A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

a. 0.7510 and 0.8290
b. 0.6592 and 0.7044
c. 0.6659 and 0.6941
d. 0.6795 and 0.6805

b. 0.6592 and 0.7044

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. This means that

90% confidence interval

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

So the correct answer is:

b. 0.6592 and 0.7044

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

answer: x=3 and y=4

step-by-step explanation: