/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Ward Doering Auto Sales is consi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Ward Doering Auto Sales is considering offering a special service contract that will cover the total cost of any service work required on leased vehicles. From experience, the company manager estimates that yearly service costs are approximately normally distributed, with a mean of \(\$ 150\) and a standard deviation of \(\$ 25\) a. If the company offers the service contract to customers for a yearly charge of \(\$ 200\) what is the probability that any one customer's service costs will exceed the contract price of \(\$ 200 ?\) b. What is Ward's expected profit per service contract?

Short Answer

Expert verified
a. The probability is 0.0228. b. Ward's expected profit per contract is $50.

Step by step solution

01

Understanding the Normal Distribution

The yearly service costs are normally distributed with a mean (\(\mu\)) of \\(150 and a standard deviation (\(\sigma\)) of \\)25. We can use this information to find the probability of the service cost exceeding \$200.
02

Standardizing the Value

To find the probability that the service cost exceeds \\(200, we first need to find the corresponding \(z\)-score. The formula for a \(z\)-score is: \[ z = \frac{X - \mu}{\sigma} \] where \(X\) is the value of interest (\\)200), \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Substituting the values: \[ z = \frac{200 - 150}{25} = 2 \]
03

Finding the Probability

Using the standard normal distribution table (or a calculator), we can find the probability that a \(z\)-score is less than 2. This is typically about 0.9772. Therefore, the probability that a service cost exceeds \$200 is \(1 - 0.9772 = 0.0228\).
04

Calculating Expected Profit

The expected profit per contract is the difference between the contract charge (\\(200) and the expected service cost (\\)150). Thus, the expected profit for each contract sold is \(\\(200 - \\)150 = \$50\).
05

Considering the Probability of Loss

Since there is a 2.28% probability that a customer’s service costs will exceed \\(200, there is also a possibility of a loss depending on exact costs. But since the average is \\)150, it is more likely that the profit remains positive per contract.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-Score Calculation
A Z-score is a way to quantify the position of a specific data point within a normal distribution. It tells us how many standard deviations a particular value is from the mean. This is particularly important when dealing with normally distributed data, such as the service costs in our example.
To calculate the Z-score, we use the formula:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
  • \(X\) is the value of interest, here it is \\(200 (the cost threshold).
  • \(\mu\) is the mean of the distribution, which is \\)150.
  • \(\sigma\) is the standard deviation, which is \\(25.
By inserting these values, we calculate:
\[ z = \frac{200 - 150}{25} = 2 \]
This tells us that \\)200 is 2 standard deviations above the mean, which can help us find the probability of service costs exceeding the contract threshold.
Probability
Probability in statistics helps us understand how likely an event is going to occur. In the context of our exercise, we're interested in the probability that a customer's service costs will exceed \\(200, which is the price of the service contract.
Given the Z-score of 2 that we calculated, we can use the standard normal distribution table to find how likely it is to be below this Z-score.
The table tells us that roughly 97.72% of data points fall below a Z-score of 2. Consequently, the probability of service costs being above \\)200 is:
\[ 1 - 0.9772 = 0.0228 \]
This means there is a 2.28% chance that the service costs will exceed the contract price, allowing the company to predictively manage risk and profits.
Expected Value
Expected value is a fundamental concept in probability and statistics. It provides a measure of the center of a probability distribution and in this context, it calculates anticipations for profit or loss.
For Ward Doering Auto Sales, the expected value of the profit from one service contract can be found by subtracting the average cost of service from the contract price:
\[ \text{Expected Profit} = \text{Contract Price} - \text{Mean Service Cost} \]
Inserting the known values:
\[ \text{Expected Profit} = 200 - 150 = 50 \]
So, the expected profit per contract is \$50. This expectation is based on the average, meaning while individual results may vary, this is the typical profit seen over many contracts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. What is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals?

The time (in minutes) between telephone calls at an insurance claims office has the following exponential probability distribution. \\[ f(x)=.50 e^{-.50 x} \quad \text { for } x \geq 0 \\] a. What is the mean time between telephone calls? b. What is the probability of having 30 seconds or less between telephone calls? c. What is the probability of having 1 minute or less between telephone calls? d. What is the probability of having 5 or more minutes without a telephone call?

The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for personal holiday shopping (USA Today, November 11,2009 ). Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution. a. The study reported that there is a .53 probability that a worker uses an office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to \(5.8,6.2,6.6,\) or 7 hours? b. Using the mean time from part (a), what is the probability that a worker uses an office computer for holiday shopping more than 10 hours? c. What is the probability that a worker uses an office computer for holiday shopping between 4 and 8 hours?

A binomial probability distribution has \(p=.20\) and \(n=100\) a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of exactly 24 successes? d. What is the probability of 18 to 22 successes? e. What is the probability of 15 or fewer successes?

Is lack of sleep causing traffic fatalities? A study conducted under the auspices of the National Highway Traffic Safety Administration found that the average number of fatal crashes caused by drowsy drivers each year was 1550 ( Business Week , January 26,2004 ). Assume the annual number of fatal crashes per year is normally distributed with a standard deviation of 300 . a. What is the probability of fewer than 1000 fatal crashes in a year? b. What is the probability that the number of fatal crashes will be between 1000 and 2000 for a year? c. For a year to be in the upper \(5 \%\) with respect to the number of fatal crashes, how many fatal crashes would have to occur?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.