Question

According to a survey in a​ country,

1717​%

of adults do not own a credit card. Suppose a simple random sample of

500500

adults is obtained. Complete parts​ (a) through​ (d) below.​(a) Describe the sampling distribution of

ModifyingAbove p with caretp​,

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

ModifyingAbove p with caretp

below.

A.Approximately normal because

n less than or equals 0.05 Upper Nn≤0.05N

and np left parenthesis 1 minus p right parenthesis less than 10np(1−p)<10

B.Not normal because

n less than or equals 0.05 Upper Nn≤0.05N

and np left parenthesis 1 minus p right parenthesis less than 10np(1−p)<10

C.Not normal because

n less than or equals 0.05 Upper Nn≤0.05N

and np left parenthesis 1 minus p right parenthesis greater than or equals 10np(1−p)≥10

D.Approximately normal because

n less than or equals 0.05 Upper Nn≤0.05N

and np left parenthesis 1 minus p right parenthesis greater than or equals 10np(1−p)≥10Your answer is correct.Determine the mean of the sampling distribution of

ModifyingAbove p with caretp.

mu Subscript ModifyingAbove p with caret Baseline equalsμp=. 17 .17

​(Round to two decimal places as​ needed.)Determine the standard deviation of the sampling distribution of

ModifyingAbove p with caretp.

sigma Subscript ModifyingAbove p with caretσpequals=. 017 .017

​(Round to three decimal places as​ needed.)​(b) What is the probability that in a random sample of

500

​adults, more than

18​%

do not own a credit​ card?

The probability is

Answer 1

Solution:

Given:

p = 0.17

n = 500

Part a-1) Describe the sampling distribution of

n*p*(1-p) =500*0.17*(1-0.17) = 70.55 > 10

Thus correct answer is:

D.Approximately normal because

n ≤ 0.05N and np(1−p)≥10

Part a-2) Determine the mean of the sampling distribution of .

Part a-3) Determine the standard deviation of the sampling distribution of .

Part b) What is the probability that in a random sample of 500 ​adults, more than 18​% do not own a credit​ card?

That is find:

Thus find z score:

thus we get:

Look in z table for z = 0.5 and 0.09 and find corresponding area.

P( Z< 0.59) = 0.7224

thus