Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a mean of19 minutes and a standard deviation of 4 minutes.

​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price?

​(b) If the automotive center does not want to give the discount to more than 2​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer 1

Solution :

Given that,

mean = = 19

standard deviation = = 4

a ) P( x 20 )

P ( x - / ) ( 20 - 19 / 4 )

P ( z 1 / 4 )

P ( z 0.25 )

Using z table

= 0.5987

Probability = 0.5987 = 59.87%

b ) Using standard normal table,

P(Z > z) = 2%

1 - P(Z < z) = 0.02

P(Z < z) = 1 - 0.01 = 0.98

P(Z < 2.326) = 0.98

z = 2.05

Using z-score formula,

x = z * +

x = 2.05 * 4 + 19

= 27.2

The limit = 27.2