Question

A study conducted by an airline showed that a random sample of nine of its passengers disembarking at the Cleveland airport, took an average of 24.1 minutes to claim their luggage. From a previous survey it was willing to assume that time to claim luggage is normally distributed with a variance of 18 (min)2. A 95% confidence interval for the mean time to claim one's luggage has endpoints.


24.1 ± 8.32

24.1 ± 3.92

24.1 ± 2.77

24.1 ± 3.26

24.1 ± 9.78

Answer 1

CI95 = mean ± 1.96(sd/√n)

Substitute the values given into the formula to choose your answer.

Note: standard deviation is the square root of the variance.