214k views
4 votes
From long experience, it is known that the time it takes to do an oil change and lubrication job on a vehicle has a normal distribution with a mean of 17.8 minutes and a standard deviation of 5.2 minutes. An auto service shop will give a free lube and oil change service to any customer who must wait beyond the guaranteed time to complete the work. If the shop does not want to give more than 1% of its customers a free lube and oil change service, how long should the guarantee be

2 Answers

5 votes

Final answer:

To find the length of time the guarantee should be, we need to find the value, x, such that P(X > x) = 0.01. We can use the Z-score formula to calculate this.

Step-by-step explanation:

To find the length of time the guarantee should be, we need to find the value, x, such that P(X > x) = 0.01. In other words, we need to find the value of x that corresponds to the 99th percentile of the distribution.

To find this value, we can use the Z-score formula:

Z = (x - mean) / standard deviation

For P(X > x) = 0.01, we want to find the value of x for which the area to the right of x is 0.01. The area to the left of x is 1 - 0.01 = 0.99. Using a Z-table or calculator, we can find the Z-score that corresponds to an area of 0.99.

Once we have the Z-score, we can use the formula:

x = Z * standard deviation + mean

Substituting in the Z-score we found, we can solve for x to find the length of time the guarantee should be.

User Jakub Bujny
by
4.0k points
4 votes

Answer: 29.916 minutes

Step-by-step explanation:

Given the following :

Mean of distribution = 17.8 minutes

Standard deviation of distribution = 5.2 minutes

Percentage of customers to give free lube and oil change service should not exceed 1% if the customer waits beyond the guaranteed time ;

Length of guarantee :

Since % of customers should not exceed 1%

Then probability of not being chosen = (1 - 0.01) = 0.99

Using the z table to find the Zscore of 0.99,

Zscore of 0.99 = 2.33

This is 2.33 standard deviations above the mean

(2.33 × standard deviation) + mean value

(2.33 × 5.2 minutes) + 17.8 minutes

12.116 + 17.8

= 29.916 minutes

User Haddr
by
4.0k points