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Most four-year automobile leases allow up to 60,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follows the normal distribution. The mean is 52,000 miles and the standard deviation is 5,000 miles. a. What percent of the leases will yield a penalty because of excess mileage? b. If the automobile company wanted to change the terms of the lease so that \(25 \%\) of the leases went over the limit, where should the new upper limit be set? c. One definition of a low-mileage car is one that is 4 years old and has been driven less than 45,000 miles. What percent of the cars returned are considered low-mileage?

Short Answer

Expert verified
a) 5.48% of leases incur a penalty. b) Set limit to 55,370 miles. c) 8.08% of cars are low-mileage.

Step by step solution

01

- Understand the Problem

The problem involves calculating probabilities and cutoff values based on the normal distribution of miles driven. We have to use the properties of the normal distribution (mean and standard deviation) to find answers to the three parts.
02

- Calculate Z-score for Limit

For part (a), determine the Z-score for the 60,000 mile limit using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \(X = 60,000\), \(\mu = 52,000\), and \(\sigma = 5,000\). Thus, \( Z = \frac{60,000 - 52,000}{5,000} = 1.6 \).
03

- Determine Percentage for Part A

Using a standard normal distribution table or calculator, find the probability that \(Z\) is greater than 1.6. This gives the percentage of leases with penalties. \( P(Z > 1.6) \approx 0.0548 \) or approximately \(5.48\% \).
04

- Calculate New Lease Limit for Part B

To ensure that 25% of the leases are over the limit, find the Z-score such that 75% of the distribution is below it. The Z-score corresponding to 0.75 (for \(P(Z < X) = 0.75\)) is approximately 0.674. Use the formula: \[ X = \mu + Z \cdot \sigma \]\( X = 52,000 + 0.674 \times 5,000 = 55,370 \) miles.
05

- Determine Low Mileage Percentage for Part C

Find the Z-score for 45,000 miles using:\[ Z = \frac{45,000 - 52,000}{5,000} = -1.4 \].Find the probability that \(Z\) is less than -1.4. Using the normal distribution table or calculator, \( P(Z < -1.4) \approx 0.0808 \) or \(8.08\%\).

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Key Concepts

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

Z-score calculation
In statistics, the Z-score is a crucial concept that helps us understand where a data point lies in relation to the mean of a distribution. The Z-score formula is given by: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the value of the data point.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation of the distribution.

The Z-score expresses how many standard deviations a data point is from the mean. A positive Z-score indicates the value is above the mean, while a negative Z-score shows it is below. For example, in part (a) of our problem, we calculated the Z-score for 60,000 miles as 1.6, meaning it is 1.6 standard deviations above the mean. Knowing this helps us find probabilities associated with the normal distribution, which is particularly useful for determining penalties.
probability distribution
A probability distribution describes how the probabilities of various outcomes are distributed for a random variable. The normal distribution, or Gaussian distribution, is one of the most common probability distributions in statistics. It is characterized by its bell-shaped curve and is defined by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).
In our exercise, the distribution of miles driven falls under the normal distribution with a mean of 52,000 miles and a standard deviation of 5,000 miles. This information allows us to calculate specific probabilities, such as those needed for excess mileage penalties or defining low-mileage returns.
When we refer to a value like 60,000 miles, we can use the Z-score to determine what percentage of the total data lies beyond this point. Using a standard normal distribution table, we found that approximately 5.48% of leases will have penalties, corresponding to the excess miles driven.
standard deviation
The standard deviation is a measure of how spread out the numbers in a data set are. In a normal distribution, most data points will lie within one standard deviation of the mean. Standard deviation helps us to quantify the amount of variation or dispersion in a set of values.
In the context of our exercise, the standard deviation of 5,000 miles tells us how much the mileage on the leases typically varies from the mean of 52,000 miles. A smaller standard deviation would indicate that the mileages are closely packed around the mean, whereas a larger one would show more spread out values.
  • It’s crucial for Z-score calculation, as the Z-score tells us how many standard deviations away a particular value is from the mean.
  • The standard deviation directly impacts the shape of the normal distribution; a larger standard deviation results in a "flatter" bell curve.
Understanding standard deviation helps us make sense of the variability in the leases and their related probabilities, such as determining what constitutes a low-mileage return.

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Most popular questions from this chapter

A normal distribution has a mean of 50 and a standard deviation of 4. a. Compute the probability of a value between 44.0 and 55.0 . b. Compute the probability of a value greater than 55.0 . c. Compute the probability of a value between 52.0 and 55.0 .

A normal population has a mean of 20.0 and a standard deviation of \(4.0 .\) a. Compute the \(z\) value associated with 25.0 . b. What proportion of the population is between 20.0 and \(25.0 ?\) c. What proportion of the population is less than \(18.0 ?\)

The annual commissions earned by sales representatives of Machine Products Inc., a manufacturer of light machinery, follow the normal probability distribution. The mean yearly amount earned is \(\$ 40,000\) and the standard deviation is \(\$ 5,000 .\) a. What percent of the sales representatives earn more than \(\$ 42,000\) per year? b. What percent of the sales representatives earn between \(\$ 32,000\) and \(\$ 42,000 ?\) c. What percent of the sales representatives earn between \(\$ 32,000\) and \(\$ 35,000 ?\) d. The sales manager wants to award the sales representatives who earn the largest commissions a bonus of \(\$ 1,000\). He can award a bonus to \(20 \%\) of the representatives. What is the cutoff point between those who earn a bonus and those who do not?

The net sales and the number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is \(\$ 180\) million and the standard deviation is \(\$ 25\) million. For the number of employees, the mean is 1,500 and the standard deviation is \(120 .\) Clarion Fabricators had sales of \(\$ 170\) million and 1,850 employees. a. Convert Clarion's sales and number of employees to \(z\) values. b. Locate the two \(z\) values. c. Compare Clarion's sales and number of employees with those of the other fabricators.

A normal distribution has a mean of 80 and a standard deviation of 14 . Determine the value above which \(80 \%\) of the values will occur.

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