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 85th percentile for the sum of the 100 wait times.

Answer #1

Given that,

mean = = 30

standard deviation = = 10

n = 100

= 30

= / n = 10 /100 = 1

Using standard normal table,

P(Z < z) = 85%

= P(Z < z) = 0.85

= P(Z <1.04 ) = 0.85

z = 1.04 Using standard normal table,

Using z-score formula

= z * +

= 1.04 *1+30

= 31.04

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