Question

Consider this problem: Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car, and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 505 discount on the charges. The company wants to limit this discount to at most 5% of the customers. What should the maximum guaranteed waiting time be

Answer 1

Answer:

The maximum time guaranteed = 19.04 minutes.

Explanation:

From the given problem data, we have:

Let Y be the random variable which follows the normal distribution.

So,

Y ~ N(u = 15, SD = 2.4

Where, u = mean and SD = Standard Deviation

Let the maximum time guaranteed is = M

So,

P (Y > M) = 0.05   equation 1

Convert this equation 1 into standard normal variable, that is,

P(Y> M) = 0.05

1 -  P(Y $\leq$ M) = 0.05

P(Y $\leq$ M) = 1 - 0.05

P(Y $\leq$ M) = 0.95

P$(\frac{Y-u}{SD} \leq \frac{M-u}{SD} )$ = 0.95

P ( Z $\leq \frac{M - 15}{2.4}$ )  = 0.95     Equation 2

From the equation 2, we have,

$\frac{M-15}{2.4}$ = 1.644853627  

1.644853627 value is from using the function of Excel

( =NORSINV(0.95)) = 1.644853627

So,

M = 1.644853627 + 2.4 + 15

M = 19.04

Hence, the maximum time guaranteed = 19.04 minutes.