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%.
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.220 |
sample size =n= | 300 |
std error of proportion=σ_{p}=√(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%.
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