/*! 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 29 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 \(X_{5}\) 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 ry \(Y\) 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 \(P(-1 \leq Y \leq 1)\). \(\left[\right.\) Hint: \(Y=a_{1} X_{1}+\cdots+a_{5} X_{5}\), with \(a_{1}=\frac{1}{2}, \ldots,\), \(\left.a_{5}=-\frac{1}{3} .\right]\)

Short Answer

Expert verified
\( P(0 \leq Y) \approx 0.319 \) and \( P(-1 \leq Y \leq 1) \approx 0.33. \)

Step by step solution

01

Define the Expression for Y

The random variable \( Y \) is given by \( Y = \frac{X_1 + X_2}{2} - \frac{X_3 + X_4 + X_5}{3} \). This can be expressed in terms of a linear combination of the \( X_i \) variables as \( Y = \frac{1}{2}X_1 + \frac{1}{2}X_2 - \frac{1}{3}X_3 - \frac{1}{3}X_4 - \frac{1}{3}X_5 \).
02

Calculate the Mean of Y

The expected value of \( Y \), \( E(Y) \), can be computed using the linearity of expectation. This gives us:\[ E(Y) = \frac{1}{2}(\mu_1 + \mu_2) - \frac{1}{3}(\mu_3 + \mu_4 + \mu_5) = \frac{1}{2}(20 + 20) - \frac{1}{3}(21 + 21 + 21) = 20 - 21 = -1. \] Thus, \( E(Y) = -1. \)
03

Calculate the Variance of Y

Here, we calculate \( \text{Var}(Y) \) using the formula for the variance of a linear combination of independent random variables:\[ \text{Var}(Y) = \left(\frac{1}{2}\right)^2 \sigma_1^2 + \left(\frac{1}{2}\right)^2 \sigma_2^2 + \left(-\frac{1}{3}\right)^2 (\sigma_3^2 + \sigma_4^2 + \sigma_5^2). \]Substituting \( \sigma_1^2 = \sigma_2^2 = 4 \) and \( \sigma_3^2 = \sigma_4^2 = \sigma_5^2 = 3.5 \), we get:\[ \text{Var}(Y) = 2 \left(\frac{1}{2}\right)^2 \times 4 + 3 \left(\frac{1}{3}\right)^2 \times 3.5 = 1 + \frac{1}{3} \times 10.5 = 1 + 3.5 = 4.5. \]
04

Standardize Y

To compute probabilities involving \( Y \), we need to standardize it. Therefore, we define:\[ Z = \frac{Y - E(Y)}{\sqrt{\text{Var}(Y)}} = \frac{Y + 1}{\sqrt{4.5}}. \] Thus, \( Y \sim N(-1, 4.5) \), and \( Z \sim N(0, 1) \).
05

Calculate \( P(0 \leq Y) \)

For this, we consider:\[ P(0 \leq Y) = P\left( \frac{0 + 1}{\sqrt{4.5}} \leq Z \right) = P\left( \frac{1}{\sqrt{4.5}} \leq Z \right). \]Approximating \( \frac{1}{\sqrt{4.5}} \approx 0.4714 \), we look up the standard normal distribution table to find \( P(Z \leq 0.4714) \), which is approximately \( 0.681 \).Thus, \( P(0 \leq Y) \approx 1 - 0.681 = 0.319. \)
06

Calculate \( P(-1 \leq Y \leq 1) \)

This probability is calculated as:\[ P(-1 \leq Y \leq 1) = P\left( \frac{-1 + 1}{\sqrt{4.5}} \leq Z \leq \frac{1 + 1}{\sqrt{4.5}} \right). \]Since \( \frac{-1 + 1}{\sqrt{4.5}} = 0 \) and \( \frac{1 + 1}{\sqrt{4.5}} \approx 0.9428 \), we look up standard normal table values:\[ P(0 \leq Z \leq 0.9428) \approx 0.33. \]Therefore, \( P(-1 \leq Y \leq 1) \approx 0.33. \)

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Key Concepts

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

Normal Distribution
In statistics, the normal distribution is a key concept often referred to as a bell curve due to its shape. A normal distribution is symmetric, with most values clustering around a central mean, tapering off on either side. This distribution is defined completely by two parameters: the mean (\( \mu \))) and the variance (\( \sigma^2 \))). The mean indicates the central point of the distribution, while the variance tells us how spread out the values are.

The normal distribution is significant in statistics because it frequently appears in natural and social phenomena. In this exercise, the fuel efficiencies of cars using different types of gas are assumed to follow a normal distribution. Economists, statisticians, and engineers often assume normal distribution properties due to the central limit theorem, which states that, with sufficient sample size, data will tend to follow this pattern.
Variance Calculation
Variance is a measure of how much the values in a data set deviate from the mean. More specifically, it's the average of the squared differences from the mean. In mathematical terms, for a random variable \( X \) with a mean \( \mu \), the variance \( \text{Var}(X) \) is calculated as: \[ \text{Var}(X) = \frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2 \] Variance provides an insight into the data's reliability and stability. A larger variance means that data points spread farther from the mean, indicating more variability.

For the cars in the exercise, variances differ slightly between economy and name-brand gasoline, affecting the overall measurement \( Y \), which combines car efficiencies. It's important to accurately calculate variance, as it affects both the expected value and the probability outcomes.
Expected Value
The expected value, often symbolized as \( E(X) \), is the mean or average of all possible values of a random variable. For continuous random variables, it's calculated as the integral of the product of a probability density function and its value. For discrete variables, the formula simplifies to: \[ E(X) = \sum_{i=1}^{N} (x_i \times P(x_i)) \]. In contexts like this exercise, it represents what you predict on average after numerous trials.

To determine the expected value of \( Y \), a linear combination of the fuel efficiencies, expectations are summed up based on weights attributed to \( X_i \) terms in the function of \( Y \). The result gives insights into the overall efficiency difference between gas types.
Standardization
Standardization is a crucial technique in statistics that allows comparison of scores on different scales. By transforming variables into a standard scale, often with mean zero and standard deviation one, it simplifies probability calculations. For a random variable \( X \), this is done through: \[ Z = \frac{X - \mu}{\sigma} \].

In the given problem, standardizing \( Y \) is necessary for determining probabilities because it facilitates the use of standard normal distribution tables. This transformation turns \( Y \sim N(-1, 4.5) \) into the standard normal variable \( Z \sim N(0, 1) \), simplifying the computation of cumulative probabilities, crucial for \( P(0 \leq Y) \) and \( P(-1 \leq Y \leq 1) \).
Probability Calculation
Calculating probabilities with normal distributions involves understanding how likely certain outcomes are, given a random variable's characteristics. First, the variable must often be standardized, converting it to a standard normal form. With this transformation, probabilities for certain ranges can be deduced using standard normal distribution tables.

For instance, to find \( P(0 \leq Y) \) and \( P(-1 \leq Y \leq 1) \), this exercise makes use of the cumulative distribution function of the normal distribution, which gives the probability that a normal random variable is within a given range. Table values are referred to for known intervals, facilitating the approximation of these probabilities in standardized form. This approach combines the power of theoretical mathematics with practical tools for real-world problem-solving.

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Most popular questions from this chapter

It is known that \(80 \%\) of all brand A DVD players work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that \(n=10\) players are randomly selected. Let \(X=\) the number of successes in the sample. The statistic \(X / n\) is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of \(X / n\) is \(.3\), corresponding to \(X=3\). What is the probability of this value (what kind of random variable is \(X)\) ?]

The National Health Statistics Reports dated Oct. 22,2008 stated that for a sample size of 277 18year-old American males, the sample mean waist circumference was \(86.3 \mathrm{~cm}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values: \(\begin{array}{lllllll}5 \text { th } & 10 \mathrm{th} & 25 \mathrm{th} & 50 \mathrm{th} & 75 \mathrm{th} & 90 \mathrm{th} & 95 \mathrm{th} \\ 69.6 & 70.9 & 75.2 & 81.3 & 95.4 & 107.1 & 116.4\end{array}\) a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution. b. Suppose that the population mean waist size is \(85 \mathrm{~cm}\) and that the population standard deviation is \(15 \mathrm{~cm}\). How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least \(86.3 \mathrm{~cm}\) ? c. Referring back to (b), suppose now that the population mean waist size is \(82 \mathrm{~cm}\) (closer to the median than the mean). Now what is the (approximate) probability that the sample mean will be at least \(86.3\) ? In light of this calculation, do you think that 82 is a reasonable value for \(\mu\) ?

For males the expected pulse rate is \(70 / \mathrm{m}\) and the standard deviation is \(10 / \mathrm{m}\). For women the expected pulse rate is \(77 / \mathrm{m}\) and the standard deviation is \(12 / \mathrm{m}\). Let \(X=\) the sample average pulse rate for a random sample of 40 men and let \(\bar{Y}=\) the sample average pulse rate for a random sample of 36 women a. What is the approximate distribution of \(X\) ? Of \(Y\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) the probability \(P(-2 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate (approximately) \(P(X-Y \leq-15)\). If you actually observed \(X-Y \leq-15\), would you doubt that \(\mu_{1}-\mu_{2}=-7\) ? Explain.

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

The lifetime of a type of battery is normally distributed with mean value \(10 \mathrm{~h}\) and standard deviation \(1 \mathrm{~h}\). There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?

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