/*! 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 91 A retail dealer sells three bran... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 91

A retail dealer sells three brands of automobiles. For brand A, her profit per sale, \(X\) is normally distributed with parameters \(\left(\mu_{1}, \sigma_{1}^{2}\right) ;\) for brand \(\mathrm{B}\) her profit per sale \(Y\) is normally distributed with parameters \(\left(\mu_{2}, \sigma_{2}^{2}\right) ;\) for brand \(C\), her profit per sale \(W\) is normally distributed with parameters \(\left(\mu_{3},\right.\) \(\sigma_{3}^{2}\) ). For the year, two-fifths of the dealer's sales are of brand \(\mathrm{A}\), one-fifth of brand \(\mathrm{B}\), and the remaining two- fifths of brand C. If you are given data on profits for \(n_{1}, n_{2},\) and \(n_{3}\) sales of brands \(A\) B, and \(C\), respectively, the quantity \(U=.4 \bar{X}+.2 \bar{Y}+.4 \bar{W}\) will approximate to the true average profit per sale for the year. Find the mean, variance, and probability density function for \(U .\) Assume that \(X, Y,\) and \(W\) are independent.

Short Answer

Expert verified
U's mean is \(0.4\mu_1 + 0.2\mu_2 + 0.4\mu_3\) and variance is \((0.4)^2\frac{\sigma_1^2}{n_1} + (0.2)^2\frac{\sigma_2^2}{n_2} + (0.4)^2\frac{\sigma_3^2}{n_3}\). U is normally distributed.

Step by step solution

01

Mean of U

The mean of a linear combination of random variables is the linear combination of their means. Since \(U = 0.4 \bar{X} + 0.2 \bar{Y} + 0.4 \bar{W}\), we have:\[E(U) = 0.4 E(\bar{X}) + 0.2 E(\bar{Y}) + 0.4 E(\bar{W})\]Given that the means of \(\bar{X}, \bar{Y}, \) and \(\bar{W}\) are \(\mu_1, \mu_2, \) and \(\mu_3\) respectively, the mean of \(U\) becomes:\[E(U) = 0.4 \mu_1 + 0.2 \mu_2 + 0.4 \mu_3\]
02

Variance of U

The variance of a linear combination of independent random variables is the sum of the variances of each term multiplied by the square of their coefficients. Thus, we find \(Var(U)\) as:\[Var(U) = (0.4)^2 Var(\bar{X}) + (0.2)^2 Var(\bar{Y}) + (0.4)^2 Var(\bar{W})\]For each \(\bar{X}, \bar{Y}, \bar{W}\), the variance is given by \(\frac{\sigma_i^2}{n_i}\):\[Var(U) = (0.4)^2 \cdot \frac{\sigma_1^2}{n_1} + (0.2)^2 \cdot \frac{\sigma_2^2}{n_2} + (0.4)^2 \cdot \frac{\sigma_3^2}{n_3}\]
03

Probability Density Function of U

Since \(X, Y,\) and \(W\) are normally distributed and they are independent, the linear combination \(U = 0.4 \bar{X} + 0.2 \bar{Y} + 0.4 \bar{W}\) is also normally distributed. Therefore, the probability density function (PDF) of \(U\) is:\[U \sim N\left(0.4 \mu_1 + 0.2 \mu_2 + 0.4 \mu_3, \ (0.4)^2 \cdot \frac{\sigma_1^2}{n_1} + (0.2)^2 \cdot \frac{\sigma_2^2}{n_2} + (0.4)^2 \cdot \frac{\sigma_3^2}{n_3}\right)\]

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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, often termed as the Gaussian distribution, is fundamental in statistics. It is described by the bell curve shape and is defined by its mean (\(\mu\)) and variance (\(\sigma^2\)). These two parameters dictate the distribution's location and spread.
  • The mean determines the center of the distribution.
  • The variance measures how spread out the values are around the mean.
The normal distribution is symmetrical, meaning the left and right sides of the curve are mirror images. Most real-world data, like profits from sales, tend to fit this distribution, which is why it is so widely used.
Understanding that each brand's profit (\(X, Y, W\)) in our exercise is normally distributed helps us determine the overall profit by looking at these qualities.
Linear Combination
A linear combination involves combining quantities by multiplying them by coefficients and then adding the products. In statistics, this is crucial when dealing with random variables.
In our context, the average profit (\(U\)) is a linear combination:
  • It uses proportions like 0.4 and 0.2 to weigh the importance of each brand's average profit.
  • \(U = 0.4 \bar{X} + 0.2 \bar{Y} + 0.4 \bar{W}\), where each term represents the proportion of total sales and the average profit from each brand.
By understanding these weights, we can effectively estimate the overall average profit for the dealership. It balances how much each brand contributes based on their presence in total sales.
Probability Density Function
A probability density function (PDF) of a continuous random variable describes the likelihood of different outcomes. In the normal distribution, the PDF follows a bell curve.
For \(U\), a combination of independent normal variables, its PDF is also normal. This is because a linear combination of normal variables remains normal.
  • The PDF for \(U\) is determined by its mean and variance.
  • It gives us insights into the probability of \(U\) taking specific values.
Mathematically, for our exercise: \(U \sim N\left(0.4 \mu_1 + 0.2 \mu_2 + 0.4 \mu_3, \ (0.4)^2 \cdot \frac{\sigma_1^2}{n_1} + (0.2)^2 \cdot \frac{\sigma_2^2}{n_2} + (0.4)^2 \cdot \frac{\sigma_3^2}{n_3}\right)\).
This tells us that, like the variables it combines, \(U\) is also well-behaved in terms of being predictable and analyzable.
Variance Calculation
Variance is a measure of how much values deviate from the mean. For any linear combination of independent variables, calculating variance involves several steps.
  • Each term of the combination contributes to the variance based on its coefficient squared and its own variance.
  • In our example, we consider \(Var(U) = (0.4)^2 \cdot \frac{\sigma_1^2}{n_1} + (0.2)^2 \cdot \frac{\sigma_2^2}{n_2} + (0.4)^2 \cdot \frac{\sigma_3^2}{n_3}\).
  • This formula shows how different brands, through their variance and the number of sales, affect \(U\).
This measure allows us to understand the spread of potential average profits, which is crucial for assessing risk and variability in expected profits from each brand.

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Most popular questions from this chapter

An airline finds that \(5 \%\) of the persons who make reservations on a certain flight do not show up for the flight. If the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

The Environmental Protection Agency is concerned with the problem of setting criteria for the amounts of certain toxic chemicals to be allowed in freshwater lakes and rivers. A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50 (lethal concentration killing 50\% of the test species). In many studies, the values contained in the natural logarithm of LC50 measurements are normally distributed, and, hence, the analysis is based on \(\ln (\mathrm{LC} 50)\) data. Studies of the effects of copper on a certain species of fish (say, species A) show the variance of In(LC50) measurements to be around. 4 with concentration measurements in milligrams per liter. If \(n=10\) studies on \(L C 50\) for copper are to be completed, find the probability that the sample mean of \(\ln (\text { LC50 })\) will differ from the true population mean by no more than .5.

Shear strength measurements for spot welds have been found to have standard deviation 10 pounds per square inch (psi). If 100 test welds are to be measured, what is the approximate probability that the sample mean will be within 1 psi of the true population mean?

The result in Exercise 7.58 holds even if the sample sizes differ. That is, if \(X_{1}, X_{2}, \ldots, X_{n_{1}}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n_{2}}\) constitute independent random samples from populations with means \(\mu_{1}\) and \(\mu_{2}\) and variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), respectively, then \(\bar{X}-\bar{Y}\) will be approximately normally distributed, for large \(n_{1}\) and \(n_{2},\) with mean \(\mu_{1}-\mu_{2}\) and variance \(\left(\sigma_{1}^{2} / n_{1}\right)+\left(\sigma_{2}^{2} / n_{2}\right)\).The flow of water through soil depends on, among other things, the porosity (volume proportion of voids) of the soil. To compare two types of sandy soil, \(n_{1}=50\) measurements are to be taken on the porosity of soil \(\mathrm{A}\) and \(n_{2}=100\) measurements are to be taken on soil \(\mathrm{B}\). Assume that \(\sigma_{1}^{2}=.01\) and \(\sigma_{2}^{2}=.02 .\) Find the probability that the difference between the sample means will be within. 05 unit of the difference between the population means \(\mu_{1}-\mu_{2}\).

One-hour carbon monoxide concentrations in air samples from a large city average 12 ppm (parts per million) with standard deviation 9 ppm. a. Do you think that carbon monoxide concentrations in air samples from this city are normally distributed? Why or why not? b. Find the probability that the average concentration in 100 randomly selected samples will exceed 14 ppm.
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