison.

b. Find the median salary, and explain what it tells you about the salaries.
Answer:
The median salary ¡s $3, 500,000 for Chris Duhon. Half of the players make more than $3, 500,000, and half of the players make less than $3, 500, 000.

c. Find the median of the lower half of the salaries and the median of the upper half of the salaries.
Answer:
$962, 195 is the median for the bottom half of the salaries. $8, 700,000 is the median for the top half of the salaries.

d. Find the width of each of the following intervals. What do you notice about the size of the interval widths, and what does that tell you about the salaries?

i. Minimum salary to the median of the lower half:
Answer:
$758, 824

ii. Median of the lower half to the median of the whole data set:
Answer:
$2,537,805

iii. Median of the whole data set to the median of the upper half:
Answer:
$5, 200, 000

iv. Median of the upper half to the highest salary:
Answer:
$19,149,149

The largest width is from the median of the upper half to the highest salary. The smaller salaries are closer together than the larger ones.

Question 6.
Use the salary table from the previous page to answer the following.

a. If you were to find the mean salary, how do you think it would compare to the median? Explain your reasoning.
Answer:
Possible answer: The mean will be a lot larger than the median because when you add in the really big salaries, the size of the mean will increase a lot.

b. Which measure do you think would give a better picture nf a typical salary for the Lakers, the mean or the median? Explain your thinking.
Answer:
Possible answer: The median seems better, os ¡t ¡s more typical of most of the salaries.

Eureka Math Grade 6 Module 6 Lesson 12 Exit Ticket Answer Key

Question 1.
What is the median age for the following data set representing the ages of students requesting tickets for a summer band concert? Explain your reasoning.
13 14 15 15 16 16 17 18 18
Answer:
The median is the 5th value in the ordered list, or 16 years, as there are 4 values less than 16 and 4 values greater than or equal to 16 (excluding the 5 value).

Question 2.
What ¡s the median number of diseased trees from a data set representing the numbers of diseased trees on each of 12 city blocks? Explain your reasoning.
11 3 3 4 6 12 9 3 8 8 8 1
Answer:
To find the median, the values first need to be ordered: 1 3 3 3 4 6 8 8 8 9 11 12.
Because there are an even number of data values, the median would be the mean of the 6th and 7th values: \(\frac{6+8}{2}\) or 7 diseased trees.

Question 3.
Describe how you would find the median for a set of data that has 35 values. How would this be different if there were 36 values?
Answer:
Answers will vary. First, you would order the data from kast to greatest. Because there are 35 values, you would look for the 18th value from the top or bottom in the ordered list. This would be the median with 17 values above and 17 values below, if the set hod 36 values, you would find the average of the middle two data values, which would be the average of the 18th and the 19th values in the ordered list.

Engage NY Eureka Math Grade 6 Module 6 Lesson 11 Answer Key

Eureka Math Grade 6 Module 6 Lesson 11 Example Answer Key

Example 1: Comparing Distributions with the Same Mean

In Lesson 10, data distribution was characterized mainly by its center (mean) and variability (MAD). How these measures help us make a decision often depends on the context of the situation. For example, suppose that two classes of students took the same test, and their grades (based on loo points) are shown in the following dot plots. The mean score for each distribution is 79 points. Would you rather be in Class A or Class B if you had a score of 79?

Exercises 1 – 6:

Exercise 1.
Looking at the dot plots, which class has the greater MAD? Explain without actually calculating the MAD.
Answer:
Class A. The data for Class A have a much wider spread. Thus, it has greater variability and a larger MAD.

Exercise 2.
If Liz had one of the highest scores in her class, in which class would she rather be? Explain your reasoning.
Answer:
She would rather be in Class A. This doss had higher scores in the 90’s, whereas Class B had a high score of only 81.

Exercise 3.
If Logan scored below average, in which class would he rather be? Explain your reasoning.
Answer:
Logan would rather be in Class B. The low scores in Class B were in the 70 ‘s, whereas Class A had low scores in the 60’s.

Your little brother asks you to replace the battery in his favorite remote control car. The car is constructed so that it is difficult to replace its battery. Your research of the lifetimes (in hours) of two different battery brands (A and B) shows the following lifetimes for 20 batteries from each brand:

Exercise 4.
To help you decide which battery to purchase, start by drawing a dot plot of the lifetimes for each brand.
Answer:

Exercise 5.
Find the mean battery lifetime for each brand, and compare them.
Answer:
The mean of Brand A is 20 hours.
The mean of Brand B is 20 hours.
Both Brand A and Brand B have the same mean lifetime.

Exercise 6.
Looking at the variability in the dot plot for each data set, give one reason you might choose Brand A. What is one reason you might choose Brand B? Explain your reasoning.
Answer:
Answers will vary.
If I choose Brand A, I might get a battery that lasts a lot longer than 20 hours, on might get a battery that has a much shorter lifetime. If I choose Brand B, I would always geta battery that lasts approximately 20 hours.

Example 2: Comparing Distributions with Different Means

You have been comparing distributions that have the same mean but different variability. As you have seen, deciding whether large variability or small variability is best depends on the context and on what is being asked. If two data distributions have different means, do you think that variability will still play a part in making decisions?
Answer:
Yes, because considering variability in addition to center provides us with more information about the distributions and allows us to make more informed decisions.

Exercises 7 – 9:

Suppose that you wanted to answer the following question: Are field crickets better predictors of air temperature than katydids? Both species of insect make chirping sounds by rubbing their front wings together.

The following data are the number of chirps (per minute) for 10 insects of each type. All the data were taken on the same evening at the same time.

Exercise 7.
Draw dot plots for these two data distributions using the same scale, going from 30 to 70. Visually, what conclusions can you draw from the dot plots?
Answer:

Visually, you can see that the value for the mean number of chirps is higher for the katydids. The variability looks to be similar.

Exercise 8.
Calculate the mean and MAD for each distribution.
Answer:
Crickets: The mean is 35 chirps per minute.
The sum of all the distances from the mean is 12 because 0 + 3 + 0 + 2 + 1 + 1 + 3 + 0 + 1 + 1 = 12. Therefore, the MAD is 1.2 chirps per minute because \(\frac{12}{10}\) = 1. 2.

Katydids: The mean is 64 chirps per minute.

The sum of all the distances from the mean is 16 because 2 + 2 + 3 + 0 + 1 + 2 + 4 + 0 + 2 + 0 = 16. Therefore, the MAD is 1.6 chirps per minute because \(\frac{16}{10}\) = 1.6.

Exercise 9.
The outside temperature T, in degrees Fahrenheit, can be predicted by using two different formulas. The formulas include the mean number of chirps per minute made by crickets or katydids.

a. For crickets, T is predicted by adding 40 to the mean number of chirps per minute. What value of T is being predicted by the crickets?
Answer:
The predicted temperature is 35 + 40, or 75 degrees.

b. For katydids, T is predicted by adding 161 to the mean number of chirps per minute and then dividing the sum by 3. What value of T is being predicted by the katydids?
Answer:
The predicted temperature is \(\frac{(64+161)}{3}\), or 75 degrees.

c. The temperature was 75 degrees Fahrenheit when these data were recorded, so using the mean from each data set gave an accurate prediction of temperature. If you were going to use the number of chirps from a single cricket or a single katydid to predict the temperature, would you use a cricket of a katydid? Explain how variability in the distributions of number of chirps played a role in your decision.
Answer:
The crickets had a smaller MAD. This indicates that an indiuidual cricket is more likely to have a number of chirps that is close to the mean.

Eureka Math Grade 6 Module 6 Lesson 11 Problem Set Answer Key

Question 1.
Two classes took the same mathematics test. Summary measures for the two classes are as follows:

a. Suppose that you received the highest score in your class. Would your score have been higher if you were in Class A or Class B? Explain your reasoning.
Answer:
My score would have been higher if I had been in Class B because the means are the same, and the variability,
as measured by the MAD, is higher in that class than it is in Class A.

b. Suppose that your score was below the mean score. In which class would you prefer to have been? Explain your reasoning.
Answer:
I would prefer to have been in Class A because the variability, as measured by the MAD, indicates a more compact distribution around the mean. In contrast, a score below the mean in Class B could be far lower than in Class A.

Question 2.
Eight of each of two varieties of tomato plants, LoveEm and Wonderful, are grown under the same conditions. The numbers of tomatoes produced from each plant of each variety are shown:

a. Draw dot plots to help you decide which variety is more productive.
Answer:

b. Calculate the mean number of tomatoes produced for each variety. Which one produces more tomatoes on average?
Answer:
The mean number of LoveEm tomatoes is 28, and the mean number of Wonderful tomatoes is 32. Wonderful produces more tomatoes on average.

c. If you want to be able to accurately predict the number of tomatoes a plant is going to produce, which variety should you choose – the one with the smaller MAD or the one with the larger MAD? Explain your reasoning.
Answer:
LoveEm produces fewer tomatoes on average but is far more consistent. Looking at the dot plots, its variability is far less than that of Wonderful tomatoes. Based on these data sets, choosing LoveEm should yield numbers in the high 20’s consistently, but the number from Wonderful could vary wildly from lower yields in the low 20’s too huge yields around 50.

d. Calculate the MAD of each plant variety.
Answer:
The sum of the distances from the mean for LoveEm is 8 because 1 + 1 + 1 + 0 + 3 + 1 + 0 + 1 = 8.
Therefore, the MAD for LoveEm is 1 tomato because \(\frac{8}{8}\) = 1.

The sum of the distances from the mean for Wonderfulis 74 because 1 + 12 + 7 + 18 + 0 + 7 + 10 + 19 = 74. Therefore, the MAD for Wonderful is 9.25 tomatoes because \(\frac{74}{8}\) = 9. 25.

Eureka Math Grade 6 Module 6 Lesson 11 Exit Ticket Answer Key

Question 1.
You need to decide which of two brands of chocolate chip cookies to buy. You really love chocolate chip cookies. The numbers of chocolate chips in each of five cookies from each brand are as follows:

a. Draw a dot plot for each set of data that shows the distribution of the number of chips for that brand. Use the same scale for both of your dot plots (one that covers the span of both distributions).
Answer:

b. Find the mean number of chocolate chips for each of the two brands. Compare the means.
Answer:
Mean for ChocFull: 18 chocolate chips
Mean for AliChoc: 18 chocolate chips
The means for the two different brands are the same.

c. Looking at your dot plots and considering variability, which brand do you prefer? Explain your reasoning.
Answer:
Students could argue either way:
1. Students who prefer ChocFull may argue that they are assured of getting 18 chips most of the time, with no fewer than 17 chips, and a bonus once in a while of 19 chips. With AliChoc, they may sometimes get more than 20 chips but would sometimes get only 14or 15 chips.

2. Students who prefer AllChoc are the risk-takers who are willing to tolerate the chance of getting only 14 or 15 chips for the chance of getting 21 or 22 chips.