Posted by Blerp on Friday, November 16, 2012 at 11:36am.
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

Statistics  MathGuru, Friday, November 16, 2012 at 7:05pm
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.