/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Five automobiles of the same typ... [FREE SOLUTION] | 91Ó°ÊÓ

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Five automobiles of the same type are to be driven on a 300 mile trip. The first two will use an economy brand of gasoline, and the other three will use a name brand. Let \(X_{1}, X_{2}\), \(X_{3}, X_{4}\), and \(X_{3}\) be the observed fuel efficiencies (mpg) for the five cars. Suppose these variables are independent and normally distributed with \(\mu_{1}=\mu_{2}=20, \mu_{3}=\mu_{4}=\mu_{5}=21\), and \(\sigma^{2}=4\) for the economy brand and \(3.5\) for the name brand. Define an rv \(Y\) by $$ Y=\frac{X_{1}+X_{2}}{2}-\frac{X_{3}+X_{4}+X_{5}}{3} $$ so that \(Y\) is a measure of the difference in efficiency between economy gas and name-brand gas. Compute \(P(0 \leq Y)\) and \(P(-1 \leq Y \leq 1)\). [Hint: \(Y=a_{1} X_{1}+\ldots+a_{5} X_{5}\), with \(\left.a_{1}=\frac{1}{2}, \ldots, a_{5}=-\frac{1}{3} .\right]\)

Short Answer

Expert verified
\(P(0 \leq Y) \approx 0.2869\), \(P(-1 \leq Y \leq 1) \approx 0.3685\).

Step by step solution

01

Define Y in terms of X

The random variable \( Y \) can be expressed as a linear combination of \( X_1, X_2, X_3, X_4, \) and \( X_5 \). Specifically, \( Y = \frac{1}{2}X_1 + \frac{1}{2}X_2 - \frac{1}{3}X_3 - \frac{1}{3}X_4 - \frac{1}{3}X_5 \) according to the problem statement.
02

Calculate the Expectation of Y

The expected value of \( Y \) can be found using the linearity of expectation: \[ \mu_Y = \frac{1}{2}\mu_1 + \frac{1}{2}\mu_2 - \frac{1}{3}\mu_3 - \frac{1}{3}\mu_4 - \frac{1}{3}\mu_5. \]Substitute the given means: \( \mu_1 = \mu_2 = 20 \) and \( \mu_3 = \mu_4 = \mu_5 = 21 \). Thus, \[ \mu_Y = \frac{1}{2}(20) + \frac{1}{2}(20) - \frac{1}{3}(21) - \frac{1}{3}(21) - \frac{1}{3}(21) = 20 - 21 = -1. \]
03

Calculate the Variance of Y

Using the property of variance for a linear combination of independent variables, \[ \sigma_Y^2 = \frac{1}{4}\sigma_1^2 + \frac{1}{4}\sigma_2^2 + \frac{1}{9}(\sigma_3^2 + \sigma_4^2 + \sigma_5^2), \]where \( \sigma_1^2 = \sigma_2^2 = 4 \) and \( \sigma_3^2 = \sigma_4^2 = \sigma_5^2 = 3.5 \). Thus,\[ \sigma_Y^2 = \frac{1}{4}(4) + \frac{1}{4}(4) + 3\times\frac{1}{9}(3.5) = 1.0 + 1.0 + 1.167 = 3.167. \]
04

Calculate Standard Deviation of Y

The standard deviation \( \sigma_Y \) is the square root of the variance calculated: \[ \sigma_Y = \sqrt{3.167} \approx 1.78. \]
05

Compute P(0 ≤ Y) using Normal Distribution

Since \( Y \) is normally distributed with \( \mu_Y = -1 \) and \( \sigma_Y \approx 1.78 \), we standardize \( Y \) to find the probability. \[ P(Y \geq 0) = P\left( \frac{Y - (-1)}{1.78} \geq \frac{0 - (-1)}{1.78} \right) = P(Z \geq 0.5621) \] where \( Z \) is a standard normal variable. The probability can be found from standard normal tables or software: \( P(Z \geq 0.5621) = 1 - P(Z \leq 0.5621) \approx 0.2869. \)
06

Compute P(-1 ≤ Y ≤ 1) using Normal Distribution

To find \(P(-1 \leq Y \leq 1)\), calculate \[ P\left( \frac{-1 - (-1)}{1.78} \leq Z \leq \frac{1 - (-1)}{1.78} \right) = P(0 \leq Z \leq 1.123) \]where \(Z\) is a standard normal variable. Calculate using standard normal tables or software: \( P(0 \leq Z \leq 1.123) = P(Z \leq 1.123) - P(Z \leq 0) \approx 0.3685 - 0.5 = 0.3685. \)

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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 fundamental concept in statistics and is known for its distinctive bell-shaped curve. It describes how the values of a variable are distributed. Most of the data points in a normal distribution are close to the mean, with fewer data points appearing as you move away from the mean in both directions. This pattern creates a symmetrical curve.

A key property of the normal distribution is that it is defined by two parameters: the mean (average) and the variance (how spread out values are). The mean determines the center of the distribution, while the variance dictates its width. When a random variable follows a normal distribution, it is referred to as "normally distributed."

Many natural phenomena follow a normal distribution, including heights, test scores, and in our exercise, fuel efficiencies of automobiles. The importance of the normal distribution also shines through when we use it to make inferences about populations or calculate probabilities, as seen in steps where we compute probabilities related to the variable "Y."
Random Variables
Random variables are the building blocks of probability and statistics. Essentially, a random variable is a numerical outcome of a random phenomenon. They are used to quantify the outcomes of random processes, whether these are the results of rolling a die, measuring fuel efficiency, or counting the number of heads in coin flips.

There are two main types of random variables:
  • Discrete Random Variables: These variables can take on specific values, often whole numbers, like the number of cars passing by in an hour.
  • Continuous Random Variables: These variables can take an infinite number of possible values within a given range, such as the fuel efficiency measured in miles per gallon.
In our example, the efficiency measures for the five cars, denoted as \(X_1, X_2, X_3, X_4,\) and \(X_5\), are continuous random variables. They each follow a normal distribution but with different means and variances, depending on the type of gasoline used. The random variable \(Y\) is then defined as a linear combination of these \(X\) variables to express the difference in efficiency between the two types of gasoline.
Variance and Standard Deviation
Variance and standard deviation are two key statistical concepts that measure the spread of data around the mean. They show us how much variability there is from the average.

**Variance**: This is the average of the squared differences from the mean. It provides a measure of how much the individual data points deviate from the mean value. A high variance indicates that data points are spread out over a wider range, whereas a low variance means they are clustered closely around the mean. The variance of a random variable \(X\) is typically denoted as \(\sigma^2\).

**Standard Deviation**: The standard deviation is simply the square root of the variance, giving us a measure of spread that is in the same units as the data itself. It is more intuitive and easier to interpret than variance, making it a preferred choice for expressing variability.

In our scenario, the variance calculations for the random variable \(Y\) involved combining the variances of independent \(X\) variables linearly. Using the properties of variance, this allows us to determine the standard deviation of \(Y\), which we then employ to compute probabilities related to \(Y\).
Expectation in Statistics
Expectation in statistics refers to the expected value or mean of a random variable. It is essentially a theoretical average, indicating what we can expect from a long series of trials.

The expectation is crucial because it allows statisticians to summarize the central tendency of the random variable in one value. For a discrete random variable, it is calculated as the sum of all possible values weighted by their probabilities. For a continuous random variable, the expectation is found using an integral over all possible values.

In our problem, the expectation of \(Y\) was calculated using the property of linearity of expectation. The expected value of \(Y\) was derived from the expectations of \(X_1, X_2, X_3, X_4,\) and \(X_5\). With the assumed means for these variables, the expectation is thus a simple weighted sum which tells us the average efficiency difference between the economy and name-brand gas when driven over many trials. Understanding the expectation helps interpret \(Y\)'s central tendency confidently even prior to conducting an experiment.

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Most popular questions from this chapter

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5 \mathrm{~min}\), respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

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