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Chapter 7: Problem 42

Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles. a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? c. What percent of the Fords traveled 62,000 miles or less during the year? d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles? Explain.

Short Answer

Expert verified
a. 0.47% logged 65,200+ miles. b. 12.41% logged between 57,060 and 58,280 miles. c. 84.13% logged 62,000 miles or less. d. It is not reasonable; probability is nearly zero.

Step by step solution

01

Understand the Normal Distribution

The problem states the mileage distribution is normal with a mean (\(\mu\)) of 60,000 miles and a standard deviation (\(\sigma\)) of 2,000 miles. For such normal distributions, percentages can be calculated using the standard normal distribution (Z-distribution).
02

Calculate Z-Score for 65,200 Miles or More (Part a)

To find the percentage of trucks that logged 65,200 miles or more, calculate the Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \). Thus, \( Z = \frac{65,200 - 60,000}{2,000} = 2.6 \). Using the standard normal distribution table, find the probability corresponding to \( Z = 2.6 \). Look for \( Z = 2.6 \) in the table, which gives approximately 0.9953. Since the table gives cumulative from the left, subtract from 1: 1 - 0.9953 = 0.0047. Thus, 0.47% of the trucks logged 65,200 miles or more.
03

Calculate Z-Scores for 57,060 and 58,280 Miles (Part b)

Find the Z-score for 57,060 miles: \( Z = \frac{57,060 - 60,000}{2,000} = -1.47 \). Look up this Z-score in the table for probability, which is approximately 0.0708. Then, find the Z-score for 58,280 miles: \( Z = \frac{58,280 - 60,000}{2,000} = -0.86 \). This corresponds to approximately 0.1949 in the table. Subtract the probability of 57,060 miles from that of 58,280 miles: 0.1949 - 0.0708 = 0.1241. Thus, 12.41% of trucks logged between these miles.
04

Calculate Z-Score for 62,000 Miles or Less (Part c)

Calculate the Z-score for 62,000 miles: \( Z = \frac{62,000 - 60,000}{2,000} = 1.0 \). Using the standard normal table, the probability for \( Z = 1.0 \) is approximately 0.8413. Thus, 84.13% of the trucks logged 62,000 miles or less.
05

Evaluate Reasonability for Miles Over 70,000 (Part d)

Calculate Z-score for 70,000 miles: \( Z = \frac{70,000 - 60,000}{2,000} = 5\). The Z-score of 5 is extremely rare in normal distribution, meaning the probability is almost 0 (can confirm by checking the table). This suggests it is highly improbable that any truck exceeded 70,000 miles, thus it is not reasonable to expect this to happen.

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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, calculating the Z-score is essential when working with the normal distribution. The Z-score allows us to determine how far away a specific data point is from the mean. To find a Z-score, you use the formula: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value you're examining
  • \( \mu \) is the mean of the distribution
  • \( \sigma \) is the standard deviation
The Z-score tells us how many standard deviations away from the mean our value is.
For example, in the exercise, to figure out the Z-score of trucks logging 65,200 miles, the calculation becomes \[ Z = \frac{65,200 - 60,000}{2,000} = 2.6 \].This indicates that 65,200 miles is 2.6 standard deviations above the mean mileage. This calculation helps us find related probabilities using the normal distribution table.
Standard deviation
Standard deviation is a metric used to quantify the amount of variation or dispersion in a set of values. It's an integral part of many statistical analyses, particularly when dealing with the normal distribution.
The formula for calculating the standard deviation in a dataset is designed to give you a measure of how spread out the data points are around the mean. In practical terms:
  • If the standard deviation is small, the data points tend to be close to the mean.
  • If the standard deviation is large, the data points are spread out over a wider range.
In the exercise involving truck mileage, the standard deviation is applied to understand how mileage varies among trucks. A standard deviation of 2,000 miles indicates how truck mileages deviate from the 60,000-mile average.
Probability in statistics
Probability in statistics is the likelihood or chance of an event occurring. It's a fundamental concept that underpins many statistical methods and analyses. When looking at normal distributions, probability helps us understand how data is distributed around the mean.
When calculating probabilities related to the normal distribution:
  • We use the Z-score to represent a point within the distribution.
  • The Z-score is then used to find the corresponding probability in a standard normal distribution table.
For instance, in the exercise, to determine the probability of trucks traveling more than 65,200 miles, we computed a Z-score of 2.6.
Next, we looked up this Z-score in the table to determine the probability that trucks would travel this distance or more.
Normal distribution table
A normal distribution table, also known as a Z-table, is a valuable tool in statistics. It helps us find the probability that a statistic is observed below, above, or between values on a standard normal distribution.
The table is used as a reference for Z-scores:
  • It is structured to show the cumulative probability of a value being below a given Z-score.
  • By subtracting these probabilities from 1, we can determine the likelihood of being above a specific Z-score.
In the exercise, after calculating the Z-score for trucks logging 65,200 miles (which was 2.6), we used the normal distribution table to find a cumulative probability of approximately 0.9953.
Subtracting this from 1 gave us the percentage of trucks exceeding 65,200 miles, helping facilitate further analysis.

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Most popular questions from this chapter

According to the Insurance Institute of America, a family of four spends between \(\$ 400\) and \(\$ 3,800\) per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than \(\$ 2,000\) per year on insurance per year? d. What is the probability a family spends more than \(\$ 3,000\) per year?

A study of long distance phone calls made from the corporate offices of the Pepsi Bottling Group, Inc., in Somers, New York, showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes. a. What fraction of the calls last between 4.2 and 5 minutes? b. What fraction of the calls last more than 5 minutes? c. What fraction of the calls last between 5 and 6 minutes? d. What fraction of the calls last between 4 and 6 minutes? e. As part of her report to the president, the Director of Communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

The mean of a normal distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? b. What is the area between the mean and 395 pounds? c. What is the probability of selecting a value at random and discovering that it has a value of less than 395 pounds?

The mean of a normal probability distribution is \(500 ;\) the standard deviation is \(10 .\) a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. Practically all of the observations lie between what two values?

A recent report in USA Today indicated a typical family of four spends \(\$ 490\) per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a mean of \(\$ 490\) and a standard deviation of \(\$ 90 .\) a. What percent of the families spend more than \(\$ 30\) but less than \(\$ 490\) per month on food? b. What percent of the families spend less than \(\$ 430\) per month on food? c. What percent spend between \(\$ 430\) and \(\$ 600\) per month on food? d. What percent spend between \(\$ 500\) and \(\$ 600\) per month on food?
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