Question

According to a survey in a​ country, 18​% of adults do not own a credit card. Suppose a simple random sample of 900 adults is obtained. ​(b) What is the probability that in a random sample of 900 ​adults, more than 22​% do not own a credit​ card? The probability is what.? ​(Round to four decimal places as​ needed.) Interpret this probability. If 100 different random samples of 900 adults were​ obtained, one would expect nothing to result in more than 22​% not owning a credit card. ​(Round to the nearest integer as​ needed.).  

(c) What is the probability that in a random sample of 900 ​adults, between 17​% and 22​% do not own a credit​ card?

Interpret this probability. If 100 different random samples of 900 adults were​ obtained, one would expect ? to result in between 17​% and 22​% not owning a credit card.

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

Would it be unusual for a random sample of 900 adults to result in 153 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

The result

is notis not

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

greatergreater

than​ 5%.

B.The result

is notis not

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

lessless

than​ 5%.

C.The result

isis

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

lessless

than​ 5%.

D.The result

isis

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

greatergreater

than​ 5%.

Answer 1

P = 0.18

n = 900

= 0.18

= sqrt(p(1 - p)/n)

= sqrt(0.18 * (1 - 0.18)/900)

= 0.0128

b) P( > 0.22)

= P(( - )/ > (0.22 - )/)

= P(Z > (0.22 - 0.18)/0.0128)

= P(Z > 3.13)

= 1 - P(Z < 3.13)

= 1 - 0.9991 = 0.0009

0.0009 * 100 = 0.09

If 100 random samples of 900 adults were obtained, one would except none of them result in more than 22% not owning a credit card.

C) P(0.17 < < 0.22)

= P((0.17 - )/ < ( - )/ < (0.22 - )/ )

= P((0.17 - 0.18)/0.0128 < Z < (0.22 - 0.18)/0.0128)

= P(-0.78 < Z < 3.13)

= P(Z < 3.13) - P(Z < -0.78)

= 0.9991 - 0.2177

= 0.7814

100 * 0.7814 = 78.14 = 78

If 100 random samples of 900 adults were obtained, one would except 78 them result in between 17% and 22% do not own a credit card.

= 153/900 = 0.17

P( < 0.17)

= P(( - )/ < (0.17 - )/)

= P(Z < (0.17 - 0.18)/0.0128)

= P(Z < -0.78)

= 0.2177

The result is not unusual because the probability that is less than or equal to the sample proportion 0.17 is 0.2177 which is greater than 5%.