Question

4. [20 Points) The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2 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

Mean ($\mu$) = 17 min

Standard Deviation ($\sigma$) = 2 min

(a)

P (X > 20 ) = 1- P(X < 20)

We know that :

$Z=\frac{x-\mu}{\sigma}$

Therefore,

1- PRI < 20 – 17. 2 o

1 - P(Z <

Using the Standard Normal Distribution Table, we get:

1 - 0.9332

= 0.0668

= 6.68%

Therefore, the Percentage of Customers that receive the Service rate for Half Price is 6.68%

(b)

We want to Restrict it to not more than 2%, i.e. P(X > x) = 0.02

Therefore, P(X < x) = 1- 0.02 = 0.98

and the Corresponding Z-Value is 2.054

Therefore,

$2.054=\frac{(X-19)}{2}$

Now, we just need to solve for X

$2.054(2)=(X-19)$

4.108 = X - 19

X = 23.108 min.

Therefore, Guaranteed Time Limit Should be made 23.108 min

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