Question

According to a survey in a​ country, 35​% of adults do not own a credit card. Suppose a simple random sample of 500 adults is obtained. Complete parts​ (a) through​ (e) below.

​(a) Determine the mean of the sampling distribution of

mu Subscript ModifyingAbove p with caret Baseline equals

μp=___

​(Round to two decimal places as​ needed.)

​(b) Determine the standard deviation of the sampling distribution of

sigma Subscript ModifyingAbove p with caret equals

σp=___

​(Round to three decimal places as​ needed.)

(c) What is the probability that in a random sample of 500 ​adults, more than 38​% 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 500 adults were​ obtained, one would expect ___ to result in more than 38​% not owning a credit card.

​(Round to the nearest integer as​ needed.)

(d) What is the probability that in a random sample of 500 adults, between 33​% and 38​% 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 500 adults were​ obtained, one would expect __ to result in between 33​% and 38​% not owning a credit card.

​(Round to the nearest integer as​ needed.)

(e) Would it be unusual for a random sample of 500 adults to result in 165 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.)

A.The result is unusual because the probability that ModifyingAbove p with caretp is less than or equal to the sample proportion is ___ ​, which is less than​ 5%.

B.The result is notunusual because the probability that ModifyingAbove p with caretp is less than or equal to the sample proportion is ___, which is greater than​ 5%.

C.The result is not unusual because the probability that ModifyingAbove p with caretp is less than or equal to the sample proportion is ___​, which is less than​ 5%.

D.The result is unusual because the probability that ModifyingAbove p with caretp is less than or equal to the sample proportion is ___​, which is greater than​ 5%.

Answer 1

Given

p : Proportion of adults who own a credit card = 0.35

Number of adults in the simple random sample : n= 500

a) Determine the mean of the sampling distribution of : = p = 0.35

= 0.35

​(b) Determine the standard deviation of the sampling distribution of :

(c) What is the probability that in a random sample of 500 ​adults, more than 38​% do not own a credit​ card?

P(>0.38)=1-P(0.38)

Z-score for 0.38 = (0.38-0.35)/0.02 = 0.03/0.02 = 1.5

From standard normla tables, P(Z1.5) = 0.9332

P(0.38) = P(Z1.5) = 0.9332

P(>0.38)=1-P(0.38)=1-0.9332 = 0.0668

The probability is 0.0668.

100 x 0.0668=6.687

If 100 different random samples of 500 adults were​ obtained, one would expect __7_ to result in more than 38​% not owning a credit card.

​(Round to the nearest integer as​ needed.)

(d) What is the probability that in a random sample of 500 adults, between 33​% and 38​% do not own a credit​ card?

P(0.33 0.38)=P(0.38)-P(0.33)

Z-score for 0.38 = (0.38-0.35)/0.02 = 0.03/0.02 = 1.5

From standard normla tables, P(Z1.5) = 0.9332

P(0.38) = P(Z1.5) = 0.9332

Z-score for 0.33 = (0.33-0.35)/0.02 = 0.02/0.02 = -1

From standard normla tables, P(Z-1) = 0.1587

P(0.33) = P(Z-1) =0.1587

P(0.38)-P(0.33) = 0.9332-0.1587=0.7745

The probability is 0.7745.

100 x 0.7745=77.4577

if 100 different random samples of 500 adults were​ obtained, one would expect _77_ to result in between 33​% and 38​% not owning a credit card.

e) Would it be unusual for a random sample of 500 adults to result in 165 or fewer who do not own a credit​ card?

Sample proportion : = 165/500 = 0.33

Probability that less than or equal to 0.33 = P(0.33)

Z-score for 0.33 = (0.33-0.35)/0.02 = 0.02/0.02 = -1

From standard normla tables, P(Z-1) = 0.1587

P(0.33) = P(Z-1) =0.1587 which is 15.87% > 5%

B.The result is notunusual because the probability that ModifyingAbove p with caretp is less than or equal to the sample proportion is 0.1587 (15.87%), which is greater than​ 5%.