Question #2460

Q1. A small bank has two cashiers dealing with customers wanting to withdraw or deposit cash. For each cashier, the time to deal with a customer is a random variable having a normal distribution with mean 150s and standard deviation 45s.
A. Find the probability that the time taken for a randomly chosen customer to be dealt with by a cashier is more than 180s.
B. One of the cashier deals with two customers, one straight after the other.

Expert's answer

Let X be the time to deal with a customer.

Then Z = (X-150)/45is a random variable which has standard normal.

A. P(X>180) =

= P(& (X-150)/45>(180-150)/45 ) =

= P(& Z>30/45& ) = P(Z>0.6667) =

= 1 -& P(Z<0.6667) = 1 - f(0.6667) =

= 1 -& 0.747518 = 0.252482

Question B is not completely formulated

Then Z = (X-150)/45is a random variable which has standard normal.

A. P(X>180) =

= P(& (X-150)/45>(180-150)/45 ) =

= P(& Z>30/45& ) = P(Z>0.6667) =

= 1 -& P(Z<0.6667) = 1 - f(0.6667) =

= 1 -& 0.747518 = 0.252482

Question B is not completely formulated

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