can you show me the work of these question A random sample of 49 lunch customers

Question

can you show me the work of these question

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 33 minutes. From past experience, it is known that the population standard deviation equals 10 minutes.

a. Compute the standard error of the mean.

b. What can be said about the sampling distribution for the average amount of time customers spent in the restaurant? Be sure to explain your answer.

c. With a .95 probability, what statement can be made about the size of the margin of error?

d. Construct a 95% confidence interval for the true average amount of time customers spent in the restaurant.

e. With a .95 probability, how large of a sample would have to be taken to provide a margin of error of 2.5 minutes or less?

Explanation / Answer

a) std error =std deviaiton/(n)1/2 =1.4286

b)as sample size is greater then 30 we can assume that to be normal from central limit theorum

c)for 95% ; z=1.96

hence margin of error =z*std error =2.7999

d)confidence interval =mean +/- margin of error =30.20 ; 35.80

e) margin of error E=10

std deviation =10

for 95% ; z=1.96

hence sample size =(z*std deviation/E)2 =~62

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