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Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

Short Answer

Expert verified
a) The probability is 0.7967. b) It would be very surprising because the probability is almost 0. c) The probability is 0.0001. d) The efficiency is 28.026 mpg.

Step by step solution

01

Find the Z-Score

Calculate the Z-score for both values (29 and 31). For 29, the Z score is calculated as \( z = \frac{29-30}{1.2} = -0.833 \). For 31, the Z score is calculated as \( z = \frac{31-30}{1.2} = 0.833 \).
02

Returning the Probability

The probability that the fuel efficiency for a randomly selected car of this model is between 29 and 31 mpg is given by the area between these two Z-scores. Using the Standard Normal Table, we find the probability as 0.7967.
03

Find the Z-Score for Efficiency Less Than 25 mpg

Calculate the Z-score for the value 25. The Z-score for 25 is \( z = \frac{25-30}{1.2} = -4.167 \).
04

Find the Probability of Efficiency Less Than 25 mpg

Using the Standard Normal Table again, we find the probability as being almost 0. This indicates that it would be highly surprising to find an efficiency less than 25 mpg.
05

Find the Probability of Three Cars Having Efficiency Over 32 mpg

First, find the Z-score for 32, which is \( z = \frac{32-30}{1.2} = 1.667 \). The probability of one car having efficiency over 32 mpg is about 0.0485. However, we need the probability of three cars each having efficiency more than 32 mpg, which is \( 0.0485^3 = 0.0001 \).
06

Calculate x* for Efficiency Exceeding 95% of Cars

The Z score for the 5th percentile is approximately -1.645. By using the formula to convert z-score to x-score, \( x = x^{*} = \mu + \sigma * Z = 30 + 1.2 * -1.645 = 28.026 \), it means 95% of all cars of this model have efficiencies exceeding about \( 28.026 \) mpg

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Key Concepts

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

Normal Distribution
The Normal Distribution is a foundational concept in statistics, visualized as a bell-shaped curve. It illustrates how the values of a dataset are distributed. In the case of our car model's fuel efficiency, the normal distribution shows that most cars have fuel efficiency close to the mean value, and the farther you move from the mean, the fewer cars you find.

This distribution is characterized by two parameters: the mean and the standard deviation.
  • The mean represents the average of all data points, which for our car model is 30.0 mpg.
  • The standard deviation measures how spread out the numbers are from the mean. Here it is 1.2 mpg, meaning most cars' efficiencies are quite close to the mean.
Understanding the normal distribution allows us to make predictions about probabilities, such as the likelihood that a randomly selected car will have a particular range of fuel efficiency.
Z-Score
The Z-Score is a critical tool for understanding data relative to the normal distribution. It describes how many standard deviations a data point is from the mean.

For example, a car having 29 mpg fuel efficiency has a Z-score of \( z = \frac{29 - 30}{1.2} = -0.833 \).This tells us that 29 mpg is 0.833 standard deviations below the average of 30 mpg. Conversely, 31 mpg yields a Z-score of 0.833, positioning it symmetrically above the mean.
Z-scores help us to:
  • Determine the position of individual data points within a distribution
  • Compare data points from different normal distributions
  • Access areas under the normal curve using the Standard Normal Table
By converting raw scores into Z-scores, data from various datasets becomes comparable.
Standard Normal Table
The Standard Normal Table, also known as the Z-Table, is a vital reference tool in statistics. It lists the cumulative probabilities of standard normal distribution, allowing us to find the probability that a Z-score is less than a given value.

Using our example, let's say we wanted to find the probability for a car to have fuel efficiency between 29 and 31 mpg. Having calculated the Z-scores as -0.833 and 0.833, the Standard Normal Table provides us with probabilities:
  • For Z = -0.833, the probability is approximately 0.2023.
  • For Z = 0.833, the probability is approximately 0.7967.
To find the probability between these two Z-scores, we subtract the lower probability from the higher one, giving us the likelihood that a car's efficiency falls within this range.
Statistical Calculation Steps
Executing statistical calculations involves a clear procedure to accurately find probabilities. Let's break it down using the fuel efficiency example.

Step 1: Calculate the Z-score for the values in question. For instance, for 29 and 31 mpg, Z-scores are -0.833 and 0.833, respectively.

Step 2: Use the Standard Normal Table to find associated probabilities. These are 0.2023 for Z=-0.833 and 0.7967 for Z=0.833.

Step 3: Determine probability ranges by subtracting values. The probability between 29 and 31 mpg is 0.7967 - 0.2023 = 0.5944.

Application: This methodology applies to various scenarios, such as checking the rarity of extreme values (e.g., less than 25 mpg) or computing joint probabilities (e.g., multiple cars exceeding 32 mpg).

Following these precise statistical calculations empowers you to make meaningful inferences about a dataset, leading to better decision-making.

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