Question

Suppose the mean wait time for a bus is 30 minutes and the standard deviation is 10 minutes. Take a sample of size n = 100.

Find the 95th percentile for the sample mean wait time.

Answer 1

solution:

Given that,

mean = = 30

standard deviation = = 10

n = 100

= 30

= / n = 10/100=1

Using standard normal table,

P(Z < z) = 95%

= P(Z < z) = 0.95

z = 1.65 Using standard normal table,

Using z-score formula  

= z * +   

= 1.65*1+30

= 31.65