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Juults do not own a credit card. Suppose a simple random sample of 200 adults is obtained Complete parts (a) through (d) below. (a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p below OA. Not normal because n≤0.05N and np(1-p) < 10 OB. Approximately normal because n ≤0.05N and np(1-p) <10 OC. Not normal because n≤0.05N and np(1-p)2 10 O D. Approximately normal because n ≤0.05N and np(1-p) > 10 Determine the mean of the sampling distribution of p HO (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p G=(Round to three decimal places as needed.) Р (b) What is the probability that in a random sample of 200 adults, more than 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.)

Juults do not own a credit card. Suppose a simple random sample of 200 adults is obtained Complete parts (a) through (d) below. (a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p below OA. Not normal because n≤0.05N and np(1-p) < 10 OB. Approximately normal because n ≤0.05N and np(1-p) <10 OC. Not normal because n≤0.05N and np(1-p)2 10 O D. Approximately normal because n ≤0.05N and np(1-p) > 10 Determine the mean of the sampling distribution of p HO (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p G=(Round to three decimal places as needed.) Р (b) What is the probability that in a random sample of 200 adults, more than 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.)

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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According to a survey in a country, 39% of adults do not own a credit card Suppose a simple random sample of 200 adults is obtained Complete parts (a) through (d) below. (a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p below OA. Not normal because n≤0.05N and np(1-p) < 10 OB. Approximately normal because n ≤0.05N and np(1-p) <10 OC. Not normal because n≤0.05N and np(1-p) ≥ 10 O D. Approximately normal because n ≤0.05N and np(1-p) > 10 Determine the mean of the sampling distribution of p (Round to two decimal places as needed.). Pp Determine the standard deviation of the sampling distribution of p GA= (Round to three decimal places as needed.) P (b) What is the probability that in a random sample of 200 adults, more than 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.)
Transcribed Image Text:According to a survey in a country, 39% of adults do not own a credit card Suppose a simple random sample of 200 adults is obtained Complete parts (a) through (d) below. (a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p below OA. Not normal because n≤0.05N and np(1-p) < 10 OB. Approximately normal because n ≤0.05N and np(1-p) <10 OC. Not normal because n≤0.05N and np(1-p) ≥ 10 O D. Approximately normal because n ≤0.05N and np(1-p) > 10 Determine the mean of the sampling distribution of p (Round to two decimal places as needed.). Pp Determine the standard deviation of the sampling distribution of p GA= (Round to three decimal places as needed.) P (b) What is the probability that in a random sample of 200 adults, more than 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.)
According to a survey in a country, 39% of adults do not own a credit card. Suppose a simple random sample of 200 adults is obtained. Complete parts (a) through (d) below. The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.) (c) What is the probability that in a random sample of 200 adults, between 32% and 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in between 32% and 40% not owning a credit card. (Round to the nearest integer as needed.). (d) Would it be unusual for a random sample of 200 adults to result in 64 or fewer who do not own a credit card? Why? Select the correct choice below and fill in the answer box to complete your choice. (Round to four decimal places as needed.) OA. The result is unusual because the probability that p is less than or equal to the sample proportion is, which is less than 5%. which is greater than 5%. OB. The result is not unusual because the probability that p is less than or equal to the sample proportion is OC. The result is not unusual because the probability that p is less than or equal to the sample proportion is, which is less than 5%. OD. The result is unusual because the probability that p is less than or equal to the sample proportion is, which is greater than 5%.
Transcribed Image Text:According to a survey in a country, 39% of adults do not own a credit card. Suppose a simple random sample of 200 adults is obtained. Complete parts (a) through (d) below. The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in more than 40% not owning a credit card. (Round to the nearest integer as needed.) (c) What is the probability that in a random sample of 200 adults, between 32% and 40% do not own a credit card? The probability is (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 200 adults were obtained, one would expect to result in between 32% and 40% not owning a credit card. (Round to the nearest integer as needed.). (d) Would it be unusual for a random sample of 200 adults to result in 64 or fewer who do not own a credit card? Why? Select the correct choice below and fill in the answer box to complete your choice. (Round to four decimal places as needed.) OA. The result is unusual because the probability that p is less than or equal to the sample proportion is, which is less than 5%. which is greater than 5%. OB. The result is not unusual because the probability that p is less than or equal to the sample proportion is OC. The result is not unusual because the probability that p is less than or equal to the sample proportion is, which is less than 5%. OD. The result is unusual because the probability that p is less than or equal to the sample proportion is, which is greater than 5%.
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