Question

The amount of time customers wait in line during hour at the bank follows a normal distribution with a mean of 8.5 minutes and standard deviation of 2.2 minutes.

During Peak hours at the bank, what is the probablity a randomly selected person will wait:

a. at least 12 minutes?

b.no more than 6 minutes?

c.between 7 and 10 minutes?

d. How many minutes should the bank advertise for service so that no more than 4% of its customers will wait longer than the time advertised?

Answer 1

Solution :

Let X be the amount of time customers wait in line during hour at the bank. So we have X follows a normal distribution with a mean of 8.5 minutes and standard deviation of 2.2 minutes.

A. During Peak hours at the bank, the probability a randomly selected person will wait at least 12 minutes is given by; P(X ≥ 12).

The Z score for 12 is, Z1= (12-8.5) / 2.2 = 1.59 and hence from the Z table we get, P(Z ≥ Z1) = P( Z ≥ 1.59 ) = 1 - P( Z < 1.59) = 1 - 0.9441 = 0.0559.

Hence we get, P(X ≥ 12) =0.0559.

B. During Peak hours at the bank, the probability a randomly selected person will wait no more than 6 minutes is given by; P(X ≤ 6).

The Z score for 6 is, Z2= (6-8.5) / 2.2 = -1.14 and hence from the Z table we get, P(Z ≤ Z2) = P( Z ≤ -1.14 ) = 0.1271.

Hence we get, P(X ≤ 6) =0.1271.

C. During Peak hours at the bank, the probability a randomly selected person will wait between 7 and 10 minutes is given by; P( 7 ≤ X ≤ 10).

The Z score for 7 and 10 are, Z3= (7-8.5) / 2.2 = -0.68; Z4 = (10-8.5) / 2.2 = 0.68 and hence from the Z table we get, P( -0.68 ≤ Z ≤ 0.68 ) = P( Z ≤ 0.68 ) - P( Z ≤ -0.68 ) = 0.7517 - 0.2483 = 0.5034.

Hence we get, P( 7 ≤ X ≤ 10) =0.5034.

D. Let X0 be the minutes should the bank advertise for service so that no more than 4% of its customers will wait longer than the time advertised. And hence we get, P(X ≥ X0) = 0.04.

From the Z table we have, P( Z ≤ 1.75) = 0.9599 0.96 this gives P(Z ≥ 1.75) = 0.04. So the Z score corresponding to the value X0 is 1.75.

This gives, 1.75 = (X0-8.5) / 2.2 this gives, X0 = 1.75*2.2+8.5 = 12.35.

The bank should advertise 12.35 minutes for service so that no more than 4% of its customers will wait longer than the time advertised.