Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics<\/h2>\n
We provide the weightage key of each topic by which you can figure out the important topics and prepare a schedule according to it. The schedule helps you to prepare each topic within the time period. Know the various methods of solving each problem. Become perfect in the concept of statistics with the given solved examples, guide, exploration, and also the formulae, etc.<\/p>\n
There are various topics involved in Statistics like Samples and Populations, Using Random Samples to Describe Populations, Comparing Populations, Using Random Samples to Compare Populations, Statistics, etc. Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics pdf download<\/strong> link is available in the next sections. You can also get the video links to understand the concept more clearly.<\/p>\n Performance Task<\/strong><\/p>\n Lesson: 1 Samples and Populations<\/strong><\/p>\n Lesson: 2 Using Random Samples to Describe Populations<\/strong><\/p>\n Lesson: 3 Comparing Populations<\/strong><\/p>\n Lesson: 4 Using Random Samples to Compare Populations<\/strong><\/p>\n Chapter: 8 – Statistics\u00a0<\/strong><\/p>\n STEAM Video<\/strong><\/p>\n Comparing Dogs Answer: 2. Dogs, wolves, and Dingo’s are all the same species. This species is called Canis lupus. b. If you want to know the average height of an animal in the Canis lupus species you should\u00a0 use a sample to gather data because they are grand children of extinct dogs species having same D.N.A yet different in color, shapes, hair because they having been changing from one generation to another.<\/p>\n c. The entire Canis lupus species is a sample of what population of mammals\u00a0<\/i>and under this genus he listed the dog-like carnivores including domestic dogs, wolves, and jackals.<\/p>\n Performance Task<\/strong><\/p>\n Estimating Animal Populations Chapter Exploration<\/strong> a) In the given picture, the school bus students is the population and the class room students\u00a0 is sample because individual class strength is used for collecting total school population.<\/p>\n b) In the given picture, the grizzly bear with GPS collars in a park is used as sample to collect the population of grizzly bear in a park.<\/p>\n c) In the given picture, 150 quarters coins are used as sample for the calculating all the total\u00a0 collection in all quarters circulation .<\/p>\n d) In the given picture, 10 fiction books in the library are used as sample to get the total count of all fiction books present in the library.<\/p>\n 2. Work with a partner. When a sample is random, each member of the population is equally likely to be selected. You want to know the favorite activity of students at your school. Tell whether each sample is random. Explain your reasoning. Answer: Explanation: b) Students in your math class is a not random sample because it gives the data of the students who like math.<\/p>\n c) Students who enter your school in a morning is a random sample because every student data entering into the school is collected and can help to know the favorite activity of them.<\/p>\n d) School newspaper readers is not a random sample because here it only collects the data of the students who are in that respective activity of newspaper readers not other students.<\/p>\n Vocabulary<\/strong> A population is an entire group of people or objects. A sample is a part of a population. You can gain information about a population by examining samples of the population.<\/p>\n EXPLORATION 1<\/strong><\/p>\n Using Samples of Populations Answer: Explanation: An unbiased sample\u00a0 is representative of a population. It is selected at random unbiased sample and is large enough to provide accurate data. A biased sample is not representative of a population. One or more parts of the population are favored over others Try It<\/strong><\/p>\n Question 1. Explanation: Question 2. Answer: Explanation: Question 3. Explanation: Self-Assessment for Concepts & Skills<\/strong> Question 4. Explanation: Question 5. Explanation: Self-Assessment for Problem Solving<\/strong> Question 6. Explanation: Question 7. Explanation: Review & Refresh<\/strong><\/p>\n Design a simulation that you can use to model the situation. Then use your simulation to find the experimental probability. Answer: Explanation: Question 2. Answer: Explanation: Solve the inequality. Graph the solution.<\/strong> Explanation: Question 4. Explanation: Question 5. Explanation: Concepts, Skills, & Problem Solving<\/strong> Explanation: Question 7. Answer: Explanation: IDENTIFYING POPULATIONS AND SAMPLES<\/strong> Identify the population and the sample. Question 9. IDENTIFYING BIASED AND UNBIASED SAMPLES<\/strong> Determine whether the sample is biased or unbiased. Explain. Answer: Explanation: Question 11. Answer: Explanation: Question 12. Explanation: Question 13. Answer: Explanation: Sample B -A random sample of 500 pencils from 1 machines\u00a0 is better than Sample A – A random sample of 500 pencils from 20 machines\u00a0 because its gives us the accurate and sufficient data of pencils produced and how many are defective among them. Its a biased sample with correct data.<\/p>\n CONDUCTING SURVEYS<\/strong> Determine whether you should survey the population or a sample. Explain.<\/p>\n Question 14. Explanation: Question 15. Explanation: Question 16. Answer: Explanation: Question 17. Explanation: Question 18. Explanation: Question 19. Explanation: Question 20. Answer: Explanation: b) Explanation: EXPLORATION 1<\/strong><\/p>\n Exploring Variability in Samples Try It<\/strong><\/p>\n Question 1. Answer: Explanation: Question 2. Explanation: Self-Assessment for Concepts & Skills<\/strong> Question 3. Answer: Explanation: Question 4. Explanation: Self-Assessment for Problem Solving<\/strong> Question 5. Answer: Explanation: Question 6. Explanation: Review & Refresh<\/strong><\/p>\n You ask 100 randomly chosen high school students whether they support a new college in your town. Determine whether your conclusion is valid. Explanation: <\/p>\n Question 2. Explanation: Write and solve a proportion to answer the question.<\/strong> Explanation: Question 4. Explanation: Concepts, Skills, & Problem Solving<\/strong> Explanation: Question 6. Explanation: Question 7. Answer: Explanation: b. Describe the variation of the estimates. Question 8. b) Variation of the estimates = 1.04<\/p>\n Explanation: b) Mean of estimates = 11 + 10 +9 + 10 +12 \u00f7 5 = 52 \u00f7 5 = 10.4 Question 9. Answer: b) One estimate for the mean number of nights students plan to camp next summer = 3.3.<\/p>\n Explanation: Mean value of Sample B= 14+0+0+6+5+0+1+2+2+5 \u00f7 10 = 35 \u00f7 10 = 3.5 Mean value of Sample C = 8+8+2+3+4+1+0+0+0+6 \u00f7 10 = 32 \u00f7 10 = 3.2 Mean value = 10+10+5+6+1+0+0+0+4+0 \u00f7 10 = 36 \u00f7 10 = 3.6 b) One estimate for the mean number of nights students plan to camp next summer Question 10. Explanation: Sample B = 14+0+0+6+5+0+1+2+2+5 Sample C = 8+8+2+3+4+1+0+0+0+6 Sample D = 10+10+5+6+1+0+0+0+4+0 Question 11. Explanation: Question 12. Explanation: Question 13. Explanation: Question 14. Explanation: Question 15. Explanation: EXPLORATION 1<\/strong><\/p>\n Comparing Two Data Distributions Answer: b) The data distribution hours for sleeping for both female and male students is commonly same and in the heights both of them are completely different as they is no common height point in them as same. The female students and male students data is overlapping as they are having same common sleeping hours in them\u00a0 from 8.5 to 10.5 hours. Here, the data distribution gets overlapping from 59 to 63 inches.<\/p>\n c) There is a lot of difference in the graph of age chart data distribution that of\u00a0 the first and second data distribution because in the center of 4, there is a big difference of 25 in this set.<\/p>\n Explanation: b) In the first data distribution, Female students and male students maximum sleep for 4 hours. Female students sleeping hours are maximum from 5 to 9 hours and male students sleeping hours are maximum from 6 to 8.5 hours. In the second distribution, Female students height starts from 57 to 63 inches. Male students height starts from 59 to 65 inches. Here, the data distribution gets overlapping from 59 to 63 inches. Maximum height of male students is 62 inches whereas female students is 60 inches.<\/p>\n c) The data distribution in the center is more significant because it shows almost same difference. the center difference in data set 1 of\u00a0 9.5 – 8.5 = 1hour\u00a0 for female students and male students 9 – 7 = 2hours. And maximum height of the both female and male students and who attended the class more. In the chart, the center of difference\u00a0 of exercises at \u00a0the center of 4, class of 10am has total 9+7+3+2+2+2= 25 and the 8pm class strength is zero, which is a big difference in this set.<\/p>\n Explanation: Question 2. Explanation: Question 3. Explanation: Question 4. Explanation: Total value of Data Set B = 1+1+1+1+2+2+2+2+3+3+3+3 = 24 Question 5. Explanation: Question 6. Explanation: Question 7. Explanation: Review & Refresh<\/strong><\/p>\n Twenty percent of all seventh graders have watched a horse race. Explain whether the sample closely estimates the percentage of seventh graders who have watched a horse race. Explanation: Question 2. Explanation: <\/p>\n Find the unit rate. Explanation: Question 4. Explanation: <\/p>\n Concepts, Skills, & Problem Solving<\/p>\n COMPARING TWO DATA DISTRIBUTIONS<\/strong> The double box-and-whisker plot represents the values in two data sets. (See Exploration 1, p. 337.) Explanation: Data of set B = 10 11 14 15 16 Question 6. Explanation: <\/p>\n COMPARING POPULATIONS Two data sets contain an equal number of values. The double box-and-whisker plot represents the values in the data sets. Explanation: Data of set B = 0.3 0.4 0.6 0.9 1.4 Comparison: Question 8. Explanation: Question 9. Explanation: Question 10. Explanation: Question 11. Explanation: Question 12. Explanation: Question 13. Explanation: Question 14. Explanation: EXPLORATION 1<\/strong><\/p>\n Using Random Samples b) Yes, the samples likely to be representative of all male and female students in your state because they are randomly selected students allover the state.<\/p>\n c) The data given in the table and the data given in the double box-and-whisker plot are both completely different and there is a lot of variation in the values. The double box-and-whisker plot has greater values than in the table.<\/p>\n d) According to the data in double box-and-whisker plot, the medians of male students is less than female students. So, the hours of male students spent are less than female students.<\/p>\n Explanation: Data of Female students = 4\u00a0 \u00a00\u00a0 3\u00a0 1\u00a0 1\u00a0 5\u00a0 1\u00a0 3\u00a0 5.5\u00a0 10\u00a0 2\u00a0 0\u00a0 6\u00a0 3.5\u00a0 2 b) Data of Male students = 1.5\u00a0 3\u00a0 0\u00a0 2.5\u00a0 1\u00a0 8\u00a0 2.5\u00a0 1\u00a0 3\u00a0 0\u00a0 6.5\u00a0 1\u00a0 5\u00a0 0\u00a0 5 c)\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data in the Table: d)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data in the double box-and-whisker plot: <\/p>\n Try It<\/p>\n Question 1. Explanation: Data of school B = 0\u00a0 2\u00a0 3\u00a0 3\u00a0 3\u00a0 \u00a04\u00a0 4\u00a0 4 5\u00a0 7 Question 2. Explanation: Question 3. Explanation: Data of Box B = 0\u00a0 0\u00a0 0 1\u00a0 2\u00a0 3\u00a0 3\u00a0 3 Question 4. Explanation: Question 5. Explanation: Question 6. Explanation: Review & Refresh<\/strong><\/p>\n The double dot plot shows the values in two data sets. Explanation: Mean of set B= 30 + 40 + 40 + 40 + 50 + 50 + 50 + 50 + 60 + 60 + 70 + 70 \u00f7 12 Question 2. Explanation: <\/p>\n Solve the equation. Check your solution. Explanation: Check: substitute b =\u00a0 5 in 5b – 3 = 22. Question 4. Explanation: Question 5. Explanation: Concepts, Skills, & Problem Solving<\/p>\n USING RANDOM SAMPLES<\/strong> You want to compare the numbers of hours spent on recreation each week by teachers and non-teachers in your state. You take100 random samples of 15 teachers and 100 random samples of 15 non-teachers throughout the state and record the median value of each sample. The double box-and-whisker plot shows the distributions of sample medians. (See Exploration 1, p. 343.) Question 6. Answer: Question 7. Answer: Explanation: Question 8. Explanation: Data of College B = 3\u00a0 4\u00a0 5\u00a0 5\u00a0 5\u00a0 6\u00a0 6\u00a0 7\u00a0 7\u00a0 7\u00a0 8\u00a0 9 Question 9. Answer: Explanation: Question 10. Answer: Explanation: Data of Research Lab B = 12 9 10 14 12 11 14 11 11 15 Question 11. Explanation: Question 12. Answer: b) The variability of the sample is similar\u00a0 to the variability of the sample numbers of words in part(a).<\/p>\n c) Number of words per sentence in political magazine are more than the number of words per sentence in\u00a0 sports magazine.<\/p>\n Explanation: b) c) Question 13. b) Number of students who spent most time on homework\u00a0 are the seventh grades students.<\/p>\n Explanation: b) Average of students spent on homework of sixth grades = 26 + 33 +30 \u00f7 3 = 89 \u00f7 3= 29.67 Statistics Connecting Concepts<\/p>\n Using the Problem-Solving Plan<\/strong><\/p>\n Question 1. Answer: Explanation: Question 2. Explanation: Question 3. Explanation:<\/p>\n Performance Task<\/strong><\/p>\n Estimating Animal Populations Review Vocabulary<\/strong><\/p>\n Write the definition and give an example of each vocabulary term. Answer: Graphic Organizers<\/strong> Answer: Its a double box-and-whisker plot. Chapter Self-Assessment<\/strong><\/p>\n As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal. Question 1. Question 2. Question 3. Explanation: Question 4. Answer: Question 5. Explanation: Question 6. Explanation: 8.2 Using Random Samples to Describe Populations (pp. 331\u2013336) Question 7. Answer: b.\u00a0 Yes, the factory pass inspection because not more than 2.5% of products can contain major defects, and not more than 4% of products can contain minor defects.<\/p>\n Explanation: Number of products with major defects at the factory = 2\u00a0 1 3 5 Number of products with critical defects at the factory = 0 0 0 0 b. Percent of minor defects in the factory = 20 \u00f7 125 = 0.16 Question 8. Answer: Explanation: Total number of Wheat food product = 34 + 29 + 28 = 91 Total number of Milk food product = 32 + 21 +25 = 78 Total number of Cereals food product = 22 +20 +15 = 8.3 Comparing Populations (pp. 337\u2013342) Question 9. Explanation: b. In the data given, 28 points are given in team A not in Team B.<\/p>\n Question 10. Answer: Explanation: b. Because the quotient is less than 2, the difference in the medians is not significant.<\/p>\n 8.4 Using Random Samples to Compare Populations (pp. 343\u2013348) Question 11. Explanation: Question 12. Explanation: Question 1. Answer: Explanation: Question 2. Answer: Explanation: Question 3. Explanation: Question 4. Answer: Explanation: Data of Show B = 20\u00a0 \u00a025\u00a0 \u00a0 35\u00a0 \u00a0 40\u00a0 \u00a055 Question 5. Answer: Explanation: Question 6. Explanation: Explanation: Question 2. Explanation: Question 3. Explanation: Question 4. Explanation: Question 5. Explanation: Question 6. Explanation: Question 7. Answer: Explanation: Question 8. Explanation: b) 8\/3 = X Question 9. Explanation: Question 10. Answer: Explanation: Conclusion:<\/em><\/p>\n We hope that the information given about Big Ideas Math Answers Grade 7 Chapter 8 Statistics satisfies you. You can compare all the questions in real-time and can understand the concept easily. Stick to our website to get the solution key for all chapters of BIM Grade 7 Chapter 8. Clear all your doubts from the below-given comment section and share the article with your friends.<\/p>\n","protected":false},"excerpt":{"rendered":" To enhance your knowledge you have to practice all the problems of Big Ideas Math Answers Grade 7 Chapter 8 Statistics. You can get the pdf’s of Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics from bigideasmathanswers.com for free of cost. 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<\/a>Statistics STEAM Video\/Performance Task<\/h3>\n
\nAlthough dogs and wolves are the same species, they can have very different characteristics. How are dogs and wolves similar?
\nWatch the STEAM Video \u201cComparing Dogs.\u201d Then answer the following questions.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-1.png\")
\n1. In the video, the dogs Devo and Etta are walking in a population park. Describe the of the dogs shown in the sample video. Then describe a of the dogs shown in the video. Explain your reasoning.
\n2. Dogs, wolves, and dingos are all the same species. This species is called Canis lupus
\na. Describe one possible sample of the species. Explain your reasoning.
\nb. You want to know the average height of an animal in the Canis lupus species. Would you use the entire population of the species or would you use a sample to gather data? Explain.
\nc. The entire Canis lupus species is a sample of what population? Explain.<\/p>\n
\n1. In the video, the dogs Devo and Etta are walking in a population park. The population of the dogs is very big. They may differ from color, size, hair, size and skin yet they all are the same species. Here they have shown Canis Lupus bread dogs. They all are the same bone body structured dogs yet\u00a0 are differently living for their specification of work performance.<\/p>\n
\na. A wolf with a grey coat living in forested northern regions of North America is one possible sample of the species for Canis Lupus because they resemble like wolfs yet they match to dogs bone structure and near to dogs D.N.A.<\/p>\n
\nAfter completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given a double box-and-whisker plot that represents the weights of male and female gray wolves.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-2.png\")
\nYou will be asked to compare the weights of male and female gray wolves. Why might a researcher want to compare data from two different groups of wildlife?<\/p>\n<\/a>Statistics Getting Ready for Chapter 8<\/h3>\n
\nA population is an entire group of people or objects. A sample is a part of the population. You can use a sample to make an inference, or conclusion about a population.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-3.png\")
\n1. Work with a partner. Identify the population and the sample in each pair.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-4.png\")
\nAnswer:
\nA population is an entire group of people or objects. A sample is a part of the population.<\/p>\n
\na. members of the school band
\nb. students in your math class
\nc. students who enter your school in a morning
\nd. school newspaper readers<\/p>\n
\nc) Students who enter your school in a morning is a random sample because every student data entering into the school is collected and can help to know the favorite activity of them.<\/p>\n
\n2. a) Members of the school band is a not random sample because it gives only the data of the students in the school band not others.<\/p>\n
\nThe following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
\npopulation
\nunbiased sample
\nsample
\nbiased sample<\/p>\n<\/a>Lesson 8.1 Samples and Populations<\/h3>\n
\nWork with a partner. You want to make conclusions about the favorite extracurricular activities of students at your school.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-1.png\")
\na. Identify the population. Then identify five samples of the population.
\nb. When a sample is selected at random, each member of the population is equally likely to be selected. Are any of the samples in part (a) selected at random? Explain your reasoning.
\nc. How are the samples below different? Is each conclusion valid? Explain your reasoning.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-2.png\")
\nd. Write a survey question about a topic that interests you. How can you choose people to survey so that you can use the results to make a valid conclusion?<\/p>\n
\nOption B is correct to make conclusions about the favorite extracurricular activities of students at your school.<\/p>\n
\nOption B is correct because the information is collected randomly and gives the accurate result conclusions about the favorite extracurricular activities of students at your school.<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-3.png\")
<\/p>\n
\nWHAT IF?<\/strong>
\nYou want to estimate the number of twelfth-grade students in a high school who ride a bus to school. Which sample is unbiased? Explain.
\nAnswer:
\nOption A is correct.<\/p>\n
\nOption A is correct to collect the twelfth- grade population. Then identity five samples of the population n make a conclusion of the students who come in bus to school.<\/p>\n
\nYou want to estimate the number of eighth-grade students in your school who find it relaxing to listen to music. You consider two samples.<\/p>\n\n
\nWhich sample is unbiased? Explain.<\/li>\n<\/ul>\n
\nEvery fifth student whose name appears on an alphabetical list of eighth-grade students.<\/p>\n
\nEvery fifth student whose name appears on an alphabetical list of eighth-grade students is a unbiased sample because here it covers large data n gives us the correct data about the students who are relaxed while listening to music.<\/p>\n
\nFour out of five randomly chosen teenagers support the new land fill. So, you conclude that 80% of the residents of your town support the new land fill. Is the conclusion valid? Explain.
\nAnswer:
\nYes, its a valid conclusion.<\/p>\n
\nYes, its a valid conclusion because 4 out of 5 people in the town means 80% of the people are supporting the new land bill mathematically.<\/p>\n
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nWRITING<\/strong>
\nYou want to estimate the number of students in your school who play a school sport. You ask 40 honors students at random whether they play a school sport. Is this sample biased or unbiased? Explain.
\nAnswer:
\nYea, its a unbiased sample.<\/p>\n
\nYes, its a unbiased sample because it gives the information about the number of students who play a school sport in the school randomly and accurate data.<\/p>\n
\nANALYZING A CONCLUSION<\/strong>
\nYou survey 50 randomly chosen audience members at a theater about whether the theater should produce a new musical. The diagram shows the results. You conclude that 80% of the audience members support production of a new musical. Is your conclusion valid? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-4.png\")
\nAnswer:
\nYes, its a valid conclusion.<\/p>\n
\nYes, its a valid conclusion because 80% of 50 people in the audience members at a theater is equal to 40 people, they support production of a new musical. Out of which 20% of 50 people is equal to 10 people, who are against the production of a new musical.<\/p>\n
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nYou want to estimate the mean photo size on your cell phone. You choose 30 photos at random from your phone. The total size of the sample is 186 megabytes. Explain whether you can use the sample to estimate the mean size of photos on your cell phone. If so, what is your estimate?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-5.png\")
\nAnswer:
\nYes, we can use it for sample to estimate the mean size of photos on your cell phone.<\/p>\n
\nYes, we can use it for sample to estimate the mean size of photos on your cell phone because it allows to estimate the size of the cell phone.
\nMean size of photos = total size of the sample \/ photos chose
\n= 186 \u00d7 30
\n= 6.2 bytes.<\/p>\n
\nDIG DEEPER!<\/strong>
\nYou ask 50 randomly chosen employees of a company how many books they read each month. The diagram shows the results. There are 600 people employed by the company. Estimate the number of employees who read at least one book each month.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-6.png\")
\nAnswer:
\nNumber of employees who read at least one book each month = 12<\/p>\n
\nTotal number of chosen employee = 50
\nTotal number of people employed = 600
\nNumber of employees who read at least one book each month = Total number of chosen employee \/ Total number of people employed
\n= 600 \u00d7 50
\n= 12.<\/p>\n<\/a>Samples and Populations Homework & Practice 8.1<\/h3>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-7.png\")
\nQuestion 1.
\nThe probability that a meal at a restaurant is overcooked is 10%. Estimate the probability that exactly 1 of the next 2 meals is overcooked.<\/p>\n
\nThe probability that exactly 1 of the next 2 meals is overcooked = 3.34%<\/p>\n
\nThe probability that a meal at a restaurant is overcooked is 10%
\nThe probability that exactly 1 of the next 2 meals is overcooked = out of 3\u00a0 meals 1 meal is overcooked of 10%
\n= 1 \u00d7 3 \u00d7 10%
\n= 10 \u00d7 3 %
\n= 3.34 %<\/p>\n
\nThe probability that you see a butterfly during a nature center tour is 80%. The probability that you see a turtle is 40%. What is the probability of seeing both?<\/p>\n
\nThe probability of seeing both = 60%<\/p>\n
\nThe probability that you see a butterfly during a nature center tour = 80%.
\nThe probability that you see a turtle = 40%
\nThe probability of seeing both = 80 % + 40%\/2
\n= 120 \u00d7 2%
\n= 60%<\/p>\n
\nQuestion 3.
\n2x – 5 < 9
\nAnswer:
\nGraph:
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve-the-inequality.-Graph-the-solution3.png\")
<\/p>\n
\n2x – 5 < 9
\nadd 5 on both sides
\n=> 2x – 5 + 5 < 9 + 5
\n=>2x < 14
\n=> Divide both sides by 2
\n=> 2x \u00f7 2 < 14 \u00f7 2
\n=> x < 7.<\/p>\n
\n5q + 2 \u2265 – 13
\nAnswer:
\nGraph:
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve-the-inequality.-Graph-the-solution4-2.png\")
<\/p>\n
\n5q + 2 \u2265 – 13
\nSubtract 2 from both sides
\n=> 5q + 2 – 2 \u2265 – 13 – 2
\n=> 5q \u2265 –\u00a0 15
\nDivide both sides by 5
\n=> 5q \u00f7\u00a0 5 \u2265 – 15 \u00f7 5
\n=> q \u2265 – 3<\/p>\n
\n2 > 6 – 3r
\nAnswer:
\nGraph :
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve-the-inequality.-Graph-the-solution5.png\")
<\/p>\n
\n2 > 6 – 3r
\nsubtract 6 from both sides.
\n2 – 6 > 6 – 3r – 6
\n=> -4 > -3r
\n=> -4 \u00f7 -3 > r
\n=> 1. 333 > r<\/p>\n
\nUSING SAMPLES OF POPULATIONS<\/strong> You ask 50 randomly chosen artists in your town about their favorite art form. Determine whether your conclusion is valid. Justify your answer. Favorite Art Form(See Exploration 1, p. 325.)
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-8.png\")
\nQuestion 6.
\nYou conclude that drawing is the favorite art form of 60% of artists in your town.
\nAnswer:
\nThe conclusion made is not valid that drawing is the favorite art form of 60% of artists in your town.<\/p>\n
\nTotal strength of town people = 50
\nNumber of people interested in drawing = 20
\nPercentage of people interested in drawing = Total strength of town people \\times Number of people interested in drawing
\n= 20 \u00d7 50 \u00d7100
\n= 40%
\nHence, the conclusion made is not valid that drawing is the favorite art form of 60% of artists in your town.<\/p>\n
\nYou conclude that ceramics is the favorite art form of 10% of people in your town.<\/p>\n
\nThe conclusion made is\u00a0 valid that ceramics is the favorite art form of 10% of people in your town.<\/p>\n
\nTotal strength of town people = 50
\nNumber of people interested in ceramics = 5
\nPercentage of people interested in drawing = Total strength of town people \/ Number of people interested in ceramics
\n= 5 \u00d7 50 \u00d7 100
\n= 10%
\nHence, the conclusion made is\u00a0 valid that\u00a0ceramics is the favorite art form of 10% of people in your town.<\/p>\n
\nQuestion 8.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-9.png\")
\nAnswer:
\nResidents of New Jersey is the population.
\nResidents of Ocean Country is the sample.<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-10.png\")
\nAnswer:
\n4 cards is the sample.
\nAll cards in a desk is the population.<\/p>\n
\nQuestion 10.
\nYou want to estimate the number of books students in your school read over the summer. You survey every fourth student who enters the school.<\/p>\n
\nIts a unbiased sample.<\/p>\n
\nYou can estimate the number of students who are reading books over the summer by the survey every fourth student who enters the school\u00a0 because its a unbiased sample which gives you the accurate rate of information.<\/p>\n
\nYou want to estimate the number of people in a town who think that a park needs to be remodeled. You survey every 10th person who enters the park.<\/p>\n
\nIts a biased sample.<\/p>\n
\nYou cannot estimate the number of people in a town who think that a park needs to be remodeled by the survey asking every 10th person who enters the park because its a biased sample as every person entering into the park would like it to be remodeled and its not going to be accurate.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nYou want to determine the number of students in your school who have visited a science museum. You survey 50 students at random. Twenty have visited a science museum, and thirty have not. So, you conclude that40% of the students in your school have visited a science museum. Is your conclusion valid? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-11.png\")
\nAnswer:
\nIts a valid Conclusion that 40% of the students in your school have visited a science museum.<\/p>\n
\nTotal number of students the survey taken = 50
\nNumber of students in visited science museum = 20
\nNumber of students who did not visit science museum = 30
\nPercentage of students visited a science museum = Total number of students the survey taken – Number of students in visited science museum
\n= 20 \u00d750 \u00d7 100
\n= 40%<\/p>\n
\nUSING A SAMPLE<\/strong>
\nWhich sample is better for making an estimate? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-12.png\")
<\/p>\n
\nSample B -A random sample of 500 pencils from 1 machines\u00a0 is better than Sample A – A random sample of 500 pencils from 20 machines<\/p>\n
\nSample A = A random sample of 500 pencils from 20 machines.
\nSample B = A random sample of 500 pencils from 1 machines.<\/p>\n
\nYou want to know the average height of seventh graders in the United States.
\nAnswer:
\nSurvey on sample of the seventh graders should be taken.<\/p>\n
\nSurvey on sample of\u00a0 the seventh graders should be taken because the total population in United States is very large and its very difficult to survey on the population. Sample survey will be easy and we can get the information\u00a0 regarding the average height of seventh graders in the United States accurately.<\/p>\n
\nYou want to know the favorite types of music of students in your homeroom.
\nAnswer:
\nSurvey on Population should be taken.<\/p>\n
\nSurvey on the population in your homeroom should be taken because its a very small area of some limited people and its better to take than sample survey to know the favorite types of music of students.<\/p>\n
\nCRITICAL THINKING
\n<\/strong>
\nDoes increasing the size of a sample necessarily make the sample more representative of a population? Give an example to support your explanation.<\/p>\n
\nYes, increasing the size of the sample necessarily makes the sample more representative of a population. A representative sample is a group or set chosen from a larger statistical population according to specified characteristics. A random sample is a group or set chosen in a random manner from a larger population.<\/p>\n
\nYes, increasing the size of the sample necessarily makes the sample more representative of a population lead to more accurate or representative results at the same time when it comes to surveying large populations, bigger isn\u2019t always better.
\nFor Example:
\nYou want to know the average height of students in the UK .<\/p>\n
\nLOGIC<\/strong>
\nA person surveys residents of a town to determine whether a skateboarding ban should be overturned. Describe how the person can conduct the survey so that the sample is biased toward overturning the ban.
\nAnswer:
\nYes, the sample is biased toward overturning the ban.<\/p>\n
\nIf the person takes the survey on a skateboarding ban should be overturned from every tenth person living in a town then the person surveys is\u00a0 going to be a biased\u00a0 sample toward overturning the ban because every tenth person wishes to overturning on the ban.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nYou ask 20 randomly chosen environmental scientists from your state to name their favorite way to eliminate waste. There are 200 environmental scientists in your state. Estimate the number of environmental scientists in your state whose favorite way to eliminate waste is recycling.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-13.png\")
\nAnswer:
\nNumber of environmental scientists in your state whose favorite way to eliminate waste is recycling = 20<\/p>\n
\nTotal number of scientists in your state = 200
\nNumber of scientists randomly ask in your state = 20
\nOut of 20 randomly asked scientists whose favorite way to eliminate waste is recycling = 2
\n=> 2 \u00d7 20 = 10%
\nNumber of environmental scientists in your state whose favorite way to eliminate waste is recycling = Total number of scientists in your state x Out of 20 randomly asked scientists whose favorite way to eliminate waste is recycling \\times100
\n= 200\u00a0 \u00d7 10 \u00d7 100 = 20.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nTo predict the result of a mayoral election, you survey 50 likely voters at random. The diagram shows the results. Describe whether the sample can be used to predict the outcome of the election. If so, what is your prediction for the number of votes received by the winner assuming that500 people vote?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-14.png\")
\nAnswer:
\nYes,\u00a0 the sample can be used to predict the outcome of the election because\u00a0 it lets us know how voters are going to vote and whose going to be the winner.
\nNumber of votes candidate A gets\u00a0 = 330<\/p>\n
\nNumber of voters randomly taken in the survey = 50
\nTotal number of people voting = 500
\nYes,\u00a0 the sample can be used to predict the outcome of the election because\u00a0 it lets us know how voters are going to vote and whose going to be the winner.
\nAccording to survey candidate A gets 33 out of 50.
\n=> 33 \u00f7 50 \u00d7100 = 66 %
\nAccording to survey candidate B gets 12 out of 50.
\n=> 12 \u00f7 50 \u00d7100 = 24%
\nAccording to survey others gets 5 out of 50.
\n=> 5 \u00f7 50 \u00d7100 = 10%
\nSo, the winner according to survey prediction is candidate A.
\nNumber of votes candidate A gets\u00a0 = 66% of 500
\n=> 330 votes.<\/p>\n
\nDIG DEEPER!<\/strong>
\nYou ask 100 randomly chosen dog owners in your town how many dogs they own. The results are shown in the table. There are 500 dog owners in your town.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-8.1-15.png\")
\na. Estimate the median number of dogs per dog owner in your town. Justify your answer.
\nb. Estimate the mean number of dogs per dog owner in your town. Justify your answer.<\/p>\n
\na) The median number of dogs per dog owner in your town = 4.
\nb) The mean number of dogs per dog owner in your town = 20.<\/p>\n
\na) The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average.
\nThe sorted order of the frequency = 1 3 4 38 54
\nThe median number of dogs per dog owner in your town = 4<\/p>\n
\nThe mean is the average \u00a0of the numbers.
\nNumber of\u00a0 listed dogs = 5
\nTotal sum of the frequency of the dogs listed = 54 + 38 + 3 + 1 + 4 = 100
\nThe mean number of dogs per dog owner in your town = 100 \u00f7 5 = 20<\/p>\n<\/a>Lesson 8.2 Using Random Samples to Describe Populations<\/h3>\n
\nWork with a partner. Sixty percent of all seventh graders have visited a planetarium.
\na. Design a simulation using packing peanuts. Mark 60% of the packing peanuts and put them in a paper bag. What does choosing a marked peanut represent?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-1.png\")
\nb. Simulate a sample of 25 students by choosing peanuts from the bag, replacing the peanut each time. Record the results.
\nc. Find the percent of students in the sample who have visited a planetarium. Compare this value to the actual percent of all seventh graders who have visited a planetarium.
\nd. Record the percent in part(c) from each pair in the class. Use a dot plot to display the data. Describe the variation in the data.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-2.png\")
<\/p>\n
\nUse each sample to make an estimate for the number of students in your school who prefer rap music. Describe the center and the variation of the estimates.<\/p>\n
\nThe Variance is defined as the average of the squared <\/span>differences from the Mean.
\nThe center of the estimates = 2
\nThe variation of the estimates = 0.64
\nThe Standard Deviation of the estimates = 0.8<\/span><\/p>\n
\nSurvey randomly took in the school = 20
\nTotal number of students in the school = 840
\nNumber of students who choose rap music by me =\u00a0 1
\nNumber of students who choose rap music by friend A = 2
\nNumber of students who choose rap music by friend B = 1
\nNumber of students who choose rap music by friend C = 1
\nNumber of students who choose rap music by friend D = 3
\nThe center of the estimates =\u00a0 1 2 3 = 2
\nMean of the estimates =\u00a0 1+ 2 + 1 + 1 + 3 \u00f75= 8 \u00f7 5 = 1.6
\nThe variation of the estimates =\u00a0\u00a0(1- 1.6)^2 + (2 – 1.6)^2 +(1- 1.6)^2 + (1- 1.6)^2 + (3 – 1.6)^2 \u00f7 5
\n= (-0.6)^2 + (0.4)^2 + (-0.6)^2 + (-0.6)^2 + (1.4)^2 \u00f7 5
\n= 0.36 + 0.16 + 0.36 + 0.36 + 1.96\u00a0 \u00f7\u00a0 5
\n= 3.20 \u00f7 5
\n= 0.64
\nThe Standard Deviation of the estimates = \\sqrt{0.64} = 0.8<\/p>\n
\nRepeat Example 2, but estimate the medians instead of the means.
\nAnswer:
\nMedian of the students who choose pop music = 5<\/p>\n
\nThe six estimates are that students with part time jobs work = 5 7 7 3 7 7
\n8 and 9 hours per week. The estimates have an average\u00a0 range from 9 – 5 = 4 hours.
\nSorted estimates that students with part time jobs work =\u00a0 3 5 7
\nMedian of the students who choose pop music = 5<\/p>\n
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nUSING MULTIPLE RANDOM SAMPLES<\/strong>
\nUse each sample in Example 1 to make an estimate for the number of students in your school who prefer rock music. Describe the variation of the estimates.<\/p>\n
\nThe variation of the estimates = 1.84<\/p>\n
\nEstimate for the number of students in your school who prefer rock music = 4 7 5 4 3
\nSorted estimate for the number of students in your school who prefer rock music = 3 4 5\u00a0 7
\nMean of the estimate for the number of students in your school who prefer rock music = 4 +7 +5 +4 +3 \u00f7\u00a0 5
\n= 4.6
\nThe variation of the estimates = (4-4.6)^2 + ( 7-4.6)^2 + ( 5 – 4.6)^2 + ( 4-4.6)^2 + ( 3 – 4.6)^2 \u00f7 5
\n= (-0.6)^2 + ( 2.4)^2 + ( 0.4)^2 + (-0.6)^2\u00a0 + (- 1.6)^2 \u00f7 5
\n= 0.36 + 5.76 + 0.16 + 0.36 + 2.56 \u00f7 5
\n= 9.2 \u00f7 5
\n= 1.84<\/p>\n
\nESTIMATING AN AVERAGE OF A POPULATION<\/strong>
\nYou want to know the mean number of hours music students at your school practice each week. At each of three music classes you randomly survey 10 students. Your results are shown. Use all three samples to make one estimate for the mean number of hours music students practice each week.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-3.png\")
\nAnswer:
\nMean value of hours for number of students practiced for each week for all three classes A,B,C
\n= 4.93<\/p>\n
\nMean value\u00a0 of hours for number of students practiced for each week of class A
\n= 6 + 5 + 5 + 6 + 4 + 6 + 8 + 5 + 5 + 2+ 6 \\div 10
\n= 58 \u00f7 10
\n= 5.8
\nMean value of hours for number of students practiced for each week of class B
\n= 0 + 6 + 6 + 5 + 4 + 5 + 6 + 3 + 4 + 9 \u00f7 10
\n= 48 \u00f7 10
\n=4.8
\nMean value of hours for number of students practiced for each week of class C
\n= 4 + 5 + 6 + 4 + 3 + 2 + 2 + 3 + 12 + 1 \u00f7 10
\n=\u00a0 42 \u00f710
\n= 4.2
\nMean value of hours for number of students practiced for each week for all three classes A,B,C
\n= 5.8 + 4.8 + 4.2 \u00f73
\n= 14.8 \u00f7 3
\n= 4.93<\/p>\n
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nRepeat Example 3 with the assumption that 50% of all seventh graders have visited a planetarium.<\/p>\n
\nSo, most of the samples are within 10% of the actual population that of all seventh graders have visited a planetarium.<\/p>\n
\nThe actual percentage of 60% , the number of samples is 200 and the sample size is 50.
\nThe estimates are clustered around 50% , most are between 45 and 60.
\nSo, most of the samples are within 10% of the actual population.<\/p>\n
\nForty percent of all seventh graders have visited a state park. How closely do 200 random samples of 50 students estimate the percent of seventh graders who have visited a state park? Use a simulation to support your answer.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-4.png\")
\nAnswer:
\nIts a biased sample of 200 random samples of 50 students for\u00a0 estimating\u00a0 the percent of seventh graders who have visited a state park.<\/p>\n
\n40% of the all seventh graders have visited a state park.
\nTotal samples taken = 200
\nNumber of students randomly taken = 50
\nIts a biased sample of 200 random samples of 50 students for estimating the percent of seventh graders who have visited a state park because it does not gives the correct information of the seventh graders and they\u00a0 may or may not be the visitors of the park.<\/p>\n<\/a>Using Random Samples to Describe Populations Homework & Practice 8.2<\/h3>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-5.png\")
\nQuestion 1.
\nYou conclude that 85% of high school students in your town support the new college.
\nAnswer:
\nNo, its not a valid conclusion.<\/p>\n
\nBecause it is randomly chosen for only 100 people for one set.
\nIt can be 85\u00a0 do not support for another set of 100 people who are randomly chosen.
\nYou support the statement if total 85 % of the people in the town support the new college .<\/p>\n
\nYou conclude that 15% of residents in your town do not support the new college.
\nAnswer:
\nNo, its not a valid conclusion.<\/p>\n
\nNo, its not a valid conclusion because this percentage which is taken is only for respected chosen students not on the total population of the town who does not prefer new colleges in the town.<\/p>\n
\nQuestion 3.
\nWhat percent of 30 is 12?
\nAnswer:
\n3.6 is the percent of 30 is 12.<\/p>\n
\n30 = 100 %
\nX = 12
\nCross multiply the equation
\n=> X \u00d7 100 = 30 \u00d7\u00a0 12
\n=> X \u00d7\u00a0 100 = 360
\n=> X = 360 \u00d7 100
\n=> X = 3.6<\/p>\n
\n17 is what percent of 68?
\nAnswer:
\n25 percent of 68 is 17.<\/p>\n
\n68 = 100%
\n17 = X%
\nCross multiply the equation
\nX \u00d7 68 = 17 \u00d7 100
\n=> X \u00d7 68 = 1700
\n=> X = 1700 \u00f7 68
\n=> X = 25<\/p>\n
\nEXPLORING VARIABILITY IN SAMPLES<\/strong> Thirty percent of all seventh graders own a bracelet. Explain whether the sample closely estimates the percentage of seventh graders who own a bracelet. (See Exploration 1, p. 331.)
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-6.png\")
\nQuestion 5.
\n50 seventh graders, 14 own a bracelet
\nAnswer:
\nIts\u00a0 not a valid conclusion that 50 seventh graders 14 own a bracelet.<\/p>\n
\nThirty percent of all seventh graders own a bracelet.
\n30 percent of 50 seventh graders.
\n=> 30 \u00d7 50 \u00f7 100
\n=> 1500 \u00f7 100
\n=> 15.
\nIts\u00a0 not a valid conclusion that 50 seventh graders 14 own a bracelet because 15 out of 30 own a bracelet.<\/p>\n
\n30 seventh graders, 3 own a bracelet
\nAnswer:
\nIts\u00a0 not a valid conclusion that 30 seventh graders 3 own a bracelet.<\/p>\n
\n30 percent of 30 seventh graders.
\n30 \u00d7 30 \u00f7 100
\n=> 900 \u00f7 100
\n=> 9.<\/p>\n
\nUSING MULTIPLE RANDOM SAMPLES<\/strong>
\nA store owner wants to know how many of her 600 regular customers prefer canned vegetables. Each of her three cashiers randomly surveys 20 regular customers. The table shows the results.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-7.png\")
\na. Use each sample to make an estimate for the number of regular customers of the store who prefer fresh vegetables.
\nb. Describe the variation of the estimates.<\/p>\n
\na) The Estimation for the number of regular customers of the store who prefer fresh
\nvegetables = 47.
\nb) The variation of the estimates = 12.6889.<\/p>\n
\na) Total number of regular customers to the store = 600
\nTotal number of cashier who takes the survey = 3
\nNumber of customers they have taken survey = 3 \u00d7 20 = 60
\nAccording to the three survey, number of customers who willing to take fresh vegetables
\n= (11 + 14 + 12) = 47<\/p>\n
\nMean of estimates = (11 + 14 + 12) \u00f7 3 = 47 \u00f7 3 = 15.67
\nVariation of the estimates = (11 – 15.67)^2 + (14 – 15.67)^2\u00a0 + (12 – 15.67)^2 \u00f7 3
\n= ( -4.67)^2 + (-1.67)^2 + (-3.67)^2 \u00f7 3
\n= 21.8089 + 2.7889 + 13.4689 \u00f7 3
\n= 38.0667 \u00f7 3
\n= 12.6889<\/p>\n
\nUSING MULTIPLE RANDOM SAMPLES<\/strong>
\nAn arcade manager wants to know how many of his 750 regular customers prefer to visit in the winter. Each of five state members randomly surveys 25 regular customers. The table shows the results.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-8.png\")
\na. Use each sample to make an estimate for the number of regular customers who prefer to visit in the winter.
\nb. Describe the variation of the estimates.
\nAnswer:
\na) An estimate for the number of regular customers who prefer to visit in the winter according to survey sample = 52.<\/p>\n
\na) Total number of regular customers who prefer to visit in the winter taken in the five survey sample = 25 \u00d7 5 = 125
\nAn estimate for the number of regular customers who prefer to visit in the winter according to survey sample = 11 + 10 +9 + 10 +12
\n= 52<\/p>\n
\nVariation of the estimates = (11 – 10.4)^2 + ( 10 -10.4)^2 + (9\u00a0 – 10.4)^2 + ( 10 – 10.4)^2 + ( 12 -10.4)^ 2 \u00f75
\n= ( 0.6)^2 + ( -0.4)^2 + ( -1.4)^2 + ( -0.4)^2 + (1.6)^2 \u00f75
\n= 0.36 + 0.16 + 1.96 + 0.16 + 2.56 \u00f75
\n= 5.2 \u00f75
\n= 1.04<\/p>\n
\nESTIMATING A MEAN OF A POPULATION<\/strong>
\nA park ranger wants to know the mean number of nights students in your school plan to camp next summer. The park ranger randomly surveys 10 students from each class. The results are shown.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-9.png\")
\na. Use each sample to make an estimate for the mean number of nights students in your school plan to camp next summer. Describe the variation of the estimates.
\nb. Use all four samples to make one estimate for the mean number of nights students plan to camp next summer.<\/p>\n
\na) The variation of the estimates is quite increasing from one to other sample.<\/p>\n
\na)\u00a0 Mean value of\u00a0 Sample A= 0+5+2+3+0+6+0+10+3+0 \u00f7 10 =\u00a0 29 \u00f7 10 = 2.9
\nVariation of Sample A= (0-2.9)^2 + (5-2.9)^2 + (2-2.9)^2 + (3-2.9)^2 + (0-2.9)^2 + (6-2.9)^2 + (0-2.9)^2 + (10-2.9)^2 + (3-2.9)^2 + (0-2.9)^2\u00a0 \u00f7 10
\n= (-2.9)^2 + (2.1)^2 + (-0.9)^2 + (0.1)^2 + (-2.9)^2 + (3.1)^2 + (-2.9)^2 + (3.1)^2 + (0.1)^2 + (-2.9)^2\u00a0 \u00f7 10
\n= 8.41 + 4.41 + 0.81 + 0.01 + 8.41 + 9.61 + 8.41 + 9.61 +0.01 + 8.41 \u00f7 10
\n= 58.1 \u00f7 10
\n=5.81<\/p>\n
\nVariation of Sample B= (14-3.5)^2 + (0-3.5)^2 + (0-3.5)^2 + (6-3.5)^2 + (5-3.5)^2 + (0-3.5)^2 + (1-3.5)^2 + (2-3.5)^2 + (2-3.5)^2 + (5-3.5)^2 \u00f7 10
\n= (10.5)^2 + (-3.5)^2 + (-3.5)^2 + (2.5)^2 + (1.5)^2 + (-3.5)^2 + (-2.5)^2 + (-1.5)^2 + (-1.5)^2 + (1.5)^2 \u00f7 10
\n= 110.25 + 12.25 + 12.25 + 6.25 + 2.25 + 12.25 + 6.25 +2.25 + 2.25 +2.25 \u00f7 10
\n= 168.5 \u00f7 10
\n= 16.85<\/p>\n
\nVariation of Sample C= (8-3.2)^2 + (8-3.2)^2 + (2-3.2)^2 + (3-3.2)^2 + (4-3.2)^2 + (1-3.2)^2 + (0-3.2)^2 + (0-3.2)^2 + (0-3.2)^2 + (6-3.2)^2 \u00f7 10
\n= (4.8)^2 + (4.8)^2 + (-1.2)^2 + (-0.2)^2 + (0.8)^2 + (-2.2)^2 + (-2.2)^2 + (-2.2)^2 + (-3.2)^2 + (2.8)^2 \u00f7 10
\n= 23.04 + 23.04 + 1.44 + 0.04 + 0.64 + 4.84 + 4.84 + 4.84 + 10.24 + 7.84 \u00f7 10
\n= 80.8 \u00f7 10
\n= 8.08<\/p>\n
\nVariation of Sample D= (10-3.6)^2 + (10-3.6)^2 + (5-3.6)^2 + (6-3.6)^2 + (1-3.6)^2 + (0-3.6)^2 + (0-3.6)^2 + (0-3.6)^2 + (4-3.6)^2 + (0-3.6)^2 \u00f7 10
\n= (6.4)^2 + (6.4)^2 + (1.4)^2 + (2.4)^2 + (-2.6)^2 + (-3.6)^2 + (-3.6)^2 + (-3.6)^2 + (0.4)^2 + (-3.6)^2 \u00f7 10
\n= 40.96 + 40.96 + 1.96 + 5.76 + 6.76 + 12.96 + 12.96 + 12.96 + 0.16 +12.96 \u00f7 10
\n= 148.4 \u00f7 10
\n= 14.84<\/p>\n
\n= 2.9 +3.5+3.2+3.6 \u00f7 4
\n= 13.2 \u00f7 4
\n= 3.3<\/p>\n
\nESTIMATING A MEDIAN OF A POPULATION<\/strong>
\nRepeat Exercise 9, but estimate the medians instead of the means.
\nAnswer:
\nMedian value of Sample A = 3 + 5 \u00f72 = 8 \u00f7 2 = 4.
\nMedian value of Sample B = 2 + 5 \u00f72 = 7 \u00f7 2 = 3.5.
\nMedian value of Sample C = 3 + 4 \u00f7 2 = 7 \u00f7 2 = 3.5.
\nMedian value\u00a0 of Sample C = 4 + 5 \u00f7 2 = 9 \u00f7 2 = 4.5.<\/p>\n
\nSample A= 0+5+2+3+0+6+0+10+3+0
\nAscending order of\u00a0 Sample A= 0 2 3 5 6 10
\nMedian value of Sample A = 3 + 5 \u00f72 = 8 \u00f7 2 = 4<\/p>\n
\nAscending order of Sample B = 0 1 2 5 6 14
\nMedian value of Sample B = 2 + 5 \u00f72 = 7 \u00f7 2 = 3.5<\/p>\n
\nAscending order of Sample C = 0 1 2 3 4 6 8
\nMedian value of Sample C = 3 + 4 \u00f7 2 = 7 \u00f7 2 = 3.5<\/p>\n
\nAscending order of Sample D = 0 1 4 5 6 10
\nMedian value\u00a0 of Sample C = 4 + 5 \u00f7 2 = 9 \u00f7 2 = 4.5<\/p>\n
\nDESCRIBING SAMPLE VARIATION<\/strong>
\nFifty-five percent of doctors at a hospital prescribe a particular medication. A simulation with 200 random samples of 50 doctors each is shown. Describe how the sample percentages vary.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-10.png\")
\nAnswer:
\nThere is a lot of change in the range of the sample percentages with respective to the frequency. The graph increases from 0 range and reaches to the highest range and later it declines to zero in the last.<\/p>\n
\nThe range of sample variation\u00a0 of frequency from 0.30 to 0.50 is from 0 to 18, here the graph range constantly increases from low to some extent.
\nThe range of the sample variation of frequency from 0.50 to 0.60 is from 18 to 22, here the graph range continuously increased and reached to the highest range and declines to middle range.
\nThe range of the sample variation of frequency from 0.60 to 0.80 is from\u00a0 18 to 0 , here the graph declines and in between its increases and in the last it declines to zero completely.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nSixty percent of vacationers enjoy water parks. Use technology to generate 20 samples of size 100. How closely do the samples estimate the percent of all vacationers who enjoy water parks?
\nAnswer:
\nSamples vary slightly in terms of their estimation of\u00a0 the percent of all vacationers who enjoy water parks compared to the people who does not enjoy the water parks. Out of 2000 total vacationers,\u00a0 50.65% of\u00a0 people loves the water parks that compared to 49.35% of\u00a0 the people who does not like water parks.<\/p>\n
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-12\"](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-12.png\")
\nTotal number of vacationers = 20 \u00d7 100 = 2000
\nTotal number of vacationers who enjoys the water parks = 48 + 53+ 65 + 79+ 87 +11 + 54 + 25 + 36 + 68 + 51 + 45 + 85 + 73 + 44 + 66 + 33 + 28 + 2 + 60 = 1013
\nPercent of Total number of vacationers who enjoys the water parks = 1013 \u00f72000 \u00d7100
\n= 50. 65 %
\nTotal number of vacationers who enjoys the water parks\u00a0 = 52 + 47 + 35 + 21 + 13 + 89 + 46 + 75 + 64 + 32 + 49 + 55 + 15 + 27 + 56 + 34 + 67 + 72 + 98 + 40 = 987.
\nPercent of Total number of vacationers who enjoys the water parks\u00a0 = 987 \u00f72000 \u00d7100
\n= 49.35%<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nThirty percent of all new wooden benches have a patch of chipped paint. Use technology to simulate 100 random samples of 10 wooden benches. How closely do the samples estimate the percent of all wooden benches with a patch of chipped paint?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-11.png\")
\nAnswer:
\nThe samples estimate the percent of all wooden benches with a patch of chipped paint are completely differ because almost 79 % are having chipped paint on the wooden bench and 21 % does not have chipped paint on the wooden bench. Majority goes with yes.<\/p>\n
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-13\"](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-13.png\")
\nThis means 79% of the students are having patch of chipped paint and 21% of the students are not having patch of chipped paint.<\/p>\n
\nDIG DEEPER!<\/strong>
\nYou want to predict whether a proposal will be accepted by likely voters. You randomly sample 3 different groups of 100 likely voters. The results are shown. Do you expect the proposal to be accepted? Justify your answer.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-8.2-12.png\")
\nAnswer:
\nNo, I think proposal is not to be accepted because out of 300 people opposing people strength is more than supporting people the proposal.<\/p>\n
\nNumber of people in the each group = 100
\nTotal number of people in the three groups = 3 \u00d7 100 = 300
\nNumber of people who are supporting the proposal in three samples A, B and C
\n= 48 + 52 +47 = 147
\nNumber of people who are opposing the proposal in three samples A, B and C
\n= 52 + 48 + 53 = 153
\nNo, I think proposal is not to be accepted because out of 300 people opposing people strength is more than supporting people the proposal.<\/p>\n
\nCRITICAL THINKING<\/strong>
\nExplain why public opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size.
\nAnswer:
\nPublic opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size because sampling more than 1000 people normally wont add much to the accuracy\u00a0 given the extra time an money it would cost.<\/p>\n
\nOpinion polls are usually designed to represent the opinions of a population by conducting a series of questions and then extrapolating generalities in ratio or within confidence intervals.
\nPublic opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size because sampling more than 1000 people normally wont add much to the accuracy\u00a0 given the extra time an money it would cost.
\nFor\u00a0example, in a\u00a0population of 5000, 10% would be 500. In a\u00a0population\u00a0of 200,000, 10% would be 20,000. This exceeds\u00a01000, so in this case the maximum would be\u00a01000. Even in a\u00a0population\u00a0of 200,000,\u00a0sampling 1000 people will normally give a fairly accurate result.<\/p>\n<\/a>Lesson 8.3 Comparing Populations<\/h3>\n
\nWork with a partner.
\na. Does each data display show? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-1.png\")
\nb. How can you describe the overlap of two data distributions using words? How can you describe the overlap numerically?
\nc. In which pair of data sets is the difference in the measures of center the most significant? Explain your reasoning.<\/p>\n
\na)\u00a0 Yes, the female students and male students data is overlapping as they are having same common sleeping hours in them from 8.5 to 10.5 hours and other\u00a0data distribution gets overlapping from 59 to 63 inches.<\/p>\n
\na) Overlap means to have something in common with.\u00a0Female students sleeping hours starts from 5 to 10.5 hours, whereas male students sleeping hours starts from 6 to 10.5 hours. The female students and male students data is overlapping as they are having same common sleeping hours in them\u00a0 from 8.5 to 10.5 hours.<\/p>\n
\nTry It
\n<\/strong>Question 1.
\nWhich data set is more likely to contain a value of 70?
\nAnswer:
\nNo data sets has more likely to contain a value of 70 in them.<\/p>\n
\nAbout 25% in the data set A data values are between 80 to 130.
\nAbout 50%\u00a0 in the data set B data values are between\u00a0 80 to 100.<\/p>\n
\nWhich data set is more likely to contain a value that differs from the center by no more than 3?
\nAnswer:
\nData set B has more likely to contain a value that differs from the center by no more than 3.<\/p>\n
\nData set A center is 60.
\nData set B center is 90.
\nDifference in the intervals from center of set A = 60 to 30 = 3
\nDifference in the intervals from center of set B = 90 to 80=1<\/p>\n
\nWHAT IF?<\/strong>
\nEach value in the dot plot for Data set A increases by 30. How does this affect your answers? Explain.
\nAnswer:
\nIf each value in the dot plot for Data set A increases by 30 there is going to be a difference of 2 increased value in the mean value and MAD Value of both the Data Sets.<\/p>\n
\nData Set A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data Set B
\nMean = 810 \u00f7 15 = 54\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Mean = 420 \u00f7 15 = 28
\nMAD = 244 \u00f7 15 = 16\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 MAD = 236 \u00f7 15 = 16
\nDifference in Mean \u00f7 MAD = 26 \u00f7 16 = 1.6
\nIf 30 increased in Data Set A:
\n=> Mean = 810+30 \u00f7 15 = 840 \u00f7 15 = 56
\n=> MAD = 244 + 30 \u00f7 15 = 274 \u00f7 15 = 18.267<\/p>\n
\nSelf-Assessment for Concepts & Skills<\/strong>
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nCOMPARING POPULATIONS<\/strong>
\nThe double dot plot shows two data sets. Compare the data sets using measures of center and variation. Then express the difference in the measures of center as a multiple of the measure of variation.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-2.png\")
\nAnswer:
\nSo, the difference in the means is about 1.25 times the MAD.<\/p>\n
\nTotal value of Data Set A = 1 +2+2+3+3+3+3+4+4+5 = 30
\nMean of Data Set A = 30 \u00f7 10 = 3
\nMAD of Data Set A = |\u2223\u22123\u2223 + |2\u22123\u2223 +|2\u22123\u2223 +|3\u22123\u2223 +|3\u22123\u2223\u00a0+ |3\u22123\u2223\u00a0+ |3\u22123\u2223 + |4\u22123\u2223\u00a0+ |4\u22123\u2223\u00a0+ |5\u22123\u2223 \u00f7 10
\n= 2 + 1 + 1 + 0 + 0 + 0 + 0 + 0+ 1 + 1 +2 \u00f7 10
\n= 8 \u00f7 10
\n= 0.8
\nVariance of Data Set A :
\n= (1 – 3 )^2 + (2 – 3 )^2 + (2 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (4 – 3)^2 + (4 – 3)^2 +
\n(5 -3)^2\u00a0 \u00f710
\n= (-2)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (2)^2\u00a0 \u00f710
\n= 4 + 1 + 1 + 0 + 0 + 0 + 0 + 1 +1 + 4 \u00f710
\n= 12 \u00f710
\n= 1.2<\/p>\n
\nMean of Data Set B = 24 \u00f7 12 = 2
\nMAD of\u00a0 Data Set B = |\u2223\u22122\u2223 + |1\u22122\u2223 +|1\u22122\u2223 +|1\u22122\u2223 +|2\u22122\u2223 + |2\u22122\u2223 + |2\u22122\u2223 + |2\u22122\u2223 + |3\u22122\u2223 + |3\u22122\u2223 +\u00a0 |3\u22122\u2223 + |3\u22122\u2223 \u00f7 12
\n= 1 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 \u00f7 12<\/span><\/span><\/span>
\n= 8 \u00f7 12<\/span><\/span><\/span>
\n= 0.8
\nVariance of Data Set B :
\n(1 -2)^2 + (1 -2)^2 + (1 -2)^2 + (1 -2)^2 + (2 – 2)^2 + (2 – 2)^2 + (2 – 2)^2 + (2 – 2)^2 + (3 -2)^2 +
\n(3 -2)^2 + (3 -2)^2 + (3 -2)^2\u00a0 \u00f712
\n= (-1)^2 + (-1)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (1)^2 + (1)^2 \u00f712
\n= 1 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 \u00f7 12
\n= 8 \u00f7 12
\n= 0.67
\nDifference in mean \u00f7 MAD = 3 – 2 \u00f7 0.8 = 1 \u00f7 0.8 = 1.25
\nSo, the difference in the means is about 1.25 times the MAD.<\/p>\n
\nWHICH ONE DOESN\u2019T BELONG?<\/strong>
\nYou want to compare two populations represented by skewed distributions. Which measure not does belong with the other three? Explain your reasoning.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-3.png\")
\nAnswer:
\nIQR of first data set does not belong to the other three measurements because its the difference between the first and third quartiles, whereas median is related to the center value of the data and MAD of a data set is the average distance between each data value and the mean.<\/p>\n
\nData of first set = 1 2 4
\nMedian of first data =\u00a0 2
\nData of second set = 3
\nMedian of second data = 3
\nMAD of second data set = 0.8
\nIQR of first data set = 4 – 1 = 3<\/p>\n
\nSelf-Assessment for Problem Solving<\/strong>
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nThe double box-and-whisker plot represents the weights of cats at two shelters. Are the cats significantly heavier at one shelter than at the other? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-4.png\")
\nAnswer:
\nThe cats are\u00a0 not significantly heavier at one shelter than at the other.<\/p>\n
\nData of cats weights in shelter A =\u00a0 4 5 7 9 12 pounds
\nData of cats weights in shelter B =\u00a0 6\u00a0 7 8 11 12 pounds
\nMedian of shelter A cats = 7\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of shelter B cats = 8
\nIQR of shelter A cats = 9 – 5 = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of shelter B cats =11 – 7 = 4
\nBecause the variables are same, you can describe the vision overlap by the expression of\u00a0 the difference in the medians as a multiple of the IQR .
\nDifference in Median \u00f7 IQR = 8 – 7 \u00f7 4 = 1 \u00f7 4 = 0.25
\nBecause the quotient is less than 2, the difference in the medians is not significant.
\nThe cats are\u00a0 not significantly heavier at one shelter than at the other.<\/p>\n
\nDIG DEEPER!<\/strong>
\nTornadoes in Region A travel significantly farther than tornadoes in Region B. The tornadoes in Region A travel a median of 10 miles. Create a double box-and-whisker plot that can represent the distances traveled by the tornadoes in the two regions.
\nAnswer:
\nDouble box-and-whisker plot that can represent the distances traveled by the tornadoes in the two regions.<\/p>\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.3-DIG](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.3-DIG-DEEPER7.png\")
<\/p>\n
\nData of Tornadoes in Region A = 4 6 10 14 20
\nMedian of Tornadoes in Region A = 10 miles
\nData of Tornadoes in Region B = 8 10 12 18 20
\nMedian of Tornadoes in Region B =12<\/p>\n<\/a>Comparing Populations Homework & Practice 8.3<\/h3>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-5.png\")
\nQuestion 1.
\nIn a sample of 15 seventh graders, 4 have watched a horse race.
\nAnswer:
\nNo, the sample\u00a0 estimation of\u00a0 the percentage of 15 seventh graders\u00a0 who have watched a horse race is not valid.<\/p>\n
\nNumber of seventh graders = 15
\nTwenty percent of all seventh graders have watched a horse race.
\n=> 20 % of the number of seventh graders = 20 % \u00d7 15 = 3
\nHence, 3 only watched the horse race not 4.<\/p>\n
\nIn a sample of 10 seventh graders, 6 have watched a horse race.
\nAnswer:
\nNo, the sample\u00a0 estimation of\u00a0 the percentage of 10 seventh graders\u00a0 who have watched a horse race is not valid.<\/p>\n
\nNumber of seventh graders = 10
\nTwenty percent of all seventh graders have watched a horse race.
\n=> 20 % of the number of seventh graders = 20 % \u00d7 10 = 2
\nHence, 2 only watched the horse race not 6.<\/p>\n
\nQuestion 3.
\n60 kilometers in 2 hours
\nAnswer:
\n60 kilometers in 2 hours = 30 kilometer per hour.<\/p>\n
\nTime taken to cover 60 kilometers = 2 hours
\nTime taken to cover 1 kilometer = 60kilometers \u00f7 2 hours
\n= 30 kilometer per hour<\/p>\n
\n$11.40 for 5 cans
\nAnswer:
\nCost for 1 can = $2.28.<\/p>\n
\nCost for 5 cans = $11.40
\nCost for 1 can = $11.40 \u00f7 5
\n= $2.28<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-6.png\")
\nQuestion 5.
\nDoes the data display show overlap? Explain.
\nAnswer:18.2
\nYes, the data display show overlap.<\/p>\n
\nData of set A = 1 3 5 7 9
\nMean\u00a0 of set A = 1 + 3 +\u00a0 5 + 7 + 9\u00a0 \u00f7 5
\n= 25 \u00f7 5 = 5
\nMAD of set A = |\u2223\u22125\u2223 + |3\u22125\u2223 +|5\u22125\u2223 +|7\u22125\u2223 + |9\u22125\u2223\u00a0 \u00f7 5
\n= 4 + 2 + 0 + 2 + 4 = 20 \u00f7 5
\n= 20 \u00f7 5 =4<\/p>\n
\nMean\u00a0 of set B = 8.2 + 6.7 + 4.2 + 3.2 + 2.2 \u00f7 5
\n=\u00a0 24.5 \u00f7 5 = 4.9
\nMAD of set B = |\u22230\u221218.2\u2223 + |11.5\u221218.2\u2223 +|14\u221218.2\u2223 +|15\u221218.2\u2223 + |16\u221218.2\u2223\u00a0 \u00f7 5
\n= 8.2 + 6.7 + 4.2 + 3.2 + 2.2 \u00f7 5
\n= 24.5 \u00f7 5 = 4.9
\nDifferences in means \u00f7 MAD = 18.2 – 4.9 \u00f7 5
\n= 13.3 \u00f7 5 = 2.66
\nSo, the differences in means is about\u00a0 2.66 times the MAD.<\/p>\n
\nIs there a significant difference in the measures of center for the pair of data sets? Explain.
\nAnswer:
\nNo, there is no significant difference in the measures of center for the pair of data sets because the quotient is less than 2, the difference in the medians is not significant.<\/p>\n
\nData of set A = 1 3 5 7 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Data of set B = 10 11 14 15 16
\nMedian of set A = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of set B = 14
\nIQR of set A = 7 – 3 = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of set B = 15 – 11.5 = 3.5
\nDifferences in Median \u00f7 IQR of set A = 5 – 4 \u00f7 4 = 1 \u00f7 4 =\u00a0 0.25
\nDifferences in Median \u00f7 IQR of set B = 5 – 4 \u00f7 3.5 = 0.286<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-7.png\")
\nQuestion 7.
\nCompare the data sets using measures of center and variation.
\nAnswer:
\nComparison:
\nVariation of set A\u00a0 and Variation of set B
\n=> 0.5432\u00a0 is greater than 0.1576.<\/p>\n
\nData of set A = 0.8 1.0 1.1 1.2 1.5
\nMean of set A = 0.8 + 1.0 + 1.1 + 1.2 + 1.5 \u00f7 5 = 5.6 \u00f7 5 = 1.12
\nVariation of set A = (0.8 -1.12)^2 + (1.0-1.12)^2 + (1.1 -1.12)^2 + (1.2 – 1.12)^2 + (1.5 – 1.12)^2 \u00f7 5
\n= (-0.32)^2 + (-0.12)^2\u00a0 + (-0.02)^2\u00a0 + (0.08)^2\u00a0 + (0.38)^2\u00a0 \u00f7 5
\n= 0.1024 + 0.0144 + 0.004 + 0.0064 + 0.1444 \u00f7 5
\n= 0.2716 \u00f7 5
\n= 0.5432<\/p>\n
\nMean of set B = 0.3 + 0.4 + 0.6 + 0.9 + 1.4 \u00f7 5 = 3.6 \u00f7 5 = 0.72
\nVariation of set B = (0.3 -0.72)^2 + (0.4 – 0.72)^2 + (0.6 -0.72)^2 + (0.9 – 0.72)^2 + (1.4 – 0.72)^2 \u00f7 5
\n= (-0.42)^2 + (-0.32)^2\u00a0 + (-0.12)^2\u00a0 + (0.18)^2\u00a0 + (0.68)^2\u00a0 \u00f7 5
\n= 0.1764 + 0.1024 + 0.0144 + 0.0324 + 0.4624 \u00f7 5
\n= 0.788\u00a0 \u00f7 5
\n= 0.1576<\/p>\n
\nVariation of set A\u00a0 and Variation of set B
\n=> 0.5432\u00a0 is greater than 0.1576.<\/p>\n
\nWhich data set is more likely to contain a value of 1.1?
\nAnswer:
\nData of set A is more likely to contain a value of 1.1.<\/p>\n
\nData of set A = 0.8 1.0 1.1 1.2 1.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of set B = 0.3 0.4 0.6 0.9 1.4
\nMedian of set A = 1.1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of set B = 0.6<\/p>\n
\nWhich data set is more likely to contain a value that differs from the center by 0.3?
\nAnswer:
\nNo, data set is more likely to contain a value that differs from the center by 0.3.<\/p>\n
\nData of set A = 0.8 1.0 1.1 1.2 1.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of set B = 0.3 0.4 0.6 0.9 1.4
\nMedian of set A = 1.1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of set B = 0.6<\/p>\n
\nDESCRIBING VISUAL OVERLAP<\/strong>
\nThe double dot plot shows the values in two data sets. Express the difference in the measures of center as a multiple of the measure of variation.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-8.png\")
\nAnswer:
\nBecause the quotient is less than 2, the difference in the median is not significant of data set A.
\nBecause the quotient is more than 2, the difference in the median is significant of data set B.<\/p>\n
\nData of set A = 30 35 40 45 50 55\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of set B = 10 15 20 25 30
\nMedian of set A = 40 +45 \u00f7 2 = 85 \u00f7 2 = 42.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of set B = 20
\nIQR of set A = 50 – 35 = 15\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of set B = 30 – 10 = 20
\nDifferences in medians \u00f7 IQR of set A = 42.5 – 20 \u00f7 15 = 22.5 \u00f7 15 = 1.5
\nBecause the quotient is less than 2, the difference in the median is not significant of data set A.
\nDifferences in medians \u00f7 IQR of set B = 42.5 – 20 \u00f7 10 = 22.5 \u00f7 10 = 2.25
\nBecause the quotient is more than 2, the difference in the median is significant of data set B.<\/p>\n
\nThe distributions of attendance at basketball games and volleyball games at your school are symmetric. Your friend makes a conclusion based on the calculations shown below. Is your friend correct? Explain your reasoning.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-9.png\")
\nAnswer:
\nNo, my friend is not correct.<\/p>\n
\nVolleyball Game attendance:
\nMean = 80
\nMAD = 20
\nBasketball Game attendance:
\nMean = 160
\nMAD = 20
\nThe difference in means is 4 times the MAD, so attendance at basketball games is significantly greater than the attendance at volleyball games. Since the MAD is “large”, it implies that the mean of 160 is not a reliable indicator of the other values within the data set.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nThe double box-and-whisker plot represents the goals scored per game by two hockey teams during a 20-game season. Is the number of goals scored per game significantly greater for one team than the other? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-10.png\")
\nAnswer:
\nYes, the number of goals scored per game significantly greater for one team than the other.<\/p>\n
\nData of Team A = 0 2 3 4 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Team B = 0 6 7 8 10
\nMedian of data Team A = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of data Team B = 7
\nIQR of Team A =\u00a0 4 – 2 = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0IQR of Team B = 8 – 6 = 2
\nDifferences in medians \u00f7 IQR = 7 – 3 \u00f7 2 = 4\u00a0 \u00f7 2 = 2
\nSo, the quotient is equal to 2 the difference in the medians is significant.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nThe dot plots show the test scores for two classes taught by the same teacher. Are the test scores significantly greater for one class than the other? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-11.png\")
\nAnswer:
\nNo, the test scores\u00a0 are not significantly greater for one class than the other.<\/p>\n
\nData of Class A = 70 75 80 85 90 95 100\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Class B = 70 75 80 85 90 95 100
\nMedian of data Class A = 85\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of data Class B = 85
\nIQR of Class A =\u00a0 95 – 85 = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0IQR of Class B = 80 – 70 = 10
\nBoth the variables and the center are similar. So, it is not significant.<\/p>\n
\nPROBLEM SOLVING<\/strong>
\nA scientist experiments with mold colonies of equal area. She adds a treatment to half of the colonies. After a week, she measures the area of each colony. If the areas are significantly different, the scientist will repeat the experiment. The results are shown. Should the scientist repeat the experiment? Justify your answer.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-Grade-7-Chapter-8-Statistics-8.3-12.png\")
\nAnswer:
\nYes, \u00a0the scientist should repeat the experiment because areas are significantly different from one to other.<\/p>\n
\nOrder of Data of untreated Mold = 2 3 4 5 6 7\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Order of Data of treated Mold = 0 1 2 3 6
\nMedian of untreated Mold = 4 + 5 \u00f7 2 = 9 \u00f7 2 = 4.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of treated Mold = 2
\nIQR of untreated Mold = 4 – 2 = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of treated Mold = 2 – 0 = 2
\nBecause the variables are same, you can describe the overlap in areas by expressing the difference in the medians as a multiple of the IQR.
\nDifference in Medians \u00f7 IQR = 4.5 – 2 \u00f7 2
\n= 2.5 \u00f7 2 = 1.25
\nBecause the quotient is\u00a0 less than 2, the differences in the medians is not significant.
\nThe areas are not significantly greater than one and other colony.<\/p>\n<\/a>Lesson 8.4 Using Random Samples to Compare Populations<\/h3>\n
\nWork with a partner. You want to compare the numbers of hours spent on homework each week by male and female students in your state. You take a random sample of 15 male students and 15 female students throughout the state.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-1.png\")
\na. Compare the data in each sample.
\nb. Are the samples likely to be representative of all male and female students in your state? Explain.
\nc. You take 100 random samples of 15 male students in your state and 100 random samples of 15 female students in your state and record the median of each sample. The double box-and-whisker plot shows the distributions of the sample medians. Compare the distributions in the double box-and-whisker plot with the distributions of the data in the tables.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-2.png\")
\nd. What can you conclude from the double box-and-whisker plot? Explain. d. How can you use random samples to make accurate comparisons of two populations?
\nAnswer:
\na) According to the given data in the table, the center of the Female students is greater than the center of the Male students.<\/p>\n
\na)Data of Male students = 1.5\u00a0 3\u00a0 0\u00a0 2.5\u00a0 1\u00a0 8\u00a0 2.5\u00a0 1\u00a0 3\u00a0 0\u00a0 6.5\u00a0 1\u00a0 5\u00a0 0\u00a0 5
\nArrange the data in order : 0\u00a0 1\u00a0 1.5\u00a0 2.5\u00a0 3\u00a0 5\u00a0 6.5\u00a0 8
\nMedian of Male students =\u00a0 2.5 + 3 \u00f7 2 = 5.5 \u00f7 2 = 2.75<\/p>\n
\nArrange the data in order : 0\u00a0 1\u00a0 2\u00a0 3\u00a0 3.5\u00a0 4\u00a0 5.5\u00a0 6
\nMedian of Female students = 3 + 3.5 \u00f7 2 = 6.5 \u00f7\u00a0 2 = 3.25<\/p>\n
\nData of Female students = 4\u00a0 \u00a00\u00a0 3\u00a0 1\u00a0 1\u00a0 5\u00a0 1\u00a0 3\u00a0 5.5\u00a0 10\u00a0 2\u00a0 0\u00a0 6\u00a0 3.5\u00a0 2
\nYes, the samples likely to be representative of all male and female students in your state because they are randomly selected students allover the state.<\/p>\n
\nData of Male students = 1.5\u00a0 3\u00a0 0\u00a0 2.5\u00a0 1\u00a0 8\u00a0 2.5\u00a0 1\u00a0 3\u00a0 0\u00a0 6.5\u00a0 1\u00a0 5\u00a0 0\u00a0 5
\nMedian of Male students =\u00a0 2.5 + 3 \u00f7 2 = 5.5 \u00f7 2 = 2.75
\nMean of Male students = 1.5 + 3 + 0 + 2.5 + 1 + 8 + 2.5 + 1 + 3 + 0 + 6.5 + 1 + 5 + 0 + 5 \u00f7 15
\n= 40\u00a0 \u00f7 15
\n=\u00a0 2.67
\nData of Female students = 4\u00a0 \u00a00\u00a0 3\u00a0 1\u00a0 1\u00a0 5\u00a0 1\u00a0 3\u00a0 5.5\u00a0 10\u00a0 2\u00a0 0\u00a0 6\u00a0 3.5\u00a0 2
\nMedian of Female students = 3 + 3.5 \u00f7 2 = 6.5 \u00f7\u00a0 2 = 3.25
\nMean of Female students = 4\u00a0 + 0\u00a0 +\u00a0 3 + 1 + 1 + 5\u00a0 + 1 + 3 + 5.5 + 10 + 2 + 0 + 6 + 3.5 + 2 \u00f7 15
\n= 47\u00a0 \u00f7 15
\n= 3.13
\nData in the double box-and-whisker plot:
\nData of Male students = 1.5\u00a0 2.5\u00a0 3\u00a0 3.5\u00a0 \u00a06\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Female students = 2\u00a0 3.5\u00a0 4\u00a0 5\u00a0 7
\nMean of Male students = 1.5 + 2.5 + 3 + 3.5 + 6 \u00f7 5\u00a0 \u00a0 \u00a0 \u00a0 Mean of Female students = 2 + 3.5 + 4 + 5 + 7 \u00f7 5
\n=\u00a0 16.5 \u00f7 5 = 3.3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 21.5 \u00f7 5 = 4.3<\/p>\n
\nData of Male students = 1.5\u00a0 2.5\u00a0 3\u00a0 3.5\u00a0 \u00a06\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Female students = 2\u00a0 3.5\u00a0 4\u00a0 5\u00a0 7
\nMedian of Male students = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of Female students = 4<\/p>\n
\nThe double dot plot shows the weekly reading habits of a random sample of 10 students in each of two schools. Compare the samples using measures of center and variation. Can you determine which school\u2019s students read less? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-3.png\")
\nAnswer:
\nAccording to the data of distribution, school A students are reading less compared to that of school B students.<\/p>\n
\nData of school A = 0\u00a0 0\u00a0 1\u00a0 2\u00a0 2\u00a0 3\u00a0 3\u00a0 4\u00a0 5\u00a0 5
\nMean of school A = 0 + 0 + 1 + 2 + 2 +3 + 3 + 4 + 5 + 5 \u00f7 10
\n= 25 \u00f7 10 = 2.5
\nVariation of school A = (0-2.5)^2 + (0-2.5)^2 + (1-2.5)^2 + (2-2.5)^2 + (2-2.5)^2 + (3-2.5)^2 + (3-2.5)^2 + (4-2.5)^2 + (5-2.5)^2 + (5-2.5)^2 \u00f7 10
\n= (-2.5)^2 + (-2.5)^2\u00a0 + (-1.5)^2 + (-0.5)^2 + (-0.5)^2 + (0.5)^2 + (0.5)^2\u00a0 + (-1.5)^2\u00a0 + (2.5)^2 + (2.5)^2 \u00f7 10
\n=\u00a0 6.25 + 6.25 + 2.25 + 0.25 + 0.25 + \u00a00.25 + \u00a00.25 + 2.25 + 6.25 + 6.25 \u00f7 10
\n= 30.5 \u00f7 10
\n= 3.05<\/p>\n
\nMean of school B = 0 + 2 + 3 + 3 + 3 + 4 + 4 + 4+ 5 +7 \u00f7 10
\n= 60 \u00f7 10 = 6
\nVariation of school A = (0-6)^2 + (2-6)^2 + (3-6)^2 + (3-6)^2 + (3-6)^2 + (4-6)^2 + (4-6)^2\u00a0 + (4-6)^2\u00a0 + (5-6)^2 + (7-6)^2 \u00f7 10
\n= (-6)^2 + (-4)^2 + (-3)^2 + (-3)^2 + (-3)^2 + (-2)^2 + (-2)^2 + (-2)^2 + (-1)^2 + (1)^2 \u00f7 10
\n= 36 + 16 + 9 + 9+ 9 + 4 + 4 + 4 + 1 + 1\u00f7 10
\n= 93 \u00f7 10
\n= 9.3<\/p>\n
\nWHAT IF?<\/strong>
\nEach value in the box-and-whisker plot of the sample medians for Bag A decreases by2. Does this change your answer?
\nAnswer:
\nYes, this changes the answer because which makes the medians of Bag A and Bag B same.<\/p>\n
\nData of bag A = 4\u00a0 5\u00a0 5.5\u00a0 6\u00a0 6.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of bag B = 2.5\u00a0 3\u00a0 3.5\u00a0 4\u00a0 5
\nMedian of bag A = 5.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of bag B = 3.5
\nIQR of bag A = 6 – 5 = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of bag B = 4 – 3 = 1
\nIf the sample medians for Bag A decreases by 2:
\nMedian of bag A = 5.5 – 2 = 3.5<\/p>\n
\nSelf-Assessment for Concepts & Skills<\/strong>
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nCOMPARING RANDOM SAMPLES<\/strong>
\nTwo boxes each contain 600 numbered tiles. The double dot plot shows a random sample of 8 numbers from each box. Compare the samples using measures of center and variation. Can you determine which box contains tiles with greater numbers? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-4.png\")
\nAnswer:
\nBox A contains greater numbers more than Box B because its has greater median and also in variation.<\/p>\n
\nData of Box A = 3\u00a0 4\u00a0 5\u00a0 5\u00a0 6\u00a0 6\u00a0 7\u00a0 8
\nOrder of Box A =3 4 5 6 7 8
\nMedian of Box A = 5 + 6 \u00f7 2 = 11 \u00f7 2 =5.5
\nMean of Box A = 3 + 4 + 5 + 5 + 6 + 6 + 7 + 8 \u00f7 8
\n= 44 \u00f7 8
\n= 5.5
\nVariation of Box A = (3 -5.5)^2 + (4-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (6-5.5)^2 + (6-5.5)^2 + (7-5.5)^2 + (8-5.5)^2 \u00f7 8
\n=\u00a0 (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (1.5)^2 + (2.5)^2 \u00a0 \u00f7 8
\n=\u00a0 6.25 + 2.25 + 0.25 + 0.25 + 0.25 + 2.25 + 2.25 + 6.25 \u00f7 8
\n= 20 \u00f7 8
\n= 2.5<\/p>\n
\nOrder of Box B = 0 1 2 3
\nMedian of Box B =\u00a0 1 +2 \u00f7 2 = 3 \u00f7 2 =1.5
\nMean of Box B = 0 + 0 + 0 + 1 + 2 + 3 + 3 + 3 \u00f7 8
\n= 12 \u00f7 8
\n= 1.5
\nVariation of Box B = (0-1.5)^2 + (0-1.5)^2 + (0-1.5)^2 + (1-1.5)^2 + (2-1.5)^2 + (3-1.5)^2 + (3-1.5)^2 + (3-1.5)^2\u00a0 \u00f7 8
\n=\u00a0 (-1.5)^2 + (-1.5)^2 + (-1.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (1.5)^2 + (1.5)^2\u00a0 \u00f7 8
\n=\u00a0 2.25 + 2.25 +2.25 + 0.25 +0.25 + 2.25\u00a0 + 2.25 + 2.25\u00a0 \u00f7 8
\n=\u00a0 14 \u00f7 8
\n= 1.75<\/p>\n
\nUSING MULTIPLE RANDOM SAMPLES<\/strong>
\nTwo crates each contain 750 objects. The double box-and-whisker plot shows the median weights of 50 random samples of 10 objects from each crate. Can you determine which crate weighs more? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-5.png\")
\nAnswer:
\nCrate B weighs more than Crate A.<\/p>\n
\nData of crate A = 6\u00a0 \u00a06.5\u00a0 \u00a07.5\u00a0 \u00a09\u00a0 \u00a09.5
\nMedian of crate A = 7.5
\nMean of crate A = 6 + 6.5 + 7.5 + 9 + 9.5 \u00f7 5 = 38.5 \u00f75 = 7.7
\nIQR = 9 – 6.5 = 2.5
\nData of crate B =\u00a0 9\u00a0 \u00a010\u00a0 \u00a011\u00a0 \u00a012.5\u00a0 \u00a013.5
\nMedian of crate B = 11
\nIQR = 12.5 – 10 = 2.5
\nMean of crate B = 9 + 10\u00a0 + 11\u00a0 + 12.5\u00a0 + 13.5 \u00f7 5 = 56 \u00f7 5 = 11.2
\nThe variation in the center of crate B is greater than the measure of crate A. The variation of crate A is similar to the variation of crate B.<\/p>\n
\nSelf-Assessment for Problem Solving<\/strong>
\nSolve each exercise. Then rate your understanding of the success criteria in your journal.<\/p>\n
\nThe double box-and-whisker plot represents the medians of 100 random samples of 20 battery lives for two cell phone brands. Compare the battery lives of the two brands.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-6.png\")
\nAnswer:
\nThe variation in the center of Brand B batteries is greater than the measure of Brand A batteries and the variation of Brand A batteries is similar to the variation of Brand B batteries.<\/p>\n
\nData of Brand A = 16.5\u00a0 20\u00a0 26\u00a0 28\u00a0 \u00a030\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Data of Brand B = 24\u00a0 26\u00a0 30\u00a0 34\u00a0 36
\nMedian of Brand A = 26\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of Brand B =\u00a0 30
\nIQR of Brand A = 28 – 20 = 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of Brand B = 34 – 28 = 6<\/p>\n
\nDIG DEEPER!<\/strong>
\nThe double box-and-whisker plot represents the medians of 50 random samples of 10 wait times at two patient care facilities. Which facility should you choose? Explain your reasoning.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-7.png\")
\nAnswer:
\nFacility B should be chosen than the Facility A.<\/p>\n
\nData of Facility A =\u00a0 10\u00a0 15\u00a0 20\u00a0 25\u00a0 30\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Facility B =\u00a0 5\u00a0 \u00a015\u00a0 \u00a020\u00a0 \u00a035\u00a0 45
\nMedian of Facility A = 20\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of Facility B = 20
\nIQR of Facility A = 25 – 15 = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of Facility B = 35 – 15 = 20
\nThe variation in the center of\u00a0 Facility A\u00a0 is similar to\u00a0 the measure of Facility B and the variation of Facility B\u00a0 is greater than\u00a0 the variation of Facility A .<\/p>\n<\/a>Using Random Samples to Compare Populations Homework & Practice 8.4<\/h3>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-8.png\")
\nQuestion 1.
\nCompare the data sets using measures of center and variation.
\nAnswer:
\nThe measures of center of set B is greater and the variation of it greater than that of\u00a0 the variation of set A.<\/p>\n
\nData of Set A = 10 10 20 20 20 30 30 30\u00a0 30 40 40 50\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of set B = 30 40 40 40 50 50 50 50 60 60 70 70
\nOrder: 10 20 30 40 50\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Order: 30 40 50 60 70
\nMedian of set A = 30\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of set B = 50
\nMean of set A =\u00a0 10+10 + 20 + 20 + 20 + 30 + 30 + 30 + 30 + 40 + 40 + 50\u00a0 \u00f7 12
\n=\u00a0 330\u00a0 \u00f7 12
\n= 27.5
\nVariation of set A = (10 -27.5)^2 + (10-27.5)^2 + (20-27.5)^2 + (20-27.5)^2 + (20-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (40-27.5)^2 + (40-27.5)^2 + (50-27.5)^2\u00a0 \u00f7 12
\n= (-17.5)^2 + (-17.5)^2 + (-7.5)^2 + (-7.5)^2 + (-7.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (12.5)^2 + (12.5)^2 + (22.5)^2 \u00a0\u00f7 12
\n= 306. 25 + 306.25 + 56.25\u00a0 + 56.25 + 56.25 + 6.25 + 6.25 +6.25 + 6.25 + 156.25 + 156.25 + 506.25 \u00f7 12
\n=\u00a0 1625\u00a0 \u00f7 12
\n= 135.42<\/p>\n
\n=\u00a0 610\u00a0 \u00f7 12
\n= 50.83
\nVariation of set B = (30-50.83)^2 + (40-50.83)^2 + (40-50.83)^2 + (40-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (60-50.83)^2 + (60-50.83)^2 + (70-50.83)^2 + (70-50.83)^2 \u00f7 12
\n= (-20.83)^2 + (-10.83)^2 + (-10.83)^2 + (-10.83)^2 + (-0.83)^2 + (-0.83)^2 + (-0.83)^2 + (-0.83)^2 + (10.83)^2 + (10.83)^2 + (20.83)^2 + (20.83)^2 \u00a0\u00f7 12
\n= 433.89 +\u00a0 117.29 + 117.29 + 117.29 + 0.6889\u00a0 +\u00a0 0.6889 + 0.6889 + 0.6889 + 117.29 + 117.29 +\u00a0 433.89 + 433.89 \u00f7 12
\n= 1890.8756 \u00f7 12
\n= 157.57<\/p>\n
\nAre the values of one data set significantly greater than the values of the other data set? Explain.
\nAnswer:
\nYes, the variation of center of set B is greater than the measure of set A and the variation of set B is greater than the variation of set A.<\/p>\n
\nMedian of set A = 30\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of set B = 50
\nVariation of set A = 135.42\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Variation of set B = 157.57<\/p>\n
\nQuestion 3.
\n5b – 3 = 22
\nAnswer:
\nb= 5.<\/p>\n
\n5b – 3 = 22
\n5b =\u00a0 22 + 3
\n5b = 25
\nb = 25\/5
\nb= 5.<\/p>\n
\nLHS => 5b – 3 = 22.
\n=> (5 x 5) -3 = 22
\n=> 25 – 3 = 22 = RHS<\/p>\n
\n1.5d\u00a0 + + 3 = – 4.5
\nAnswer:
\nd = -5<\/p>\n
\n1.5d\u00a0 + 3 = – 4.5
\n=>1.5d = -4.5 -3
\n=> 1.5d = -7.5
\n=> d = -7.5 \u00f7 1.5
\n=> d = – 5
\nChecking:
\nSubstitute d = – 5 in Equation:
\nLHS:\u00a0 \u00a0 1.5d\u00a0 + 3
\n=> (1.5 \u00d7 -5) + 3
\n=> -7.5 + 3
\n=> -4.5 = RHS<\/p>\n
\n4 = 9z – 2
\nAnswer:
\nz = 0.67<\/p>\n
\n4 = 9z – 2
\n=> 4 + 2 = 9z
\n=> 6 = 9z
\n=>6 \u00f7 9 = z
\n=> 0.67 = z
\nChecking:
\nSubstitute z = 0.67 in Equation:
\nRHS:\u00a0 9z – 2
\n=> (9 \u00d7 0.67) – 2
\n=> 6 – 2
\n= 4 = LHS<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-9.png\")
<\/p>\n
\nAre the samples likely to be representative of all teachers and non-teachers in your state?<\/p>\n
\nYes, the samples are likely to be representative of all teachers and non-teachers in your state because it has been given in the question.<\/p>\n
\nWhat can you conclude from the double box-and-whisker plot? Explain.<\/p>\n
\nThe number of hours spent by Non-teachers is more than the teachers.<\/p>\n
\nMedian of Non-Teachers = 7
\nMedian of Teachers = 5<\/p>\n
\nCOMPARING RANDOM SAMPLES<\/strong>
\nThe double dot plot shows the weekly running habits of athletes at two colleges. Compare the samples using measures of center and variation. Can you determine which college\u2019s athletes spend more time running? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-10.png\")
\nAnswer:
\nData of College B has the\u00a0 greater measure of center and in the measure of variation that of data of college B.
\nCollege B athletes spend more time running than college A.<\/p>\n
\nData of College A = 2\u00a0 2\u00a0 4\u00a0 4\u00a0 5\u00a0 5\u00a0 5\u00a0 5\u00a0 6\u00a0 6\u00a0 8\u00a0 8
\nOrder : 2 4 5 6 8
\nMedian of Data A =\u00a0 5
\nVariation of Data A = (2-5)^2 + (2-5)^2 + (4-5)^2 + (4-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2 + (6-5)^2 + (6-5)^2 + (8-5)^2 + (8-5)^2 \u00f7\u00a0 12
\n= (-3)^2 + (-3)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (3)^2 + (3)^2 \u00f7\u00a0 12
\n= 9 + 9 + 1 + 1 + 0+ 0 + 0 + 0\u00a0 + 1 + 1 + 9 + 9 \u00f7 12
\n= 40 \u00f7 12
\n= 3.33<\/p>\n
\nOrder : 3 4 5 6\u00a0 7 8
\nMedian of Data A =\u00a0 5 + 6 \u00f7 2 = 11 \u00f7 2 = 5.5
\nVariation of Data A = (3-5.5)^2 + (4-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (6-5.5)^2 + (6-5.5)^2 + (7-5.5)^2 + (7-5.5)^2 + (7-5.5)^2 + (8-5.5)^2 + (9-5.5)^2\u00a0 \u00f7\u00a0 12
\n= (-2.5)^2 + (-1.5)^2 + (0)^2 + (0)^2 + (0)^2 + (1.5)^2 + (1.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (3.5)^2 + (4.5)^2 \u00f7\u00a0 12
\n= 6.25 + 2.25 + 0 + 0 + 0 + 2.25 + 2.25 + 6.25 + 6.25 + 6.25 + 12.25 + 20.25 \u00f7 12
\n=\u00a0 64.25\u00a0 \u00f7 12
\n= 5.35<\/p>\n
\nUSING MULTIPLE RANDOM SAMPLES<\/strong>
\nTwo lakes each contain about 2000 fish. The double box-and-whisker plot shows the medians of 50 random samples of 14 fish lengths from each lake. Can you determine which lake contains longer fish? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-11.png\")
<\/p>\n
\nLake A is the Lake contains longer fish because the measure in the variation is comparatively greater than the measure of variation of Lake B.<\/p>\n
\nData of Lake A = 6\u00a0 7\u00a0 8\u00a0 9\u00a0 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Lake B =\u00a0 3\u00a0 4\u00a0 5\u00a0 6\u00a0 7
\nMedian of Lake A = 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of Lake B = 5
\nIQR of Lake A = 9 – 7 = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of Lake B = 6 – 4 = 2<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nTwo laboratories each produce 800 chemicals. A chemist takes 10 samples of 15 chemicals from each lab, and records the number that pass an inspection. Are the samples likely to be representative of all the chemicals for each lab? If so, which lab has more chemicals that will pass the inspection? Justify your answer.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-12.png\")
<\/p>\n
\nYes, the samples are likely to be representative of all the chemicals for each lab.
\nResearch Lab B is having the more chemicals that will pass the inspection.<\/p>\n
\nData of Research Lab A = 14 14 13 15 1515 15 13 14 12
\nOrder : 12\u00a0 13\u00a0 14\u00a0 15
\nMedian of Research Lab A =13 +14 \u00f7 2=27 \u00f7 2 = 3.5
\nMean of Research Lab A = 14 +14 +13+ 15 +15 +15 + 15 + 13 + 14 + 12 \u00f7 10
\n=\u00a0 140 \u00f7 10
\n=\u00a0 \u00a0 14<\/p>\n
\nOrder : 9\u00a0 10\u00a0 11\u00a0 12\u00a0 14\u00a0 15
\nMedian of Research Lab B = 11+12 \u00f7 2 = 23 \u00f7 2 = 11.5
\nMean of Research Lab B = 12\u00a0 + 9 + 10 +14 + 12 +11 + 14 + 11 + 11 + 15 \u00f7 10
\n= 119 \u00f7 10
\n=\u00a0 11.9
\nSo, the variation of Research Lab A is having greater values and the measure of the center\u00a0 Research Lab B is\u00a0 greater than the center of the Research Lab A.<\/p>\n
\nMODELING REAL LIFE<\/strong>
\nA farmer grows two types of corn seedlings. There are 1000 seedlings of each type. The double box-and-whisker plot represents the median growths of 50 random samples of 7 corn seedlings of each type. Compare the growths of each type of corn seedling. Justify your result.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-13.png\")
\nAnswer:
\nThe growth of Type B type of corn seedling is more that of growth of Type B of corn seedling.<\/p>\n
\nData of Type A = 1.5\u00a0 3\u00a0 3.5\u00a0 4\u00a0 5.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Type B = 2.5\u00a0 5\u00a0 6.5\u00a0 7\u00a0 9
\nMedian of Type A = 3.5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of\u00a0 Type B = 6.5
\nIQR of Type A = 4 – 3 = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0IQR of Type B = 7 – 5 = 2
\nSo, the Variation of Type B is having greater than the measure of Variation of Type A . The measure of the center\u00a0 of Type B is\u00a0 greater than the center of the Type A.<\/p>\n
\nDIG DEEPER!<\/strong>
\nYou want to compare the number of words per sentence in a sports magazine to the number of words per sentence in a political magazine.
\na. The data represent random samples of the number of words in 10 sentences from each magazine. Compare the samples using measures of center and variation. Can you use the data to make a valid comparison about the magazines? Explain.
\nSports magazine: 9, 21, 15, 14, 25, 26, 9, 19, 22, 30
\nPolitical magazine: 31, 22, 17, 5, 23, 15, 10, 20, 20, 17
\nb. The double box-and-whisker plot represents the means of 200 random samples of 20 sentences from each magazine. Compare the variability of the sample means to the variability of the sample numbers of words in part(a).
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Solutions-Grade-7-Chapter-8-Statistics-8.4-14.png\")
\nc. Make a conclusion about the numbers of words per sentence in each magazine.<\/p>\n
\na) Variation of Sports magazine is having greater measure in variation and Median of\u00a0 Sports magazine\u00a0 is having greater value of center.<\/p>\n
\na)
\nData of Sports magazine =\u00a0 9, 21, 15, 14, 25, 26, 9, 19, 22, 30
\nOrder : 9 14 15 19\u00a0 21 22 25\u00a0 26 30
\nMedian of\u00a0 Sports magazine = 21
\nMean of Sports magazine = 9 + 21 + 15 + 14 + 25 + 26 + 9 + 19 + 22 + 30 \u00f7 10 = 190 \u00f7 10 = 19
\nVariation of Sports magazine = (9-19)^2 + (21-19)^2 + (15-19)^2 + (14-19)^2 + (25-19)^2 + (26-19)^2 + (9-19)^2 + (19-19)^2 + (22-19)^2 + (30-19)^2\u00a0 \u00f7 10
\n= (-10)^2 + (2)^2 + (-4)^2 + (-5)^2 + (6)^2 + (7)^2 + (-10)^2 + (0)^2 + (3)^2 + (11)^2\u00a0 \u00f7 10
\n= 100 + 4 + 16 + 25 + 36 + 49 + 100 + 0 + 9 +121 \u00f7 10
\n= 460 \u00f7 10
\n= 46
\nData of Political magazine = 31, 22, 17, 5, 23, 15, 10, 20, 20, 17
\nOrder : 5\u00a0 10 15\u00a0 17\u00a0 20\u00a0 22\u00a0 23\u00a0 31
\nMedian of Political magazine = 17 + 20 \u00f7 2 = 37 \u00f7 2 = 13.5
\nMean of Political magazine = 31+ 22 +17 + 5 +23 + 15 +10 + 20 + 20 + 17 \u00f7 10
\n= 180\u00f7 10
\n= 18
\nVariation of Sports magazine = (31-18)^2 + (22-18)^2 + (17-18)^2 + (5-18)^2 + (23-18)^2 + (15-18)^2 + (10-18)^2 + (20-18)^2 + (20-18)^2 + (17-18)^2\u00a0 \u00f7 10
\n= (13)^2 + (4)^2 + (-1)^2 + (-13)^2 + (5)^2 + (-3)^2 + (-8)^2 + (2)^2 + (2)^2 + (-1)^2\u00a0 \u00f7 10
\n= 169 + 16 + 1 + 169 + 25 + 9 + 64 + 4 + 4 + 1\u00a0 \u00a0 \u00f7 10
\n= 462 \u00f7 10
\n= 46.2<\/p>\n
\nData of Sports magazine = 16\u00a0 19\u00a0 20\u00a0 21 23\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Political magazine = 18\u00a0 22\u00a0 23\u00a0 24\u00a0 26
\nMedian of Sports magazine = 20\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of Political magazine = 23
\nIQR of Sports magazine = 21 – 9 = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0IQR of Political magazine = 24 -22 = 2
\nMean of Sports magazine = 16 + 19 + 20 + 21 + 23\u00a0 \u00f7 5
\n= 99 \u00f7 5\u00a0 = 19.8
\nMean of Political magazine = 18 + 22 + 23 + 24 + 26 \u00f7 5
\n= 113 \u00f7 5 = 22.6<\/p>\n
\nData of Sports magazine = 16\u00a0 19\u00a0 20\u00a0 21 23
\nMaximum used words in Political magazine = 16 to 23
\n=> 23 – 16 = 7
\nData of Political magazine = 18\u00a0 22\u00a0 23\u00a0 24\u00a0 26
\nMaximum used words in Political magazine = 18 to 26.
\n=> 26 – 18 = 8<\/p>\n
\nPROJECT<\/strong>
\nYou want to compare the average amounts of time students in sixth, seventh, and eighth grade spend on homework each week.
\na. Design an experiment involving random sampling that can help you make a comparison.
\nb. Perform the experiment. Can you make a conclusion about which grade spends the most time on homework? Explain your reasoning.
\nAnswer:
\na)Number of hours :
\nTotal number of students of sixth spent = 26 + 33 +30 = 89
\nTotal number of students of seventh spent = 32\u00a0 + 46 + 40 = 118
\nTotal number of students of Eighth spent = 42 +21 +25 = 88<\/p>\n
\na) In a school, randomly chosen students are asked\u00a0 about amounts of time students in sixth, seventh, and eighth grade spend on homework each week. The school has 1000 students.\u00a0 Randomly 100 students are asked and three different sampling is taken.
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-PROJECT-13\"](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-PROJECT-13-1.png\")
<\/p>\n
\nAverage of students spent on homework of seventh grades = 32\u00a0 + 46 + 40 \u00f7 3 = 118 \u00f7 3 = 39.33
\nAverage of students spent on homework of eighth grades = 42 +21 +25 \u00f7 3 = 88 \u00f7 3 = 29.33<\/p>\n
\nIn a city, 1500 randomly chosen residents are asked how many sporting events they attend each month. The city has 80,000 residents. Estimate the number of residents in the city who attend at least one sporting event each month.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cc-1.png\")
\nUnderstand the problem.
\nYou are given the numbers of sporting events attended each month by a sample of 1500 residents. You are asked to make an estimate about the population, all residents of the city.
\nMake a plan.
\nThe sample is representative of the population because it is selected at random and is large enough to provide accurate data. So, find the percent of people in the survey that attend at least one sporting event each month, and use the percent equation to make an estimate.
\nSolve and .check.
\nUse the plan to solve the problem. Then check your solution<\/p>\n
\nNumber of residents attended one sports events = 30
\nNumber of residents in the city who attend at least one sporting event each month = 30\u00f7 1500 \u00d7 80000
\n= 0.02 \u00d7 80000 = 1600<\/p>\n
\nNumber of residents attended zero sports events = 660
\nNumber of residents attended zero sports events in the city = 660 \u00f7 1500 = 0.44 \u00d7 80000 = 35200
\nNumber of residents attended one or two sports events = 447
\nNumber of residents attended one or two sports events in the city = 447 \u00f7 1500 = 0.298 \u00d7 80000 =23840
\nNumber of residents attended three or four sports events = 276
\nNumber of residents attended three or four sports events in the city= 276 \u00f7 1500 = 0.184 \u00d7 80000 = 14720
\nNumber of residents attended five or more sports events = 87
\nNumber \u00a0of residents attended five or more sports events in the city= 87 \u00f7 1500 = 0.058 \u00d7 80000 =4640
\nTotal number of residents sample taken = Number of residents attended zero sports events + Number of residents attended one or two sports events + Number of residents attended three or four sports events + Number of residents attended five or more sports events\u00a0 + Number of residents attended one sports events
\n=>1500 = 660 + 447 + 276 + 87 + Number of residents attended one sports events
\n=> 1500 = 1470 + Number of residents attended one sports events
\n=> 1500 – 1470 = Number of residents attended one sports events
\n=> 30 = Number of residents attended one sports events
\nCheck :
\nTotal number of all events add = Total population in the city
\nLHS:
\n=> 35200 + 23840 + 14720 + 4640 + 1600
\n=> 80000
\n= RHS
\nHence, LHS = RHS<\/p>\n
\nThe dot plots show the values in two data sets. Is the difference in the measures of center for the data sets significant?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cc-2.png\")
\nAnswer:
\nYes, the difference in the measures of center for the data set of A is not significant because the quotient is not less than 2.
\nNo, the difference in the measures of center for the data set of B is significant because the quotient is less than 2.<\/p>\n
\nData of set A= – 6 – 5 -4 -4 – 4 -4 -3 -3 -3 -3
\nOrder: -3 -4 -5 -6
\nMedian of set A = -4 + – 5 \u00f7 2 = -9 \u00f7 2 = -4.5
\nMean of set A = – 6 + -5 + -4 + -4 + -4 + -4 + -3 + -3 + -3 + -3 \u00f710
\n= 35 \u00f710 = 3.5
\nIQR of set A = -6 – (-4) = -6 + 4 = -2
\nData of set B = -6 -6 -2 -2 -1 -1 -1 -1 0 0
\nOrder : 0 -1 -2 -6
\nMedian of set B = -1 + -2 \u00f7 2 = -3 \u00f7 2 = -1.5
\nMean of set B = -6 + -6 + -2 + -2 + -1 + -1 + -1 + -1 + 0 + 0 \u00f710
\n=\u00a0 20\u00a0 \u00f710 = 2.0
\nIQR of set B = -6 – (-1) = -6 +1 = -5
\nDifferences in medians \u00f7 IQR = -4.5 – (-1.5) \u00f7 -2 = -4.5 + 1.5 \u00f7 -2 = -3.0 \u00f7 -2 = 1.5
\nDifferences in medians \u00f7 IQR = -4.5 – (-1.5) \u00f7 -5 = -4.5 + 1.5 \u00f7 -5 = -3.0 \u00f7 -5 = 0.6<\/p>\n
\nYou ask 60 randomly chosen students whether they support a later starting time for school. The table shows the results. Estimate the probability that at least two out of four randomly chosen students do not support a later starting time.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cc-3.png\")
\nAnswer:<\/p>\n
\nAt the beginning of the this chapter, you watched a STEAM Video called \u201cComparing Dogs.\u201d You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cc-4.png\")
<\/p>\n<\/a>Statistics Chapter Review<\/h3>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-1.png\")
<\/p>\n
\nPopulation: Population is an entire group of objects or people.
\nSample: A Sample\u00a0 is a part of the population.
\nUnbiased Sample: Unbiased Sample is a representative of a population. It is selected random and is a large enough to provide accurate data.
\nBiased Sample: Biased Sample\u00a0is a not representative of a population. One or more parts of the population are favored over others.<\/p>\n
\nYou can use a Definition and Example Chart to organize information about a concept. Here is an example of a Definition and Example Chart for the vocabulary term sample.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-2.png\")
\nChoose and complete a graphic organizer to help you study each topic.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-3.png\")
\n1. population
\n2. shape of a distribution
\n3. mean absolute deviation (MAD)
\n4. interquartile range
\n5. double box-and-whisker plot
\n6. double dot plot<\/p>\n
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Graphic](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Graphic-Organizers.png\")
<\/p>\n
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-4.png\")
<\/p>\n
\n8.1 Samples and Populations (pp. 325\u2013330)<\/strong>
\nLearning Target: Understand how to use random samples to make conclusions about a population.<\/p>\n
\nYou want to estimate the number of students in your school whose favorite subject is biology. You survey the first 10 students who arrive at biology club. Determine whether the sample is biased or unbiased. Explain.
\nAnswer:
\nThe sample is a Biased Sample because the students in the biology club are not representative of a population. One or more parts of the students are favored over others.<\/p>\n
\nYou want to estimate the number of athletes who play soccer. Give an example of a biased sample. Give an example of an unbiased sample.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-5.png\")
\nAnswer:
\nExample of Biased sample :
\nWhat is your favorite sport? Sample is chosen from people attending a soccer game.
\nExample of Unbiased sample :
\nWhat is your favorite sport? Sample is chosen by picking names out of a register of school.<\/p>\n
\nYou want to know how the residents of your town feel about building a new baseball stadium. You randomly survey 100 people who enter the current stadium. Eighty support building a new stadium, and twenty do not. So, you conclude that80% of the residents of your town support building a new baseball stadium. Is your conclusion valid? Explain.
\nAnswer:
\nThe conclusion is valid.<\/p>\n
\nNumber of randomly chosen = 100
\nNumber of people support new stadium = 80
\nNumber of people does not support new stadium = 20
\nPercentage of people support new stadium = 80 \u00f7100 \u00d7 100= 80%<\/p>\n
\nWhich sample is better for making an estimate? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-6.png\")
<\/p>\n
\nSample B is better for making an estimate about predicting the number of students in a school who like gym class because its a unbiased sample and we get accurate data out of it.<\/p>\n
\nYou ask 125 randomly chosen students to name their favorite beverage. There are 1500 students in the school. Predict the number of students in the school whose favorite beverage is a sports drink.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-7.png\")
\nAnswer:
\nNumber of students in the school whose favorite beverage is a sports drink = 696.<\/p>\n
\nNumber of students in the school whose favorite beverage is a sports drink = Favorite Beverage sport drink \u00f7 randomly chosen students \u00d7 students in the school.
\n= 58 \u00f7 125 \u00d71500
\n= 0.464 \u00d71500
\n= 696.<\/p>\n
\nYou want to know the number of students in your state who have summer jobs. Determine whether you should survey the population or a sample. Explain.
\nAnswer:
\nYou should survey the sample.<\/p>\n
\nYou should survey the sample because a sample is the representation of population and selected at random and is a large which gives you the accurate data.<\/p>\n
\nLearning Target: Understand variability in samples of a population.<\/p>\n
\nTo pass a quality control inspection, the products at a factory must contain no critical defects, no more than 2.5% of products can contain major defects, and no more than 4% of products can contain minor defects. There are 40,000 products being shipped from a factory. Each inspector randomly samples 125 products. The table shows the results.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-8.png\")
\na. Use each sample to make an estimate for the number of products with minor defects at the factory. Describe the center and the variation of the estimates.
\nb. Use the samples to make an estimate for the percent of products with minor defects, with major defects, and with critical defects at the factory. Does the factory pass inspection? Explain.<\/p>\n
\na. The median of major defect is greater than the median of minor defect in the factory. The variation of major defects is greater than the measure of variation of minor defects in the factory.<\/p>\n
\na. Number of products with minor defects at the factory = 5 6 3 6
\nOrder: 3 5 6
\nMedian of minor defects= 5
\nMean of minor defects= 5+6+3+6 \u00f7 4 = 20 \u00f7 4 =5
\nVariation of minor defects= (5-5)^2 + (6-5)^2 + (3-5)^2 + (6-5)^2 \u00a0\u00f7 4
\n= (0)^2 + (1)^2 + (-2)^2 + (1)^2 \u00a0\u00f7 4
\n= 0 +\u00a0 1 + 4 +1 \u00f7 4
\n= 6\u00f7 4
\n= 1.5<\/p>\n
\nOrder: 1 2 3 5
\nMedian of major defects= 2 + 3 \u00f7 2 = 5 \u00f7 2 = 2.5
\nMean of major defects= 2 + 1+ 3 +5 \u00f7 4 = 11 \u00f7 4 = 2.75
\nVariation of major defects = (2-2.75)^2 + (1-2.75)^2 + (3-2.75)^2 + (5-2.75)^2 \u00a0\u00f7 4
\n= (-0.75)^2 + (-1.75)^2 + (0.25)^2 + (2.25)^2 \u00a0\u00f7 4
\n=\u00a0 0.5625 + 3.0625 +0.625 + 5.0625 \u00f7 4
\n=\u00a0 9.3125 \u00f7 4
\n= 2.33<\/p>\n
\nMedian critical defects = 0
\nMean critical defects = 0
\nVariation critical defects = 0<\/p>\n
\nPercent of major defects in the factory = 11 \u00f7125 = 0.088
\nPercent of critical defects in the factory = 0 \u00f7 125 = 0<\/p>\n
\nA scientist determines that 35% of packages of a food product contain a specific bacteria. Use technology to simulate choosing 100 random samples of 20 packages. How closely do the samples estimate the percent of all packages with the specific bacteria?<\/p>\n
\nPercent of bacteria in rice food product = 74 \u00d7 35 \u00f7 100 = 2590 \u00f7 100 = 25.9
\nPercent of bacteria in Wheat food product = 91 \u00d7 35 \u00f7 100 = 3185 \u00f7 100 = 31.85
\nPercent of bacteria in Milk food product = 78 \u00d7 35 \u00f7 100 = 2730 \u00f7 100 = 27.30
\nPercent of bacteria in Cereals food product = 57 \u00d7 35 \u00f7 100 =\u00a0 1995 \u00f7 100 = 19.95<\/p>\n
\n![\"Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.2-Using-Random-Samples-to-Describe-Populations-(pp.](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.2-Using-Random-Samples-to-Describe-Populations-pp.-331\u20133368.png\")
\nTotal number of Rice food product = 12 + 30 +32 = 74
\nPercent of bacteria in rice food product = 74 \u00d7 35 \u00f7 100 = 2590 \u00f7 100 = 25.9<\/p>\n
\nPercent of bacteria in Wheat food product = 91 \u00d7 35 \u00f7 100 = 3185 \u00f7 100 = 31.85<\/p>\n
\nPercent of bacteria in Milk food product = 78 \u00d7 35 \u00f7 100 = 2730 \u00f7 100 = 27.30<\/p>\n
\nPercent of bacteria in Cereals food product = 57 \u00d7 35 \u00f7 100 =\u00a0 1995 \u00f7 100 = 19.95<\/p>\n
\nLearning Target: Compare populations using measures of center and variation.<\/p>\n
\nThe double box-and-whisker plot represents the points scored per game by two football teams during the regular season.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-9.png\")
\na. Compare the data sets using measures of center and variation.
\nb. Which team is more likely to score 28 points in a game?
\nAnswer:
\na. Team B has greater measures of center and both the team have same measures of variation.
\nb. Team A is most likely to score 28 points in a game.<\/p>\n
\na. Data of Team A =\u00a0 0 6 14 22 28
\nMedian of Team A =\u00a0 14
\nIQR of Team A = 22 – 6 = 16
\nData of Team B = 0\u00a0 20\u00a0 32\u00a0 36\u00a0 42
\nMedian of Team B = 32
\nIQR of Team B = 36 – 20 = 16<\/p>\n
\nThe dot plots show the ages of campers at two summer camps.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-10.png\")
\na. Express the difference in the measures of center as a multiple of the measure of variation.
\nb. Are the ages of campers a tone camp significantly greater than at the other? Explain.<\/p>\n
\na. Differences in medians \u00f7 IQR = 15 – 13 \u00f7 2 = 2 \u00f7 2 = 1
\nb. No, the ages of\u00a0 campers are\u00a0 significantly greater than at the other.<\/p>\n
\na. Data of Camp A =\u00a0 13 13 13 14 14 14 14 15 15 15 15 15 15 16 16 16 16 17 17 17
\nOrder: 13 14 15 16 17
\nMedian of Camp A = 15
\nIQR of Camp A = 16 – 14 = 2
\nData of Camp B = 11 11 11 12 12 12 12 13 13 13 13 13 13 14 14 14 14 15 15 15
\nOrder: 11 12 13 14 15
\nMedian of Camp B = 13
\nIQR of\u00a0 Camp B = 14 – 12 = 2
\nDifferences in medians \u00f7 IQR = 15 – 13 \u00f7 2 = 2 \u00f7 2 = 1<\/p>\n
\nLearning Target: Use random samples to compare populations.<\/p>\n
\nThe double dot plot shows the median gas mileages of 10 random samples of 50 vehicles for two car models. Compare the samples using measures of center and variation. Can you determine which car model has a better gas mileage? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-11.png\")
\nAnswer:
\nThe Variation of Model B measures are greater compared to the measures of\u00a0 the Variation of Model A.
\nThe measures of center of Model B is greater than the measures of Model A.
\nModel B cars has a better gas mileage than that of Model A cars.<\/p>\n
\nData of Model A =\u00a0 18\u00a0 19 19\u00a0 20 20 20 21 21 22 27
\nOrder: 18 19 20 21 22 27
\nMedian of Model A =\u00a0 20 + 21 \u00f7 2 =41 \u00f7 2 = 20.5
\nMean of Model A = 18 + 19 + 19 + 20 + 20 + 20 + 21 + 21 +22 +27 \u00f7 10
\n=\u00a0 207 \u00f7 10
\n= 20.7
\nVariation of Model A = (18-20.7)^2 + (19-20.7)^2 + (19-20.7)^2 + (20-20.7)^2 + (20-20.7)^2 + (20-20.7)^2 + (21-20.7)^2 + (21-20.7)^2 +(22-20.7)^2 + (27-20.7)^2\u00a0 \u00f7 10
\n= (-2.7)^2 + (-1.7)^2 + (-1.7)^2 + (-0.7)^2 + (-0.7)^2 + (-0.7)^2 + (0.3)^2 + (0.3)^2 +(1.3)^2 + (6.3)^2\u00a0 \u00f7 10
\n= 7.29 + 2.89 + 2.89 + 0.49 + 0.49 + 0.49 + 0.09 +0.09 + 1.69 + 39.69 \u00f7 10
\n=\u00a0 56.37\u00a0 \u00f7 10
\n= 5.637
\nData of Model B = 20 21\u00a0 27\u00a0 29 29 29 30 30 31 32
\nOrder: 20 21 27 29 30 31 32
\nMedian of Model B = 29
\nMean of Model B = 20 + 21 + 27 + 29 + 29 + 29 + 30 + 30 + 31 + 32 \u00f7 10
\n=\u00a0 278 \u00f7 10
\n= 27.8
\nVariation of Model B = (20-27.8)^2 + (21-27.8)^2 + (27-27.8)^2 + (29-27.8)^2 + (29-27.8)^2 + (30-27.8)^2 + (30-27.8)^2 + (31-27.8)^2 +(31-27.8)^2 + (32-27.8)^2\u00a0 \u00f7 10
\n= (-7.8)^2 + (-6.8)^2 + (-0.8)^2 + (1.2)^2 + (1.2)^2 + (2.2)^2 + (2.2)^2 + (3.2)^2 +(3.2)^2 + (4.2)^2\u00a0 \u00f7 10
\n= 60.84 + 46.24 + 0.64\u00a0 + 1.44 + 1.44 + 4.84 + 4.84 + 10.24 + 10.24 + 17.64 \u00f7 10
\n= 158.4 \u00f7 10
\n= 15.84<\/p>\n
\nYou compare the average amounts of time people in their twenties and thirties spend driving each week. The double box-and-whisker plot represents the medians of 100 random samples of 8 people from each age group. Can you determine whether one age group drives more than the other? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answer-Key-Grade-7-Chapter-8-Statistics-cr-12.png\")
\nAnswer:
\nBecause the variables are same, you can determine the one age group drives more than the other visual overlap by expressing the differences in the medians as the multiples of IQR.<\/p>\n
\nData of Twenties = 0 5 10 15 25
\nMedian of Twenties = 10
\nIQR of Twenties = 15 – 5 = 10
\nData of Thirties = 0 10 15 20\u00a0 40
\nMedian of Thirties = 15
\nIQR of Thirties = 20 – 10 =10
\nDifferences in medians \u00f7 IQR = 15- 10 \u00f710 = 5 \u00f710 =0.5
\nBecause the quotient is less than 2, the difference is not significant.<\/p>\n<\/a>Statistics Practice Test<\/h3>\n
\nYou want to estimate the number of students in your school who prefer to bring a lunch from home rather than buy one at school. You survey five students who are standing in the lunch line. Determine whether the sample biased or unbiased. Explain.<\/p>\n
\nThe sample is biased.<\/p>\n
\nThe sample taken is biased because it does not gives you the accurate information of the number of students in your school prefer to bring a lunch from home rather than buy one at school. The sample size chosen is incorrect as taking the students in the lunch line..<\/p>\n
\nYou want to predict which candidate will likely be voted Seventh Grade Class President. There are 560 students in the seventh grade class. You randomly sample 3 different groups of 50 seventh-grade students. The results are shown.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-pt-2.png\")
\na. Use each sample to make an estimate for the number of students in seventh grade that vote for Candidate A.
\nb. Who do you expect to be voted Seventh Grade Class President? Explain.<\/p>\n
\na. Estimation for the number of students who vote for Candidate A = 64.
\nb. Candidate B is to be expected to be voted Seventh Grade Class President.<\/p>\n
\na. Estimation for the number of students who vote for Candidate A = 27 + 22 +15 = 64.
\nEstimation for the number of students who vote for Candidate B = 23 + 28 + 35 = 86.
\nb. Candidate B is to be expected to be voted Seventh Grade Class President.<\/p>\n
\nOf 60 randomly chosen students from a school surveyed, 16 chose the aquarium as their favorite field trip. There are 720 students in the school. Predict the number of students in the school who choose the aquarium as their favorite field trip.
\nAnswer:
\nNumber of students in the school who choose the aquarium as their favorite field trip = 192.<\/p>\n
\nNumber of students who chose the aquarium as their favorite field trip = 16
\nNumber of students\u00a0 chosen students from a school surveyed = 60
\nTotal number of students in the school = 720
\nNumber of students in the school who choose the aquarium as their favorite field trip = 720\/60 \u00d7 16 = 192<\/p>\n
\nThe double box-and-whisker plot shows the ages of the viewers of two television shows in a small town.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-pt-4.png\")
\na. Compare the data sets using measures of center and variation.
\nb. Which show is more likely to have a 44-year-old viewer?<\/p>\n
\na. The Show A\u00a0 center is likely to be greater than the measures of the center of Show B. Variation of show B is greater than Variation of show A.
\nb.\u00a0 Show A\u00a0 is more likely to have a 44-year-old viewer.<\/p>\n
\nData of Show A = 35\u00a0 \u00a040\u00a0 \u00a045\u00a0 \u00a0 50\u00a0 \u00a055
\nMedian of Show A = 45
\nMean of Show A = 35\u00a0 + 40 +\u00a0 45\u00a0 +\u00a0 50 +\u00a0 55 \u00f7 5 = 225 \u00f7 5 = 45
\nVariation of Show A = (35-45)^2 + (40-45)^2 + (45-45)^2 + (50-45)^2 +\u00a0 (55-45)^2 \u00f7 5
\n= (-10)^2 + (-5)^2 + (0)^2 + (5)^2 +\u00a0 (10)^2 \u00f7 5
\n= 100 + 25 + 0 + 25 +100 \u00f7 5
\n= 250 \u00f7 5
\n= 50.<\/p>\n
\nMedian of Show B = 35
\nMean of Show B = 20 + 25\u00a0 +\u00a0 35\u00a0 +\u00a0 40 +\u00a0 55 \u00f7 5 = 175 \u00f7 5 = 35.
\nVariation of show B = (20-35)^2 + (25-35)^2 + (35-35)^2 + (40-35)^2 +\u00a0 (55-35)^2 \u00f7 5
\n= (-15)^2 + (-10)^2 + (0)^2 + (5)^2 +\u00a0 (5)^2 \u00f7 5
\n= 225 + 100 + 0 + 25 + 25\u00a0 \u00a0\u00f7 5
\n= 375 \u00f7 5
\n= 75.<\/p>\n
\nThe double box-and-whisker plot shows the test scores for two French classes taught by the same teacher.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-pt-5.png\")
\na. Express the difference in the measures of center as a multiple of the measure of variation.
\nb. Are the scores for one class significantly greater than for the other? Explain.<\/p>\n
\na. The difference of the Variation of Class A is similar to the Variation of Class A and the measures of the center of Class A\u00a0 is comparably greater than the measures of the center of Class B.
\nb. Yes, the scores for one class significantly greater than for the other.<\/p>\n
\nData of Class A =\u00a0 76\u00a0 84\u00a0 88\u00a0 92\u00a0 100\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Data of Class B = 68\u00a0 76\u00a0 80\u00a0 \u00a084\u00a0 \u00a0100
\nMedian of Class A = 76 + 84 + 88 + 92 + 100 \u00f7 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Median of Class B = 68 + 76 + 80 + 84\u00a0 + 100 \u00f7 5
\n=\u00a0 440\u00a0 \u00f7 5 = 88\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 408\u00a0 \u00f7 5\u00a0 = 81.6
\nIQR of Class A = 92 – 84 = 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 IQR of Class B = 84 – 76 =\u00a0 8<\/p>\n
\nTwo airplanes each hold about 400 pieces of luggage. The double dot plot shows a random sample of 8 pieces of luggage from each plane. Compare the samples using measures of center and variation. Can you determine which plane has heavier luggage? Explain.
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-pt-6.png\")
\nAnswer:
\nThe measures of center of Plane A\u00a0 is greater than the measure of center of Plane B.
\nPlane A is likely to have heavier luggage than Plane B. The Variation of Plane A is greater than Variation of Plane B.<\/p>\n
\nData of Plane A =\u00a0 28\u00a0 36\u00a0 36\u00a0 38\u00a0 38\u00a0 40\u00a0 40\u00a0 40\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Data of plane B = 32\u00a0 32\u00a0 34\u00a0 34\u00a0 34\u00a0 36\u00a0 38 50
\nOrder = 28 36 38 40\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Order = 32\u00a0 34\u00a0 36\u00a0 38 50
\nMedian of Plane A = 36 + 38 \u00f7 2 = 74 \u00f7 2 = 37\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Median of Plane B\u00a0 = 36
\nVariation of Plane A =\u00a0 (28-37)^2 + (36-37)^2 + (36-37)^2 + (38-37)^2 +\u00a0 (38-37)^2 + (40-37)^2 + (40-37)^2 +\u00a0 (40-37)^2 \u00f7 8
\n=(-9)^2 + (-1)^2 +\u00a0 (-1)^2\u00a0 + (1)^2 + (1)^2 +\u00a0 (3)^2\u00a0 + (3)^2 + (3)^2 \u00a0\u00f7 8
\n= 81 + 1 +1 +1 +1 +9 +9 +9\u00a0 \u00f7 8
\n= 112 \u00f7 8
\n= 14.
\nVariation of Plane B =\u00a0 (32-36)^2 + (32-36)^2 +\u00a0 (34-36)^2\u00a0 + (34-36)^2 + (34-36)^2 +\u00a0 (36-36)^2\u00a0 + (38-36)^2 + (50-36)^2 \u00f7 8
\n= (-4)^2 + (-4)^2 +\u00a0 (-2)^2 + (-2)^2 + (-2)^2 +\u00a0 (0)^2\u00a0 + (2)^2 + (4)^2\u00a0 \u00f7 8
\n= 16 + 16 + 4+ 4+ 4+ 0 +4 + 16\u00a0 \u00f7 8
\n=\u00a0 64\u00a0 \u00f7 8
\n= 8.<\/p>\n<\/a>Statistics Cumulative Practice<\/h3>\n
![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-1.png\")
\nQuestion 1.
\nWhich of the ratios form a proportion?
\nA. 5 to 2 and 4 to 10
\nB. 2 : 3 and 7 : 8
\nC. 3 to 2 and 15 to 10
\nD. 12 : 8 and 8 : 42.
\nAnswer:
\nC. 3 to 2 and 15 to 10 is the \u00a0ratios form a proportion<\/p>\n
\nA) 5 to 2 and 4 to 10
\nB) 2 : 3 and 7 : 8
\nC) 3 to 2 and 15 to 10
\n=> 3 : 2 and 15 : 10
\n=> Divide by 5 on both sides
\n=> 3 : 2 and\u00a0 3 : 2
\nD) 12 : 8 and 8 : 42.
\n=> Divide by 2 on both sides
\n=> 6 : 4 and 4 : 21
\n=> Divide by 2 on both sides
\n=> 3 : 2 and\u00a0 2 :21<\/p>\n
\nA student scored 600 the first time. she took the mathematics portion of a college entrance exam. The next time she took the exam, she scored 660. Her second score represents what percent increase over her first score?
\nF. 9.1%
\nG. 10%
\nH. 39.6%
\nI. 60%
\nAnswer:
\nF = 9.1 %\u00a0 is the Percentage increase in her second score over her first score.<\/p>\n
\nNumber of marks she scored in first attempt in the exam = 600
\nNumber of marks she scored in second attempt in the exam = 660
\nDifference in the score = 660 – 600 = 60
\nPercentage increase in her second score over her first score = 60 \u00f7 660 \u00d7 100
\n= 0.9090 \u00d7 100 = 9.1%<\/p>\n
\nYou ask 100 randomly chosen students to name their favorite food. There are 1250 students in the school. Based on this sample, what is the number of students in the school whose favorite food is chicken?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-3.png\")
\nA. 100
\nB. 225
\nC. 450
\nD. 475
\nAnswer:
\nNumber of students in the school whose favorite food is chicken = 100.<\/p>\n
\nTotal number of students in the school = 1250
\nNumber of students randomly chosen = 100
\nNumber of students in the school whose favorite food is chicken = 8 \u00f7 100 \u00d7 1250
\n= 0.08 \u00d7 1250 = 100.<\/p>\n
\nWhich value of makes the equation p + 6 = 5 true?
\nF. – 1
\nG. 1
\nH. 11
\nI. 30
\nAnswer:
\nF. – 1
\np = -1 makes the equation p + 6 = 5 true.<\/p>\n
\nF) Substitute p = -1 in Equation:
\np + 6 = 5 = -1 + 6 = 5
\nG) Substitute p = 1 in Equation:
\np + 6 = 5 = 1 + 6 = 7
\nH ) Substitute p = 11 in Equation:
\np + 6 = 5 = 11 + 6 = 17
\nI) Substitute p = 30 in Equation:
\np + 6 = 5 = 30 + 6 = 36<\/p>\n
\nThe table shows the costs for four cans of tomato soup. Which can has the lowest cost per ounce?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-5.png\")
\nA. Can A
\nB. Can B
\nC. Can C
\nD. Can D
\nAnswer:
\nThe lowest cost per ounce has D. Can D.<\/p>\n
\nCost of Can A cost per ounce= 1.95 \u00f7 26 = 0.075
\nCost of Can B\u00a0 cost per ounce= 0.72 \u00f7 8 = 0.09
\nCost of Can C\u00a0 cost per ounce=0.86 \u00f7 10.75 = 0.08
\nCost of Can D\u00a0 cost per ounce= 2.32 \u00f7 23.2 = 0.1<\/p>\n
\nWhat value of y makes the equation – 3y = – 18 true?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-6.png\")
\nAnswer:
\ny = 6 makes the equation – 3y = – 18 true.<\/p>\n
\nEquation – 3y = – 18
\n=> y = -18 \u00f7 -3
\nDivide by 3 both sides.
\n=> y = 6<\/p>\n
\nThe double dot plot shows the values in two data sets. Which sentence best represents the difference in the measures of center as a multiple of the measure of variation?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-7.png\")
\nF. The difference of the means is about 3.3 times the MAD.
\nG. The difference of the means is about 3.8 times the MAD.
\nH. The difference of the means is 36 times the MAD.
\nI. The difference of the means is 48.7 times the MAD.<\/p>\n
\nG. The difference of the means is about 3.8 times the MAD.<\/p>\n
\nData of set A = 42\u00a0 43\u00a0 46\u00a0 48\u00a0 \u00a048\u00a0 51\u00a0 51\u00a0 51\u00a0 52\u00a0 55
\nMean of set A = 42 + 43 + 46 + 48 + 48 + 51 + 51 + 51 + 52 + 55 \u00f7 10
\n= 487 \u00f7 10
\n= 48.7
\nMAD of set A = |42\u221248.7\u2223 + |43\u221248.7\u2223 +|46\u221248.7\u2223 +|48\u221248.7\u2223 + |48-48.7\u2223 + |51-48.7\u2223 + |51-48.7\u2223 +|51-48.7\u2223 +|52-48.7\u2223 + |55-48.7\u2223 \u00f7 10
\n=\u00a0 |-6.7\u2223 + |-5.7\u2223 +|-2.7\u2223 +|-0.7\u2223 + |-0.7\u2223 + |2.3\u2223 + |2.3\u2223 +|2.3\u2223 +|3.3\u2223 + |6.3 \u2223 \u00f7 10
\n= 33 \u00f7 10
\n= 3.3
\nData of set B = 30\u00a0 31\u00a0 33\u00a0 34\u00a0 36\u00a0 36\u00a0 38\u00a0 40\u00a0 41\u00a0 41
\nMean of set B = 30 + 31 + 33 + 34 + 36 + 36 + 38 + 40 + 41 + 41 \u00f7 10
\n= 360\u00f7 10
\n= 36.0
\nMAD of set B = |30-36\u2223 + |31-36\u2223 +|33-36\u2223 +|34-36\u2223 + |36-36\u2223 + |36-36\u2223 + |38-36\u2223 +|40-36\u2223 +|41-36\u2223 + |41-36\u2223 \u00f7 10
\n= |-6\u2223 + |-5\u2223 +|-3\u2223 +|-2\u2223 + |0\u2223 + |0\u2223 + |2\u2223 +|4\u2223 +|5\u2223 + |5\u2223 \u00f7 10
\n=\u00a0 32\u00a0 \u00f7 10
\n= 3.2
\nDifference in Means \u00f7 MAD of set A= 48.7 – 36 \u00f7 3.3 = 12.7 \u00f7 3.3= 3.85
\nDifference in Means \u00f7 MAD of set B= 48.7 – 36 \u00f7 3.2 = 12.7 \u00f7 3.2 =3.97<\/p>\n
\nWhat is the missing value in the ratio table?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-8.png\")
\nAnswer:
\nThe missing value in the ratio table is 20.<\/p>\n
\na) 2\/3\u00a0 = 5
\n=> 2 = 5 \u00d7 3
\n=> 2 = 15<\/p>\n
\n8 = 3 \u00d7 =\u00a0 60
\n=> 60 \u00f7 3
\n=> 20<\/p>\n
\nYou are selling tomatoes. What is the minimum number of 4pounds of tomatoes you need to sell to earn at least $44?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-9.png\")
\nA. 11
\nB. 12
\nC. 40
\nD. 176
\nAnswer:
\nA. 11 is the minimum number of\u00a0 tomatoes needed to sell to earn at least $44.<\/p>\n
\nCost of the tomatoes = $44
\ncost of the tomatoes per pound = $4
\nThe minimum number of\u00a0 tomatoes needed to sell to earn at least $44 = 44 \u00f7 4 = 11.<\/p>\n
\nYou and a group of friends want to know how many students in your school prefer science. There are 900 students in your school. Each person randomly surveys 20 students. The table shows the results. Which subject do students at your school prefer?
\n![\"Big](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/01\/Big-Ideas-Math-Answers-7th-Grade-Chapter-8-Statistics-cp-10.png\")
\nPart A Use each sample to make an estimate for the number of students in your school who prefer science.
\nPart B Describe the variation of the estimates.
\nPart C Use the samples to make one estimate for the number of students who prefer science in your school.<\/p>\n
\nNumber of students who prefer science = 28
\nMean = 5.6
\nVariation of the estimate\u00a0 = 4.24<\/p>\n
\nNumber of students who prefer science = 6 + 7 + 8 +5 + 2 = 28
\nMean of number of students who prefer science = 6 + 7 + 8 +5 + 2 \u00f7 5 = 28 \u00f7 5 = 5.6
\nVariation of the estimate\u00a0 = (6-5.6)^2 + (7-5.6)^2 + (8-5.6)^2 + (5-5.6)^2 +\u00a0 (2-5.6)^2\u00a0 \u00f7 5
\n= (-0.4)^2 + (1.4)^2 + (2.4)^2 + (0.6)^2 + (-3.6)^2\u00a0 \u00f7 5
\n=\u00a0 0.16 +1.96 + 5.76 + 0.36 +\u00a0 12.96\u00a0 \u00f7 5
\n=\u00a0 21.2 \u00f7 5
\n= 4.24<\/p>\n