/*! 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 33 If \(E[X]=1\) and \(\operatornam... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 33

If \(E[X]=1\) and \(\operatorname{Var}(X)=5,\) find (a) \(E\left[(2+X)^{2}\right]\) (b) \(\operatorname{Var}(4+3 X)\)

Short Answer

Expert verified
(a) \(E\left[(2+X)^{2}\right] = 14\) (b) \(\operatorname{Var}(4+3X) = 45\)

Step by step solution

01

(a) Calculate E[((2+X)^2)] #

We start by expanding the expression within the expected value operator: \[ (2+X)^2 = 4 + 4X + X^2 \] Now, we can use the linearity of expected value, that is: \[ E[a + b\,Y] = a + b\,E[Y] \] Applying this property to our problem: \[ E[(2+X)^2] = E[4+4X+X^2] = E[4] + 4E[X] + E[X^2] \] We know that \(E[X]=1\). To find \(E[X^2]\), we will use the relation between variance and second moment: \[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 \Rightarrow E[X^2] = \operatorname{Var}(X) + (E[X])^2 \] Now, we can substitute the given values and find \(E[X^2]\): \[ E[X^2] = 5 + 1^2 = 6 \] Finally, we can find \(E[(2+X)^2]\): \[ E[(2+X)^2] = E[4] + 4E[X] + E[X^2] = 4 + 4(1) + 6 = 14 \]
02

(b) Calculate Var(4+3X) #

We will use the following property of variance, for a transformation of a random variable \(Y = a + bX\): \[ \operatorname{Var}(a + b\,X) = b^2 \operatorname{Var}(X) \] In our case, we have \(a = 4\) and \(b = 3\), so we can find \(\operatorname{Var}(4+3X)\): \[ \operatorname{Var}(4+3X) = 3^2 \operatorname{Var}(X) = 9(5) = 45 \] So the final answers are: (a) \(E\left[(2+X)^2\right] = 14\). (b) \(\operatorname{Var}(4+3X) = 45\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Cards from an ordinary deck are turned face up one at a time. Compute the expected number of cards that need to be turned face up in order to obtain (a) 2 aces; (b) 5 spades; (c) all 13 hearts.

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days' travel. The second leads to a tunnel that returns him to his cell after 4 days' travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors \(1,2,\) and 3 with respective probabilities \(.5,3,\) and \(.2,\) what is the expected number of days until the prisoner reaches freedom?

Let \(Z\) be a standard normal random variable, and, for a fixed \(x,\) set $$ X=\left\\{\begin{array}{ll} Z & \text { if } Z>x \\ 0 & \text { otherwise } \end{array}\right. $$ Show that \(E[X]=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\)

If 101 items are distributed among 10 boxes, then at least one of the boxes must contain more than 10 items. Use the probabilistic method to prove this result.

A group of 20 people consisting of 10 men and 10 women is randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consist of 10 married couples. Compute the mean and variance of the number of married couples that are paired together.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.