Question

According to a survey in a​ country, 22​% of adults do not own a credit card. Suppose a simple random sample of 300 adults is obtained

a) What is the probability that in a random sample of 300 adults, between 19​% and 24​% do not own a credit​ card? The probability is ____. (Round to four decimal places as​ needed.)

b) If 100 different random samples of 300 adults were​ obtained, one would expect nothing to result in between 19% and 24% not owning a credit card. The probability is ____. ​(Round to the nearest integer as​ needed.)

c) Would it be unusual for a random sample of 300 adults to result in 57 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 not is not unusual because the probability that p with caret is less than or equal to the sample proportion is ____​, which is less than​ 5%.

B.The result is unusual because the probability that p with caret 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 p with caret is less than or equal to the sample proportion is _____​, which is greater than​ 5%.

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

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.220
sample size       =n=	300
std error of proportion=σp=√(p*(1-p)/n)=	0.0239

a)

probability =P(0.19<X<0.24)=P((0.19-0.22)/0.024)<Z<(0.24-0.22)/0.024)=P(-1.25<Z<0.84)=0.7995-0.1056=0.6939

b)

one would expect 69 to result in between 19% and 24% not owning a credit card.

c)

probability =P(X<57/300)=(Z<(0.19-0.22)/0.024)=P(Z<-1.2544)=0.1056

C.The result is not unusual because the probability that p with caret is less than or equal to the sample proportion is 0.1056  which is greater than​ 5%.