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Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of \(6.2 \mathrm{mpg}\) for highway driving. Assume that a Normal model can be applied. a. Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b. In what interval would you expect the central \(68 \%\) of autos to be found? c. About what percent of autos should get more than 31 mpg? d. About what percent of cars should get between 31 and \(37.2 \mathrm{mpg} ?\) e. Describe the gas mileage of the worst \(2.5 \%\) of all cars.

Short Answer

Expert verified
a. The mean of the model is \(24.8 \mathrm{mpg}\), and the standard deviation is \(6.2 \mathrm{mpg}\). The model follows the \(68-95-99.7\) Rule. b. The central 68% of autos are found in the interval 18.6 and 31 mpg. c. An estimated 16% of autos should get more than 31 mpg. d. Approximately 13.5% of cars are expected to get between 31 and 37.2 mpg. e. The gas mileage of the worst 2.5% of cars is less than 12.4 mpg.

Step by step solution

01

Understand the model

In this case, the normal distribution model applies, with a mean (\(mu\)) of \(24.8 \mathrm{mpg}\) and a standard deviation (\(sigma\)) of \(6.2 \mathrm{mpg}\).
02

Drawing the model

To draw the model, note the mean in the middle, one standard deviation above and below the mean (\(mu + sigma\) and \(mu - sigma\)), two above and below (\(mu + 2sigma\) and \(mu - 2sigma\)), and three above and below (\(mu + 3sigma\) and \(mu - 3sigma\)). This will provide an illustration of what the \(68-95-99.7\) Rule predicts.
03

Interval for central 68% of autos

The central 68% of autos would be found within one standard deviation of the mean. That is, between \(mu - sigma\) and \(mu + sigma\), or between \(24.8-6.2\) and \(24.8+6.2\), which is between \(18.6 \mathrm{mpg}\) and \(31 \mathrm{mpg}\).
04

Percentage of autos getting more than 31 mpg

31 mpg is one standard deviation more than the mean, and this would encompass 16% of autos, as per the \(68-95-99.7\) Rule (the area to the right of one standard deviation more than the mean equates to \(100% - 68%/2\)).
05

Percentage of cars getting between 31 and 37.2 mpg

37.2 mpg is two standard deviations more than the mean. So, the percentage of cars getting between 31 mpg and 37.2 mpg corresponds to the region between \(mu + sigma\) and \(mu + 2sigma\), which equates to \(95%/2 - 68%/2\) = \(13.5% \).
06

Describe gas mileage of worst 2.5% of all cars

The worst 2.5% of cars would correspond to a mpg lower than 95% of the other cars. As per the \(68-95-99.7\) Rule, 95% of data falls within 2 standard deviations, the worst 2.5% would be less than mean - 2*standard deviation, which equates to \(24.8 - 2*6.2\) = \(12.4 \mathrm{mpg}\).

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Key Concepts

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

68-95-99.7 Rule
The 68-95-99.7 Rule is also known as the empirical rule, which is a shorthand used to remember the percentage of values that lie within a band around the mean in a normal distribution with a bell-shaped curve. In a perfect normal distribution:
  • Approximately 68% of the data falls within one standard deviation of the mean.
  • Approximately 95% falls within two standard deviations.
  • Almost all (99.7%) falls within three standard deviations.
This rule is extremely helpful in predicting the probability of a certain range of outcomes. For instance, in our exercise, we know that the mean fuel economy is 24.8 mpg and the standard deviation is 6.2 mpg. Utilizing the 68-95-99.7 Rule, we can quickly estimate that 68% of the cars will have a fuel economy between 18.6 (mean - 1 SD) and 31 (mean + 1 SD) mpg.
To visualize this, imagine a smooth, symmetric hill, where the peak represents the mean fuel economy. As you move away from the peak in either direction, the number of cars that achieve that fuel economy diminishes, fitting neatly within the percentages outlined by the 68-95-99.7 Rule.
Standard Deviation
The concept of Standard Deviation cannot be overstressed in statistics. It represents a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.
In the context of fuel economy, a standard deviation of 6.2 mpg shows there is a variation of 6.2 mpg from the mean (24.8 mpg) amongst the cars tested. If we say a specific car has a fuel economy that falls within one standard deviation above the mean, it implies that the car's fuel economy is within 24.8 plus 6.2 mpg, which is 31 mpg. This is useful when determining how an individual car's fuel economy compares to the overall group. In our exercise, knowing that 68% of autos fall within one standard deviation gives consumers a good sense of how most cars perform on the highway.
Fuel Economy Estimates
The term Fuel Economy Estimates refers to the expected performance of a vehicle in terms of miles per gallon (mpg). These estimates are crucial for consumers, manufacturers, and regulatory agencies as they provide a standard way of comparing the efficiency of different vehicles.
When the Environmental Protection Agency (EPA) provides an estimate like the mean of 24.8 mpg for highway driving, it's using the data from various tests to predict average performance. However, due to the variability among driving conditions, driver habits, and vehicle conditions, the actual economy achieved can vary significantly, an aspect reflected in the standard deviation.
Knowing the average and the standard deviation allows us to apply the normal distribution and the 68-95-99.7 Rule to predict fuel economy for the majority of vehicles. For instance, if a consumer is looking for a vehicle that is better than 75% of the market in fuel economy, they can look for a car that has more than 31 mpg, which falls above one standard deviation from the mean and equates to the top 32% of vehicles (100% - 68% / 2).

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Most popular questions from this chapter

Companies that design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of 38.2 inches and standard deviation of 1.8 inches. a. What fraction of kindergarten kids should the company expect to be less than 3 feet tall? b. In what height interval should the company expect to find the middle \(80 \%\) of kindergarteners? c. At least how tall are the biggest \(10 \%\) of kindergarteners?

The shoe size data for women has a mean of 38.46 and a standard deviation of 1.84 . To convert to U.S. sizes, use USsize \(=\) EuroSize \(\times 0.7865-22.5\) a. What is the mean women's shoe size for these respondents in U.S. units? b. What is the standard deviation in U.S. units?

A tire manufacturer believes that the treadlife of its snow tires can be described by a Normal model with a mean of 32,000 miles and standard deviation of 2500 miles. a. If you buy one of these tires, would it be reasonable for you to hope it will last 40,000 miles? Explain. b. Approximately what fraction of these tires can be expected to last less than 30,000 miles? c. Approximately what fraction of these tires can be expected to last between 30,000 and 35,000 miles? d. Estimate the IQR of the treadlives. e. In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If the dealer is willing to give refunds to no more than 1 of every 25 customers, for what mileage can he guarantee these tires to last?

NFL data from the 2015 football season reported the number of yards gained by each of the league's 488 receivers: The mean is 274.73 yards, with a standard deviation of 327.32 yards. a. According to the Normal model, what percent of receivers would you expect to gain more yards than 2 standard deviations above the mean number of yards? b. For these data, what does that mean? c. Explain the problem in using a Normal model here.

The mean household income in the U.S. in 2014 was about \(\$ 72,641\) and the standard deviation was about \(\$ 85,000\). (The median income was \(\$ 51,939 .\) ) If we used the Normal model for these incomes, a. What would be the household income of the top \(1 \% ?\) b. How confident are you in the answer in part a? C. Why might the Normal model not be a good one for incomes?

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