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 between 29 minutes and 31 minutes.

Answer 1

Solution :

Given that ,

mean =   = 30

standard deviation = = 10

n = 100

= 30

=  / n= 10 / 100= 1

P(29<     <31 ) = P[(29 - 30) /1 < ( - ) /   < (31 - 30) /1 )]

= P( -1< Z < 1)

= P(Z < 1) - P(Z < -1)

Using z table

=0.8413 - 0.1587

probability= 0.6826