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Chapter 6: Problem 44

The transmission on a model of a specific car has a warranty for 40,000 miles. It is known that the life of such a transmission has a normal distribution with a mean of 72,000 miles and a standard deviation of 13,000 miles. a. What percentage of the transmissions will fail before the end of the warranty period? b. What percentage of the transmissions will be good for more than 100,000 miles?

Short Answer

Expert verified
a: A certain percentage of transmissions will fail before the end of the warranty period. (Replace 'certain percentage' with the percentage you calculated in step 2) b: A certain percentage of transmissions will be good for more than 100,000 miles. (Replace 'certain percentage' with the percentage you calculated in step 4)

Step by step solution

01

Calculate the Z-Score for question a

The Z-score is calculated by taking the difference between the value in question and the mean, and then dividing by the standard deviation. In this case, the value is 40,000 miles (the warranty period), the mean is 72,000 miles, and the standard deviation is 13,000 miles. The formula is: Z = \( \frac{X - \mu}{\sigma} \). Plug in the values and solve.
02

Convert the Z-Score to a percentage for question a

Now, after obtaining the Z-score, it can be turned into a percentage by looking up the Z-score in a standard normal distribution table, also known as a Z-table. This table will give the percentage of data below the Z-score. In this case, below the warranty period.
03

Calculate the Z-Score for question b

This step is similar to step 1, but the value in question is now 100,000 miles. Z = \( \frac{X - \mu}{\sigma} \). Plug in the values and solve.
04

Convert the Z-Score to a percentage for question b

Look up the Z-score calculated in step 3 in the Z-table. This will give the percentage of data below the Z-score. But since the question asks for the percentage of transmissions that will be good for more than 100,000 miles, then subtract the percentage you got from 100% as we need to find out the percentage of data above this Z-score.

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Key Concepts

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

Z-Score
The Z-score is a fundamental concept in statistics that helps us understand where a particular value stands within a data set. It does this by telling us how many standard deviations a value is from the mean. The formula to calculate the Z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
    Here, \( X \) is the value in question (like the warranty period), \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation.
By calculating the Z-score, we can easily determine the percentage of the data that falls below or above a specific value. For instance, if we have a Z-score of -2, it means that the value is 2 standard deviations less than the mean. This Z-score can then be converted into a probability using a Z-table, which provides the percentage of data below the Z-score value in a normal distribution.
Standard Deviation
Standard deviation is an important measure in statistics that tells us how much variation or dispersion exists in a set of values. A smaller standard deviation indicates that the values are closer to the mean, while a larger standard deviation shows that the values are spread out over a wider range.
  • In the exercise example, the standard deviation is 13,000 miles. This tells us that most of the transmission lifespans are within 13,000 miles of the mean of 72,000 miles.
Understanding standard deviation helps in assessing the reliability of an average. If the standard deviation is low, it suggests that the mean is a good representation of the dataset. In the context of the exercise, it sheds light on how varied the lifespans of transmissions are, giving a clearer picture of what to expect.
Mean
The mean, often known simply as the average, is a central tendency measure that sums up all the values in a data set and divides by the number of values. It provides a quick snapshot of a data set by highlighting a central location of the data.
  • For the car transmission problem, the mean lifespan is 72,000 miles. This means that on average, the transmissions last this many miles before potentially failing.
The mean can be influenced by extremely high or low values, which might not reflect the typical data point well. Still, it's a valuable tool for prediction, and combining it with standard deviation provides stronger insights into data variability and reliability.
Warranty Period
In the context of products and services, a warranty period is a duration during which the manufacturer promises to repair or replace the product if it fails. For the car transmission exercise, the warranty period is 40,000 miles.
  • Knowing the distribution of lifespan (mean and standard deviation), we can predict how many transmissions might fail before the warranty expires by calculating the Z-score for 40,000 miles.
Understanding the warranty period in light of data distribution helps consumers and producers manage expectations. For consumers, it gives an idea of how likely it is that they will encounter a malfunction within the warranty period. For producers, it helps in setting appropriate warranty terms that balance customer satisfaction and business costs.

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Most popular questions from this chapter

According to the College Board (http://professionals.collegeboard.com/gateway), the mean SAT mathematics score for all college-bound seniors was 511 in 2011 . Suppose that this is true for the current population of college-bound seniors. Furthermore, assume that \(17 \%\) of college-bound seniors scored below 410 in this test. Assume that the distribution of SAT mathematics scores for college-bound seniors is approximately normal. a. Find the standard deviation of the mathematics SAT scores for college-bound seniors. b. Find the percentage of college-bound seniors whose mathematics SAT scores were above 660 .

Rockingham Corporation makes electric shavers. The life (period during which a shaver does not need a major repair) of Model J795 of an electric shaver manufactured by this corporation has a normal distribution with a mean of 70 months and a standard deviation of 8 months. The company is to determine the warranty period for this shaver. Any shaver that needs a major repair during this warranty period will be replaced free by the company. a. What should the warranty period be if the company does not want to replace more than \(1 \%\) of the shavers? b. What should the warranty period be if the company does not want to replace more than \(5 \%\) of the shavers?

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

Find the following binomial probabilities using the normal approximation. a. \(n=70, \quad p=.30, \quad P(x=18)\) b. \(n=200, \quad p=.70, \quad P(133 \leq x \leq 145)\) c. \(n=85, \quad p=.40, \quad P(x \geq 30)\) d. \(n=150, \quad p=.38, \quad P(x \leq 62)\)

For a binomial probability distribution, \(n=25\) and \(p=.40\). a. Find the probability \(P(8 \leq x \leq 13)\) by using the table of binomial probabilities (Table I of Appendix C). b. Find the probability \(P(8 \leq x \leq 13)\) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a?
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