Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean of 180 pounds and sample standard deviation of 30 pounds. Assume the weight distribution is approximately normal.

a.) Find a one sided 95% confidence interval with an upper bound for the mean weight of all airline passengers during the summer. Show you work.

b.) Find a 95% prediction interval for the weight of another random selected airline passenger during the summer. Show you work.

Step-by-step explanation:

Given Parameters

Mean, = 180

total samples, n = 20

Standard dev, = 30

= 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, , is given by

a).

Confidence Interval, , is given by the formula

b).

Critical Value, , is given by

Confidence Interval, , is given by

    = 180 +/- 14.0403

    = 165.9597 < < 194.0403

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^