Question

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 50% discount on the charges. The company wants to limit this discount to at most 8% of the customers. What should the maximum quaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution. Round your answer to the nearest minute. 49 The maximum guaranteed waiting time should be approximately minutes. the tolerance is +/-5%

Answer 1

Solution:-

Given that,

mean = $\mu$ = 15

standard deviation = $\sigma$ = 2.4

Using standard normal table,

P(Z > z) = 8%

= 1 - P(Z < z) = 0.08  

= P(Z < z) = 1 - 0.92

= P(Z < z ) = 0.92

= P(Z < 1.41 ) = 0.92  

z = 1.41

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.41 * 2.4 + 15

x = 18.38

x = 18 minutes