/*! 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 75 Consider Review Exercise 3.79 on... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider Review Exercise 3.79 on page \(105 .\) The random variables \(X\) and \(Y\) represent the number of vehicles that arrive at two separate street corners during a certain 2-minute period in the day. The joint distribution is $$f(x, y)-\frac{1}{4^{(x+y)}} \cdot \frac{9}{16}$$ for \(x=0,1,2, \ldots,\) and \(y=0,1,2, \ldots\) (a) Give \(E(X), E(Y): \operatorname{Var}(X),\) and \(\operatorname{Var}(Y)\). (b) Consider \(Z=X+Y,\) the sum of the two. Find \(E(Z)\) and \(\operatorname{Var}(Z)\).

Short Answer

Expert verified
The solution will yield values for \(E(X)\), \(E(Y)\), \(\operatorname{Var}(X)\), \(\operatorname{Var}(Y)\), \(E(Z)\) and \(\operatorname{Var}(Z)\) based on the calculations performed.

Step by step solution

01

Calculate the expected values E(X) and E(Y).

The expected value of a random variable is a weighted average of all possible values that this random variable can take, with weights equal to the probabilities of these outcomes. For E(X) and E(Y), the formula to calculate is: \[E(X) = \sum x * P(X = x)\] and \[E(Y) = \sum y * P(Y = y)\]. Substituting in the given joint probability function to these equations, will provide the expected values for \(X\) and \(Y\).
02

Calculate the variances Var(X) and Var(Y).

The variance of a random variable is a measure of how much values in the distribution vary on average around the expected value. The formulas for computing Variance of \(X\) and \(Y\) are: \[\operatorname{Var}(X)=E[(X-E(X))^{2}]=E[X^{2}]-E^{2}[X]\], Var(Y)=E[(Y-E(Y))^{2}]=E[Y^{2}]-E^{2}[Y]\] Here, the expected values found in step one will be used along with the given joint probability density.
03

Compute E(Z) and Var(Z).

Here, \(Z = X + Y\). To find \(E(Z)\), we have to use the property of expected values that says the expected value of a sum of random variables is the sum of their expected values: \[E(Z) = E(X+Y) = E(X) + E(Y)\]. For the variance, we must remember that the variance of the sum of two random variables is equal to the sum of their variances only if the two variables are uncorrelated. Assuming \(X\) and \(Y\) are uncorrelated which is typically the case for two independent locations like here, we get: \[\operatorname{Var}(Z) = \operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)\]
04

Final calculation.

Once the equations are set, substitute the necessary values into the equations to get the results.

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Key Concepts

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

Joint Probability
When considering probability distributions, understanding joint probability is essential. It describes the likelihood of two events occurring simultaneously. For two random variables, like our variables \(X\) and \(Y\) representing vehicle arrivals at two intersections, the joint probability distribution provides information about how these variables might interact with each other.
To compute the value of joint probability, you'll need a joint probability function, such as the one provided in the exercise: \[f(x, y) = \frac{1}{4^{(x+y)}} \cdot \frac{9}{16}\] This function evaluates the probability of different combinations of \(X\) and \(Y\). You can use it to build a probability table for precise analysis, enabling calculation of both expected values and variances for each of the variables.
Expected Value
The expected value of a random variable is like the average or mean, providing an idea of its central tendency. Formally, it is the weighted average of all possible values. For randomly arriving vehicles \(X\) and \(Y\), the expected values \(E(X)\) and \(E(Y)\) can be determined using the formula:
  • \(E(X) = \sum x \cdot P(X = x)\)
  • \(E(Y) = \sum y \cdot P(Y = y)\)
These sums involve multiplying each possible outcome by its probability and then adding them all together. The provided function allows you to substitute \(x\) and \(y\) values with their associated probabilities to find these expectations. Once calculated, they give insights into the average number of vehicles expected to arrive at each intersection during the period.
Variance
Variance gives you an idea of how much the values of a random variable spread out from the expected value. It's crucial for determining the unpredictability of the variable. For our variables \(X\) and \(Y\), variance is calculated using the formula:
\[\operatorname{Var}(X) = E[(X - E(X))^2] = E[X^2] - (E[X])^2\]\[\operatorname{Var}(Y) = E[(Y - E(Y))^2] = E[Y^2] - (E[Y])^2\] These expressions calculate how far values typically lie from the mean, squared for positive numbers, providing a sense of variability. To apply, compute each value's square, calculate its expectation, and utilize the previously found expected values.
Random Variables
In probability and statistics, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. Unlike deterministic variables, they do not have fixed values, allowing them to model the randomness of real-life situations such as traffic.Random variables can be discrete or continuous, and the exercise here focuses on discrete random variables \(X\) and \(Y\).
  • \(X\) and \(Y\) represent the number of arrivals in a given period at distinct intersections.
  • The joint probability distribution describes their simultaneous behavior.
Understanding how these random variables work and the impact of their probability distributions is key to analyzing such scenarios, allowing for better predictions and strategic decisions in real-world contexts.

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Most popular questions from this chapter

In a gambling game a woman is paid \(\$ 3\) if she draws a jack or a queen and \(\$ 5\) if she draws a king or an ace from an ordinary deck of 52 playing cards. If she draws any other card, she loses. How much should she pay to play if the game is fair?

Consider a ferry that can carry both buses and cars on a trip across a waterway. Each trip costs the owner approximately \(\$ 10 .\) The fee for cars is \(\$ 3\) and the fee for buses is \(\$ 8 .\) Let \(X\) and \(Y\) denote the number of buses and cars, respectively, carried on a given trip. The joint distribution of \(X\) and \(Y\) is given by $$\begin{array}{cccc} & & x & \\\y & 0 & 1 & 2 \\\\\hline 0 & 0.01 & 0.01 & 0.03 \\\1 & 0.03 & 0.08 & 0.07 \\ 2 & 0.03 & 0.06 & 0.06 \\\3 & 0.07 & 0.07 & 0.13 \\\4 & 0.12 & 0.04 & 0.03 \\\5 & 0.08 & 0.06 & 0.02\end{array}$$ Compute the expected profit for the ferry trip.

In business it is important to plan and carry on research in order to anticipate what will occur at the end of the year. Research suggests that the profit (loss) spectrum is as follows with corresponding probabilities. $$\begin{array}{rc}\text { Profit } & \text { Probability } \\\\\hline-\$ 15,000 & 0.05 \\\\\$ 0 & 0.15 \\\\\$ 15,000 & 0.15 \\\S 25,000 & 0.30 \\\\\$ 40,000 & 0.15 \\\\\$ 50.000 & 0.10 \\\\\$ 100.000 & 0.05 \\\\\$ 150.000 & 0.03 \\\\\$ 200,000 & 0.02\end{array}$$ (a) What is the expected profit? (b) Give the standard deviation of the profit.

A convenience store has two separate locations in the store where customers can be: checked out as they leave. These locations both have: two cash registers and have two employees that check out customers. Let \(X\) be the number of cash registers being used at a particular time for location \(L\) and \(Y\) the number being used at the same: time period for location \(2 .\) The joint probability function is given by $$\begin{array}{cccc} & {\mathrm{y}} \\\\\mathrm{x} & 0 & 1 & 2 \\\\\hline 0 & 0.12 & 0.04 & 0.04 \\\1 & 0.08 & 0.19 & 0.05 \\\2 & 0.06 & 0.12 & 0.30\end{array}$$ (a) Give the marginal density of both \(X\) and \(Y\) as well as the probability distribution of A" given \(V=2\). (b) Give \(E(X)\) and \(\operatorname{Var}(X)\). (c) Give \(E(X \mid Y=2)\) and \(\operatorname{Var}(X \mid Y=2)\).

Consider Review Exercise 3.66 on page 104 . There are two service lines. The random variables \(X\) and \(Y\) are the proportion of time that line 1 and line 2 are in use respectively. The joint probability density function for \((\mathrm{X}, Y)\) is given by $$f(x, y)=\left\\{\begin{array}{l}\frac{3}{2}\left(x^{2}+y^{2}\right), \\\0\end{array}\right.,$$ $$0 \leq x, y \leq 1$$ (a) Determine whether or not \(X\) and \(Y\) are independent. (b) It is of interest to know something about, the proportion of \(Z=X 4-Y .\) the sum of the two proportions. Find \(E(X+Y)\) Also find \(E(X Y)\). (c) Find \(\operatorname{Var}(X), \operatorname{Var}(Y),\) and \(\operatorname{Cov}(X, Y)\) (d) Find \(\operatorname{Var}(X+Y)\).

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