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 probability that the sample mean wait time is more than 31 minutes.

Answer 1

Solution :

Given that,

mean = = 30

standard deviation = =10

n=100

= =30

= / n = 10 / 100 = 10/10=1

P( > 31) = 1 - P( < 31)

= 1 - P[( - ) / < (31 - 30) /1 ]

= 1 - P(z < 1)

Using z table

= 1 - 0.8413

probability= 0.1587