/*! 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 81 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 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 \(f(x, y)\) \(=\left\\{\begin{array}{cl}k x y & x \geq 0, y \geq 0,20 \leq x+y \leq 30 \\\ 0 & \text { otherwise }\end{array}\right.\) 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 pdf of each variable. c. Compute \(P(X+Y \leq 25)\). d. What is the expected total amount of this grain on hand? e. 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
1) Draw region, find trapezoid area. 2) Find k: \(\frac{3}{98000}\). 3) Independent? No. 4) \(P(X+Y \leq 25) = \frac{99}{560}\). 5) \(E(X+Y) = 25\). 6) \(Cov(X,Y) = -\frac{25}{144}\), \(Corr = -0.21\). 7) \(Var(X+Y) = \frac{1917}{560}\).

Step by step solution

01

Draw the Region of Positive Density

The joint pdf is positive where both constraints are satisfied: \(x \geq 0, y \geq 0,\) and \(20 \leq x+y \leq 30\). Geometrically, this region is a band between the lines \(x+y=20\) and \(x+y=30\) in the first quadrant. It forms a trapezoid with corners at \((20, 0), (30, 0), (0, 30),\) and \((0, 20)\).
02

Determine the Value of k

We need \(\int_0^{30} \int_0^{30-x} kxy\, dy \, dx = 1\) over the region of positive density. Split the integral using the constraints: first, integrate \(y\) from \(20-x\) to \(30-x\), then \(x\) from \(0\) to \(20\) covering the entire region. Solving gives \(k = \frac{3}{98000}\).
03

Derive Marginal PDF of X

The marginal pdf of \(X\) is \(f_X(x) = \int_{20-x}^{30-x} kxy\, dy\). Solving the integral gives: \(f_X(x) = \frac{k}{2}(30-x)^2 - \frac{k}{2}(20-x)^2\), for \(0 \leq x \leq 10\).
04

Derive Marginal PDF of Y

The marginal pdf of \(Y\) is \(f_Y(y) = \int_{20-y}^{30-y} kxy\, dx\). Solving the integral gives: \(f_Y(y) = \frac{k}{2}(30-y)^2 - \frac{k}{2}(20-y)^2\), for \(0 \leq y \leq 10\).
05

Determine Independence of X and Y

X and Y are independent if their joint PDF equals the product of their marginals: \(f(x, y) = f_X(x) f_Y(y)\). Checking the expressions demonstrates dependence, as this equality does not hold.
06

Compute P(X+Y ≤ 25)

Calculate \(P(X+Y \leq 25) = \int_{x+y \leq 25} kxy\, dA \). Change variables to set bounds for x and y, then integrate: \(x\) from \(0\) to \(5\), \(y\) from \(20-x\) to \(25-x\), yielding \(P = \frac{99}{560}\).
07

Compute Expected Total Amount

The expected total amount \(E(X+Y) = E(X) + E(Y)\). Find expected values \(E(X) = \int_0^{10} xf_X(x)\, dx\) and \(E(Y) = \int_0^{10} yf_Y(y)\, dy\). Combine for \(E(X+Y) = 25\).
08

Compute Covariance and Correlation

Cov(X, Y) = E(XY) - E(X)E(Y). Calculate E(XY) using the joint PDF: \(E(XY) = \int \int xykxy\, dxdy\) over the range, yielding \(Cov(X, Y) = -\frac{25}{144}\). The correlation coefficient is Cov(X, Y) divided by \( \sqrt{Var(X)Var(Y)} \).
09

Find Variance of Total Amount

Using var formulas: Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y). Pre-calculate variances via their definitions, then combine to find \(Var(X+Y)\). Simplify to find \(\frac{1917}{560}\).

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Key Concepts

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

Independent Variables
Independence of random variables is a core concept in probability theory. Two variables are said to be independent if the occurrence or value of one does not affect the occurrence or value of the other.
In terms of probability density functions, this means the joint probability density function (pdf) of the two variables is equal to the product of their marginal pdfs.
For two variables, say \(X\) and \(Y\), the criterion for independence is expressed as:
  • \[ f(x, y) = f_X(x) \cdot f_Y(y) \]

  • This equation holds true when each event's probability does not influence the probability of the other.
In our problem with grain brands, we see that the expression of the joint pdf \(f(x, y)\) does not match the product of marginals \(f_X(x)\) and \(f_Y(y)\). Hence, \(X\) and \(Y\) are not independent, as their dependency can be directly derived from the inequality of these expressions.
Dependency in this context means a certain amount of one brand affects the availability of the other, either due to constraints or shared resources in a real-world application.
Marginal Probability Density Function
The marginal probability density function provides insight into the distribution of one variable irrespective of the other. It's a way to focus on a single variable within the structure of a joint distribution.
To derive the marginal pdf for a variable, you integrate the joint pdf over the entire range of the other variable. In formulaic terms, the marginal pdf of \(X\) from a joint pdf \(f(x, y)\) is:
  • \[ f_X(x) = \int f(x, y) \, dy \]
Similarly, the marginal pdf of \(Y\) is obtained by integrating \(f(x, y)\) with respect to \(x\):
  • \[ f_Y(y) = \int f(x, y) \, dx \]
In our exercise related to the grain stores, the calculation of marginals for \(X\) and \(Y\) gives insight into the individual distributions without considering the other variable. It's particularly useful in scenarios where one might want to assess risk or probability tied to a single variable independent of context.
Covariance and Correlation
Covariance is a measure that indicates the extent to which two variables change together. If the greater values of one variable tend to correspond with the greater values of another variable, the covariance is positive. Conversely, if greater values of one variable correspond with lesser values of another, the covariance is negative.
The formula for covariance \(Cov(X, Y)\) is:
  • \[ Cov(X, Y) = E(XY) - E(X)E(Y) \]
The correlation, meanwhile, is a normalized form of covariance that measures the strength and direction of a linear relationship between two variables. It is given by:
  • \[ \text{Corr}(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}} \]
In the grain inventory problem, the negative covariance indicates that as the amount of one brand increases, the other tends to decrease, reflecting a specific dependency between them. The correlation, being a standardized measure, confirms the strength and direction of this relationship.
Expected Value
The expected value, often referred to as the mean, is a fundamental concept representing the average outcome of a random variable if repeated trials are conducted. It is a measure of the center of the distribution and gives insight into the typical value you can expect from your random variable.
Mathematically, it's defined for a continuous random variable \(X\) as:
  • \[ E(X) = \int x f_X(x) \, dx \]
In multi-variable scenarios, the expected value of the sum of variables, like \(E(X + Y)\), can be straightforwardly calculated by summing their individual expected values. This is expressed as:
  • \[ E(X + Y) = E(X) + E(Y) \]
In our exercise with brand A and B grains, calculating the expected total amount shows us not only an average stock level but also assists in planning and decision-making by giving a single expected quantity over time. This value is crucial for inventory management and resource allocation in practical scenarios.

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Most popular questions from this chapter

Consider a sample of size \(n=3\) from the standard normal distribution, and obtain the expected value of the largest order statistic. What does this say about the expected value of the largest order statistic in a sample of this size from any normal distribution? [Hint: With \(\phi(x)\) denoting the standard normal pdf, use the fact that \((d / d x) \phi(x)=-x \phi(x)\) along with integration by parts.]

Suppose two identical components are connected in parallel, so the system continues to function as long as at least one of the components does so. The two lifetimes are independent of each other, each having an exponential distribution with mean \(1000 \mathrm{~h}\). Let \(W\) denote system lifetime. Obtain the moment generating function of \(W\), and use it to calculate the expected lifetime.

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda .\) a. What is the probability that all ten bulbs fail before time \(t\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(t\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

A store is expecting \(n\) deliveries between the hours of noon and 1 p.m. Suppose the arrival time of each delivery truck is uniformly distributed on this one-hour interval and that the times are independent of each other. What are the expected values of the ordered arrival times?

Two messages are to be sent. The time (min) necessary to send each message has an exponential distribution with parameter \(\lambda=1\), and the two times are independent of each other. It costs \(\$ 2\) per minute to send the first message and \(\$ 1\) per minute to send the second. Obtain the density function of the total cost of sending the two messages. [Hint: First obtain the cumulative distribution function of the total cost, which involves integrating the joint pdf.]

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