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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 percent 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 lowmileaqe?

Short Answer

Expert verified
a) 5.48% incur penalties, b) New limit: 55,370 miles, c) 8.08% are low-mileage.

Step by step solution

01

Define the Problem and Identify Known Variables

We are given that the number of miles driven on four-year leases follows a normal distribution with a mean \( \mu = 52,000 \) miles and a standard deviation \( \sigma = 5,000 \) miles. We need to calculate the probability (percent) of going over 60,000 miles and get penalties. Lastly, we need to calculate the percentage of cars classified as low-mileage (less than 45,000 miles).
02

Calculate Z-Score for 60,000 miles

The Z-score is calculated using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.For 60,000 miles: \[Z = \frac{60,000 - 52,000}{5,000} = \frac{8,000}{5,000} = 1.6\]
03

Find the Probability of Exceeding 60,000 Miles

Using standard normal distribution tables or a calculator, find the probability for \( Z \leq 1.6 \). The cumulative probability of \( Z = 1.6 \) is approximately 0.9452. Therefore, the probability of exceeding 60,000 miles is:\[P(X > 60,000) = 1 - 0.9452 = 0.0548\]Hence, approximately 5.48% of the leases result in a penalty.
04

Determine New Upper Limit for Excess Miles with 25% Penalty

We need 25% of leases to exceed the upper miles limit. So, we find the Z-score corresponding to the 75th percentile (which complements the 25% exceeding).Using Z-tables, the z-score for the 75th percentile is approximately 0.674. We then use the Z-score formula in reverse to find the mileage:\[X = \mu + Z \times \sigma = 52,000 + 0.674 \times 5,000 = 55,370 \text{ miles}\]
05

Calculate Z-Score for Low-Mileage of Less than 45,000 Miles

For low mileage under 45,000 miles:\[Z = \frac{45,000 - 52,000}{5,000} = \frac{-7,000}{5,000} = -1.4\]
06

Find the Probability of Low-Mileage Cars

Using Z-scores, the cumulative probability for \( Z = -1.4 \) is approximately 0.0808. Therefore, 8.08% of cars are considered low-mileage.

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Key Concepts

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

Z-score calculation
The Z-score is a statistical measurement that describes a value's relationship to the mean of a data set. If you're tackling problems involving normal distributions, understanding Z-scores is essential. The mean is the middle point of your data, and the standard deviation tells us how spread out the data is. Together, they help us find out how far away a specific point (like 60,000 miles driven in a lease) is from the average.

To calculate a Z-score, you use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
  • Where \( X \) is the value of interest (like those 60,000 miles), \( \mu \) is the mean (52,000 miles in our example), and \( \sigma \) is the standard deviation (5,000 miles).
Using these to calculate the Z-score for 60,000 miles, you find that:
  • \[ Z = \frac{60,000 - 52,000}{5,000} = 1.6 \]
This Z-score tells us how many standard deviations away the 60,000 miles is from the mean. Essentially, a Z-score of 1.6 indicates that 60,000 miles is 1.6 standard deviations above the average mileage.
mileage penalty probability
Once you have calculated a Z-score, you can use it to find probabilities related to the normal distribution. That's where the mileage penalty probability comes into play. By integrating the Z-score into a normal distribution table or using a calculator, you can identify the proportion of data above a certain threshold.

For the case where you exceed 60,000 miles, it's essential to calculate the chances of this happening. With a Z-score of 1.6:
  • The cumulative probability of \( Z \leq 1.6 \) is approximately 0.9452, meaning that 94.52% of all samples fall below 60,000 miles.
  • Conversely, the probability of being above 60,000 miles is \( 1 - 0.9452 = 0.0548 \) or 5.48%.
Thus, only about 5.48% of leases result in a penalty due to excess mileage, making these higher-mileage leases rather rare. Calculating this probability helps lessees understand their financial risk.
percentile for lease terms
Percentiles tell you the relative position of a particular value within a data set. They are especially useful in the context of setting targets or limits in lease terms. Let's say an automobile company wants to adjust lease limits so that only a certain percentage of cars exceed a mileage and thus incur penalties.

In our problem, if the company desires that 25% of leases surpass the mileage cap, then 25% should exceed this new limit. This means we want to find the 75th percentile of the normal distribution.
  • To find the 75th percentile, you look up the corresponding Z-score, which is approximately 0.674.
  • Reversing the Z-score formula gives us the mileage limit:\[ X = \mu + Z \times \sigma = 52,000 + 0.674 \times 5,000 \]
  • This results in a new limit of approximately 55,370 miles.
Setting the limit at 55,370 miles will ensure that 25% of the cars will exceed it, aligning with the policy intent and making it a calculable target for the lease agreement.

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Most popular questions from this chapter

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