Question

1. A bank teller serves customers one at a time so that each customer has an associated ‘service time,’ or time during which they interact with the teller. We know that the service times are not Normally distributed as some customers have very long service times. However, we do know that service times have a mean of 4 minutes and standard deviation of 1.2 minutes. What is the probability that a sample of 40 bank customers has an average service time of between 3.5 and 4.5 minutes?

Answer 1

Solution :

Given that,

mean = = 4

standard deviation = = 1.2

n = 40

= 4

= / n = 1.2 / 40 = 0.1897

P( 3.5 < < 4.5 )  

= P[(3.5 - 4) /0.1897  < ( - ) / < (4.5 - 4) /0.1897)]

= P( -2.64 < Z < 2.64 )

= P(Z < 2.64 ) - P(Z < -2.64 )

Using z table,  

= 0.9959 - 0.0041

= 0.9918

P( 3.5 < < 4.5 ) = 0.9918

Probability = 0.9918