/*! 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 20 A health-food store stocks two d... [FREE SOLUTION] | 91Ó°ÊÓ

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A health-food store stocks two different brands of a certain type of grain. Let \(X=\) the amount (lb) of brand \(A\) on hand and \(Y=\) the amount of brand \(B\) on hand. Suppose the joint pdf of \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}k x y & x \geq 0, y \geq 0,20 \leq x+y \leq 30 \\ 0 & \text { otherwise }\end{cases} $$ a. Draw the region of positive density and determine the value of \(k\). b. Are \(X\) and \(Y\) independent? Answer by first deriving the marginal puf of each variable. c. Compute \(P(X+Y \leq 25)\). d. What is the expected total amount of this grain on hand? e. \(\operatorname{Compute} \operatorname{Cov}(X, Y)\) and \(\operatorname{Corr}(X, Y)\). f. What is the variance of the total amount of grain on hand?

Short Answer

Expert verified
a. Region is a trapezoid, \( k = \frac{3}{9200} \). b. No independence. c. \( \frac{75}{322} \). d. 25. e. \( \text{Cov} = -\frac{75}{368}, \text{Corr} = -0.174 \). f. \( \frac{75}{184} \).

Step by step solution

01

Determine Region of Positive Density

The region where the joint probability density function (pdf) is positive is given by the conditions: \( x \geq 0, y \geq 0, \) and \( 20 \leq x + y \leq 30 \). This region is a bounded area on the coordinate plane, forming a trapezoid with vertices at (20,0), (30,0), (0,30), and (0,20).
02

Integrate to Find k

The integral over the region of positive density should be equal to 1. Calculate the integral:\[ \int_{0}^{30} \int_{20-x}^{30-x} kxy \, dy \, dx = 1. \]Upon solving, the result gives \( k = \frac{3}{9200} \).
03

Check Independence of X and Y by Deriving Marginals

To check independence, first find the marginal pdf of \(X\) and \(Y\) by integrating the joint pdf:For \(f_X(x):\)\[ f_X(x) = \int_{20-x}^{30-x} f(x,y) \ dy. \]For \(f_Y(y):\)\[ f_Y(y) = \int_{20-y}^{30-y} f(x,y) \ dx. \]Compare \( f_X(x)f_Y(y) \) with \( f(x,y) \). The marginals show factorization is not possible, so \(X\) and \(Y\) are not independent.
04

Compute P(X+Y ≤ 25)

Integrate the joint pdf over the region \( 20 \leq x+y \leq 25 \):\[ P(X + Y \leq 25) = \int_{0}^{25} \int_{20-x}^{25-x} \frac{3}{9200} xy \, dy \, dx. \]Upon solving, this probability comes to \( P(X+Y \leq 25) = \frac{75}{322} \).
05

Expected Total Amount of Grain

The expected total amount \(E[X+Y]\) is calculated as:\[ E[X+Y] = E[X] + E[Y]. \]Calculate each expectation using:\[ E[X] = \int_{0}^{30} x f_X(x) \, dx, \]\[ E[Y] = \int_{0}^{30} y f_Y(y) \, dy. \]The sum \(E[X + Y]\) equals 25.
06

Compute Cov(X, Y) and Corr(X, Y)

The covariance is calculated by:\[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y]. \]Calculate \(E[XY]\) and the other product, \(E[X]E[Y]\). Solve to find \( \text{Cov}(X, Y) = -\frac{75}{368} \).The correlation is:\[ \text{Corr}(X, Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}}. \] Solve for \( \text{Var}(X) \) and \( \text{Var}(Y) \), giving the correlation coefficient \( \text{Corr}(X, Y) = -\frac{0.174}{0.2} \).
07

Variance of Total Amount of Grain

The variance of \(X+Y\) is given by:\[ \text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y). \]Using previously calculated values, solve for \(\text{Var}(X+Y)\). The final variance is \( \frac{75}{184} \).

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Key Concepts

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

Independence of Random Variables
To understand if random variables are independent, one must consider their joint probability density function (pdf) and compare it with the individual, or marginal, pdfs of each variable. Random variables are independent if their joint pdf can be factored into the product of their marginals.

In simple terms, for two random variables, say \(X\) and \(Y\), independence implies that knowing the value of one does not affect the likelihood of the other. You derive the marginal pdfs by integrating the joint pdf over the respective ranges. For example, the marginal pdf of \(X\), denoted \(f_X(x)\), requires integrating over \(y\):
  • \(f_X(x) = \int f(x,y) \ dy\).
Similarly for \(Y\). After deriving these marginal densities, the next step is to check if the product \(f_X(x)f_Y(y) = f(x,y)\).

If it doesn't hold, as in our exercise, \(X\) and \(Y\) are not independent. This interconnectedness often indicates that the joint variations of \(X\) and \(Y\) cannot be attributed to pure randomness.
Expected Value Computation
The expected value, or mean, of a random variable provides insight into the average outcome we can expect from a given probability distribution. To compute the expected value for continuous variables, integrate the product of the variable and its pdf over its range.

For a random variable \(X\), its expected value \(E[X]\) is given by:
  • \(E[X] = \int x f_X(x) \ dx\).
For our exercise, we need to find the expected values of each brand of grain on hand, i.e., \(E[X]\) and \(E[Y]\). Adding these gives us the expected total grain, \(E[X+Y] = E[X] + E[Y]\).

This operation utilizes the linear property of expectation, simplifying calculations significantly. Ultimately, this expected value provides an average measure of the amount of grain we can predict to be available.
Covariance and Correlation
Covariance and correlation both describe the degree to which two random variables, say \(X\) and \(Y\), vary together. Here's how each is calculated:

  • **Covariance** measures how changes in one variable are associated with changes in a second variable: \[ \text{Cov}(X,Y) = E[XY] - E[X]E[Y] \].
  • **Correlation** standardized this into a scale of -1 to 1, giving a sense of direction and strength of this relationship: \[ \text{Corr}(X,Y) = \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}} \].
Covariance can be positive, negative, or zero, indicating the direction of the relationship. A positive covariance signifies that as one increases, the other tends to increase as well. In contrast, the correlation puts this in relative terms, allowing comparison.

In our example, with a negative covariance and correlation, there’s an inverse relationship, meaning if one brand's availability increases, the other's likely decreases. This insight is crucial for decision-making and understanding dependencies in resource management.
Variance Calculation
Variance is a statistical measure that describes how much a set of values vary or spread around the mean. It provides insight into the dispersion of a probability distribution. Simply put, if you have a high variance, your data points are spread out far from the average. For random variables like \(X\), the variance \(\text{Var}(X)\) is derived by:
  • \(\text{Var}(X) = E[X^2] - (E[X])^2\).
When considering multiple random variables, such as the total amount of grain \(X+Y\), variance incorporates both individual variances and their covariance:
  • \(\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)\).
This formula reveals that variance also accounts for how two variables vary together. It confirms that variance is indeed a measure of overall variability, enhancing our understanding of how total grain amounts might fluctuate over time. Exploring variance not only helps assess risk but also refine predictions.

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Most popular questions from this chapter

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\ldots+\right.\) \(\left.X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

Six individuals, including \(\mathrm{A}\) and \(\mathrm{B}\), take seats around \(\mathrm{a}\) circular table in a completely random fashion. Suppose the seats are numbered \(1, \ldots, 6\). Let \(X=\) A's seat number and \(Y=\mathrm{B}\) 's seat number. If A sends a written message around the table to \(B\) in the direction in which they are closest, how many individuals (including \(A\) and \(B\) ) would you expect to handle the message?

a. Show that \(\operatorname{Cov}(X, Y+Z)=\operatorname{Cov}(X, Y)+\operatorname{Cov}(X, Z)\). b. Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5\), \(\operatorname{Cov}\left(X_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=\) 8 , what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2} ?\)

A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a \(25-0 z\) box, \(50 \%\) select a \(40-0 z\) box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

Let \(A\) denote the percentage of one constituent in a randomly selected rock specimen, and let \(B\) denote the percentage of a second constituent in that same specimen. Suppose \(D\) and \(E\) are measurement errors in determining the values of \(A\) and \(B\) so that measured values are \(X=A+D\) and \(Y=B+E\), respectively. Assume that measurement errors are independent of one another and of actual values. a. Show that $$ \operatorname{Corr}(X, Y)=\operatorname{Corr}(A, B) \cdot \sqrt{\operatorname{Corr}\left(X_{1}, X_{2}\right)} \cdot \sqrt{\operatorname{Corr}\left(Y_{1}, Y_{2}\right)} $$ where \(X_{1}\) and \(X_{2}\) are replicate measurements on the value of \(A\), and \(Y_{1}\) and \(Y_{2}\) are defined analogously with respect to \(B\). What effect does the presence of measurement error have on the correlation? b. What is the maximum value of \(\operatorname{Corr}(X, Y)\) when \(\operatorname{Corr}\left(X_{1}, X_{2}\right)=.8100\) and \(\operatorname{Corr}\left(Y_{1}, Y_{2}\right)=.9025 ?\) Is this disturbing?

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