(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the cust 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 guara Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is 30.85 %. Round to two decimal places as needed.) (b) The guaranteed time limit is minutes. (Round up to the nearest minute.) Standard Normal Dis Ama z 0.0 01 0,00 0.5000 05308 0.01 0.5040 05428 0.0 0.50 05
(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the cust 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 guara Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is 30.85 %. Round to two decimal places as needed.) (b) The guaranteed time limit is minutes. (Round up to the nearest minute.) Standard Normal Dis Ama z 0.0 01 0,00 0.5000 05308 0.01 0.5040 05428 0.0 0.50 05
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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![### Analysis of Automotive Center Service Times
The time taken by an automotive center to complete an oil change service on an automobile approximately follows a normal distribution. The mean time for the service is 19 minutes, with a standard deviation of 2 minutes.
#### Problems:
1. **Service Guarantee Analysis**
- **Given:** The automotive center guarantees customers that the service will not take longer than 20 minutes.
- **Condition:** If it does take longer, the customer will receive the service for half-price.
- **Question:** What percent of customers receive the service for half-price?
2. **Customer Discount Limitation**
- **Given:** The automotive center aims to limit discounts to no more than 2% of its customers.
- **Question:** How long should the guaranteed time limit be to meet this criterion?
#### Solutions:
**(a) Percent of Customers Receiving Half-Price Service**
To determine the percentage of customers who receive the service for half-price, follow these steps:
1. Calculate the z-score for 20 minutes using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where \( X = 20 \) minutes, \( \mu = 19 \) minutes (mean), and \( \sigma = 2 \) minutes (standard deviation).
2. Substituting the values:
\[
z = \frac{20 - 19}{2} = 0.5
\]
3. Using a standard normal distribution table or a z-score calculator, find the area to the right of \( z = 0.5 \), which represents the percentage of customers receiving the half-price service.
- From the table, \( P(Z > 0.5) \approx 1 - 0.6915 = 0.3085 \).
Thus, \( 30.85\% \) of customers receive the service for half-price.
**(b) Time Limit for Not More Than 2% Discounts**
To ensure that no more than 2% of customers get the service for half-price:
1. Determine the z-score that corresponds to the top 2% of a normal distribution. This z-score is approximately \( z = 2.05 \).
2. Solve for \( X \) (the guaranteed time limit) using the z-score formula:
\[
X = z\sigma + \mu](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9885b298-0da3-45d1-bdbe-d5991ebff07e%2F52664556-9fd8-436a-8182-69df349b5472%2Fmh2p4b9_processed.png&w=3840&q=75)
Transcribed Image Text:### Analysis of Automotive Center Service Times
The time taken by an automotive center to complete an oil change service on an automobile approximately follows a normal distribution. The mean time for the service is 19 minutes, with a standard deviation of 2 minutes.
#### Problems:
1. **Service Guarantee Analysis**
- **Given:** The automotive center guarantees customers that the service will not take longer than 20 minutes.
- **Condition:** If it does take longer, the customer will receive the service for half-price.
- **Question:** What percent of customers receive the service for half-price?
2. **Customer Discount Limitation**
- **Given:** The automotive center aims to limit discounts to no more than 2% of its customers.
- **Question:** How long should the guaranteed time limit be to meet this criterion?
#### Solutions:
**(a) Percent of Customers Receiving Half-Price Service**
To determine the percentage of customers who receive the service for half-price, follow these steps:
1. Calculate the z-score for 20 minutes using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where \( X = 20 \) minutes, \( \mu = 19 \) minutes (mean), and \( \sigma = 2 \) minutes (standard deviation).
2. Substituting the values:
\[
z = \frac{20 - 19}{2} = 0.5
\]
3. Using a standard normal distribution table or a z-score calculator, find the area to the right of \( z = 0.5 \), which represents the percentage of customers receiving the half-price service.
- From the table, \( P(Z > 0.5) \approx 1 - 0.6915 = 0.3085 \).
Thus, \( 30.85\% \) of customers receive the service for half-price.
**(b) Time Limit for Not More Than 2% Discounts**
To ensure that no more than 2% of customers get the service for half-price:
1. Determine the z-score that corresponds to the top 2% of a normal distribution. This z-score is approximately \( z = 2.05 \).
2. Solve for \( X \) (the guaranteed time limit) using the z-score formula:
\[
X = z\sigma + \mu
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