A population of 1000 students spends an average of 1050 a da

A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken.
a. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean? (Hint: the expected value is the mean)
b. What is the probability that these 64 students will spend a combined total of more than $715.21?
c. What is the probability that these 64 students will spend a combined total between $703.59 and $728.45?

Solution

a.Using the central limit theorem, the sampling distribution of sample mean follows normal distribution.

Thus mean is mu=$10.50

Standard deviation(sigma) =population standard deviation/sqrt n

=3/sqrt 64

=3/8=0.375