Explanation Given: The number of young adults selected n = 1000 The probability of success on a single trial p = 0.15 Formula used: The expression for the mean and the standard deviation for binomial distribution is: μ = n p σ = n p ( 1 − p ) The z -score for any value x : z = x − μ σ Here, μ is the mean, σ is the standard deviation, n is the number of trials and p is the probability of success on a single trial Calculation: Use the formula of mean, standard deviation and z -score. Substitute 1000 for n and 0.15 for p as: μ = n p = 1000 ( 0.15 ) = 150 σ = n p ( 1 − p ) = 1000 ( 0.15 ) ( 1 − 0.15 ) ≃ 11.3 Therefore, μ = 150 and σ = 11

Pearson eText for Mathematics with Applications In the Management, Natural, and Social Sciences -- Instant Access (Pearson+)

12th Edition

ISBN: 9780137504381

Author: Margaret Lial, Thomas Hungerford

Publisher: PEARSON+

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Estimation

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Chapter 10, Problem 53RE

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To calculate: The probability between 120 and 165 (inclusive) young adults who live in their parents’ home, while a random sample of 1000 adults are considered.

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Chapter 10, Problem 52RE

Chapter 10, Problem 54RE

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Chapter 10 Solutions

Pearson eText for Mathematics with Applications In the Management, Natural, and Social Sciences -- Instant Access (Pearson+)

Ch. 10.1 - Checkpoint 1 A restaurant trade group commissioned...Ch. 10.1 - Checkpoint 2 Make a histogram and a frequency...Ch. 10.1 - Checkpoint 3 Make a stem-and-leaf plot for the...Ch. 10.1 - Checkpoint 4 List the original data for the...Ch. 10.1 - Checkpoint 5 Characterize the shape of the...Ch. 10.1 - SP 500 Stocks The data for Exercises 1 -4 consist...Ch. 10.1 - SP 500 Stocks The data for Exercises 1 -4 consist...Ch. 10.1 - SP 500 Stocks The data for Exercises 1 -4 consist...Ch. 10.1 - SP 500 Stocks The data for Exercises 1 -4 consist...Ch. 10.1 - The data for Exercises 5-10 consist of random...

Ch. 10.1 - The data for Exercises 5-10 consist of random...Ch. 10.1 - The data for Exercises 5-10 consist of random...Ch. 10.1 - The data for Exercises 5-10 consist of random...Ch. 10.1 - The data for Exercises 5-10 consist of random...Ch. 10.1 - The data for Exercises 5-10 consist of random...Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - Prob. 14E Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - For Exercises 11-20, construct a stem-and-leaf...Ch. 10.1 - Describe the shape of each of the given...Ch. 10.1 - Describe the shape of each of the given...Ch. 10.1 - Describe the shape of each of the given...Ch. 10.1 - Describe the shape of each of the given...Ch. 10.1 - Student Loan Defaults The following histogram...Ch. 10.1 - Stocks The following histogram shows the stock...Ch. 10.1 - Cat Ownership The stem-and-leaf plot below...Ch. 10.1 - Personal Bankruptcies The stem-and-leaf plot below...Ch. 10.1 - 29. 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When is the median the most appropriate...Ch. 10.2 - 21. When would the mode be an appropriate measure...Ch. 10.2 - Pet Ownership For Exercises 22 and 23, the...Ch. 10.2 - Pet Ownership For Exercises 22 and 23, the...Ch. 10.2 - 24. To predict the outcome of the next...Ch. 10.2 - Work each problem. (See Example 6.) MLB Payrolls...Ch. 10.2 - Work each problem. (See Example 6.) NFL Team...Ch. 10.2 - Work each problem. (See Example 6.) 27. Business...Ch. 10.2 - Work each problem. (See Example 6.) Sirius XM...Ch. 10.2 - Work each problem. (See Example 6.) 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The probability between 120 and 165 (inclusive) young adults who live in their parents’ home, while a random sample of 1000 adults are considered.

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To calculate: The probability between 120 and 165 (inclusive) young adults who live in their parents’ home, while a random sample of 1000 adults are considered.

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Chapter 10 Solutions