Question

A population of 1000 students spends an average of 10.50 a day on dinner. The standard deviation of the expenditure is 3.00. A simple random sample of 64 students is taken.

a. What is the probability that these 64 students will spend a combined total between 703.59 and 728.45? Show all work.

b. What is the probability that these 64 students will spend a combined total of more than 715.21? Show all work

Answer 1

a) P(10.994 < < 11.382)

= P((10.994 - )/() < ( - )/() < (11.382 - )/())

= P((10.994 - 10.5)/(3/) < Z < (11.382 - 10.5)/(3/))

= P(1.32 < Z < 2.35)

= P(Z < 2.35) - P(Z < 1.32)

= 0.9906 - 0.9066

= 0.0840

b) P( > 11.175)

= P(( - )/() > (11.175 - )/())

= P(Z > 1.8)

= 1 - P(Z < 1.8)

= 1 - 0.9641

= 0.0359