Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with the 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

Hi,

Those customers who receive half-price are those where the service takes more than 20 minutes. Since the mean is 17 minutes, the customers who receives half-price are those where the service takes 20 - 17 = 3 minutes above the mean.

Answer 2

When dealing with these kind of problems, the thing is all about converting units. You need to convert those 3 units ''of minutes'' in units of standard deviation.

Answer 3

How do I do that?

Answer 4

You are given the following information:

Answer 5

\(\large{\mu = 17\ minutes}\)

\(\large{\sigma = 2 \frac{minutes}{standard\ deviation}}\) <- It can be read as 'minutes per standard deviation'.

\(\large{X = 20\ minutes}\)

Answer 6

A unit conversion (for instance, from inches to meters) is all about cancelling units.

Answer 7

We want to find how many standard deviations away those 3 minutes lie above the mean.

Cancelling unit of minutes and leaving units of standard deviations is needed. Do you guess how it can be done?

Answer 8

I'm not sure. Cross multiplication?

Answer 9

Yep. Let me help you:

\(\large{\frac{3\ minutes}{\left(2 \frac{minutes}{standard\ deviation}\right)} = \frac{3}{2} minutes * \frac{standard\ deviation}{minutes}}\)

Answer 10

minutes cancels out and we are left with 3/2 standard deviations (sd) = 1.5 (sd)

Answer 11

We are almost done with a), because you are asked to find the percentage of customer where the service takes more than 3 minutes. We know that more than 3 minutes means more than 1.5 sd above the mean.

Answer 12

To find this out, you need a z-table or z-calculator. Do you know how to do it?

Answer 13

I have a z-table. I was curious, how did you get 1.5 sd?

Answer 14

Take a look the last expression I wrote in Latex code.

3/2 = 1.5

and minutes cancel each other out, leaving just the units of standard deviation.

The 1.5 sd (:

Answer 15

I understand. What would I have to look for on my z table?

Answer 16

Since you want to find the probability of getting a value (an oil change service) greater than 1.5 sd from the mean, you need to look at the proportion that lies 1.5 sd toward the tail.

Answer 17

I'm not sure how to read this :[

Answer 18

Usually a z-table has 4 columns. What we've found, 1.5 sd is called the z-score.

Answer 19

Take a look at the image below.

Answer 20

Proportion in tail is what we are looking for. It just means the proportion above 1.5sd and it goes all the way to the upper tail.

Answer 21

Does it seem too difficult?

Answer 22

Yes. I'm confused.

Answer 23

Notice the process we followed:

First we find the difference of the value given (X) from the mean.

Then we convert from those units to unit of standar deviation. This is generally written as:

\(\large{\frac{X-\mu}{\sigma}}\)

Then we get the z-score.

And finally we find the probability.

Having X and mean -> Difference -> unit conversion -> z-score -> probability.

Answer 24

To solve b) you need to do the whole process in reverse (following the arrow from the right):

Having mean but not X <- Difference <- unit conversion <- z-score <- probability