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III. The Big Smiles Portrait Studio is conducting a survey among their clients. One of the questions being asked is if they would recommend the studio to a friend. The studio has given the survey to a simple random sample of 65 clients during the past two weeks. a. If the true proportion of clients who are very satisfied with the Big Smiles Portrait Studio and would therefore recommend the studio to a friend is 82%, how many clients in the sample would we expect to answer Yes on the survey question? Let X represent the number of clients in the sample who answer Yes on the survey question. We know X~Binomial 65,0.82) Thus x = 65x0.82 = 53.3 b. What is the probability that more than 60 clients in the sample would answer Yes on the survey question? You can use normal approximation. a X~ N(np, √np (1- p)) np = 53.3 √np (1-p) = 3.1 P(X>60) ≈ PZ > =1-0.9846= 0.0154 60-53.3 3.1 =P(Z >2.16) c. Big Smiles has studios all across the country. Each studio surveys 65 of their clients and records the number of surveyed clients that respond Yes. If a studio scores in the top 6% of the distribution for the number of clients who answer Yes, then the studio will receive an award. Approximately how many clients had to answer Yes on the survey question before the studio will win an award? 3.1 X > 58 First we need to find the 94th percentile of a standard normal distribution, which is about 1.56. Then for a studio to win an award, it needs X - 53.3 >1.56

III. The Big Smiles Portrait Studio is conducting a survey among their clients. One of the questions being asked is if they would recommend the studio to a friend. The studio has given the survey to a simple random sample of 65 clients during the past two weeks. a. If the true proportion of clients who are very satisfied with the Big Smiles Portrait Studio and would therefore recommend the studio to a friend is 82%, how many clients in the sample would we expect to answer Yes on the survey question? Let X represent the number of clients in the sample who answer Yes on the survey question. We know X~Binomial 65,0.82) Thus x = 65x0.82 = 53.3 b. What is the probability that more than 60 clients in the sample would answer Yes on the survey question? You can use normal approximation. a X~ N(np, √np (1- p)) np = 53.3 √np (1-p) = 3.1 P(X>60) ≈ PZ > =1-0.9846= 0.0154 60-53.3 3.1 =P(Z >2.16) c. Big Smiles has studios all across the country. Each studio surveys 65 of their clients and records the number of surveyed clients that respond Yes. If a studio scores in the top 6% of the distribution for the number of clients who answer Yes, then the studio will receive an award. Approximately how many clients had to answer Yes on the survey question before the studio will win an award? 3.1 X > 58 First we need to find the 94th percentile of a standard normal distribution, which is about 1.56. Then for a studio to win an award, it needs X - 53.3 >1.56

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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III. The Big Smiles Portrait Studio is conducting a survey among their clients. One of the questions being asked is if they would recommend the studio to a friend. The studio has given the survey to a simple random sample of 65 clients during the past two weeks. a. If the true proportion of clients who are very satisfied with the Big Smiles Portrait Studio and would therefore recommend the studio to a friend is 82%, how many clients in the sample would we expect to answer Yes on the survey question? Let X represent the number of clients in the sample who answer Yes on the survey question. We know X~ Binomial 65,0.82) Thus x = 65x0.82 = 53.3 b. What is the probability that more than 60 clients in the sample would answer Yes on the survey question? You can use normal approximation. a X~ N(np, √np (1- p)) np = 53.3 |np(1-p) = 3.1 P(X>60) ≈ PZ> =1-0.9846= 0.0154 60-53.3 3.1 |=P(Z >2.16) c. Big Smiles has studios all across the country. Each studio surveys 65 of their clients and records the number of surveyed clients that respond Yes. If a studio scores in the top 6% of the distribution for the number of clients who answer Yes, then the studio will receive an award. Approximately how many clients had to answer Yes on the survey question before the studio will win an award? 3.1 X > 58 First we need to find the 94th percentile of a standard normal distribution, which is about 1.56. Then for a studio to win an award, it needs X - 53.3 ≥1.56
Transcribed Image Text:III. The Big Smiles Portrait Studio is conducting a survey among their clients. One of the questions being asked is if they would recommend the studio to a friend. The studio has given the survey to a simple random sample of 65 clients during the past two weeks. a. If the true proportion of clients who are very satisfied with the Big Smiles Portrait Studio and would therefore recommend the studio to a friend is 82%, how many clients in the sample would we expect to answer Yes on the survey question? Let X represent the number of clients in the sample who answer Yes on the survey question. We know X~ Binomial 65,0.82) Thus x = 65x0.82 = 53.3 b. What is the probability that more than 60 clients in the sample would answer Yes on the survey question? You can use normal approximation. a X~ N(np, √np (1- p)) np = 53.3 |np(1-p) = 3.1 P(X>60) ≈ PZ> =1-0.9846= 0.0154 60-53.3 3.1 |=P(Z >2.16) c. Big Smiles has studios all across the country. Each studio surveys 65 of their clients and records the number of surveyed clients that respond Yes. If a studio scores in the top 6% of the distribution for the number of clients who answer Yes, then the studio will receive an award. Approximately how many clients had to answer Yes on the survey question before the studio will win an award? 3.1 X > 58 First we need to find the 94th percentile of a standard normal distribution, which is about 1.56. Then for a studio to win an award, it needs X - 53.3 ≥1.56
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