Question

According to a survey in a​ country, 22% of adults do not have any credit cards. Suppose a simple random sample of

500 adults is obtained.

​(a) Describe the sampling distribution of

ModifyingAbove p with caretp​,

the sample proportion of adults who do not have a credit card. Choose the phrase that best describes the shape of the sampling distribution of

ModifyingAbove p with caretp

below.

A.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 10.np(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 10.np(1−p)<10.

C.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 10.np(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 less than 10.np(1−p)<10.Determine the mean of the sampling distribution of

ModifyingAbove p with caretp.

mu Subscript ModifyingAbove p with caret Baseline equalsμp=nothing

​(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=nothing

​(Round to three decimal places as​ needed.)​(b) In a random sample of

500500

​adults, what is the probability that less than

2020​%

have no credit​ cards?The probability is

nothing.

​(Round to four decimal places as​ needed.)​(c) Would it be unusual if a random sample of

500500

adults results in

130130

or more having no credit​ cards?

Answer 1

Solution:-

a) (C) Approximately normal because n less than or equals 0.05N (n ≤ 0.05N) and np(1 - p) greater than or equals 10. (np(1−p)≥10).

The mean of the sampling distribution is 110.

$\mu =n\ast p$

u = 500*0.22

u = 110

$\sigma =\sqrt{n\ast p(1-p)}$

$\sigma = 9.2628$

The standard deviation of the sampling distribution of 9.2628.

(b) The probability that less than 20​% have no credit​ cards is 0.1402.

$\hat{p}=0.20$

By applying normal distribution:-

$z= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

z = -1.0796

P(z < -1.0796) = 0.1402

(c) it would be unusual if a random sample of 500 adults results in 130 or more having no credit​ cards because probability (0.0154) is less than 0.05.

$\hat{p}=\frac{130}{500} = 0.26$

By applying normal distribution:-

$z= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

z = 2.16

P(z > 2.16) = 0.0154