Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 6.3, Problem 1E

Let Y be a random variable with probability density function given by

f ( y ) = { 2 ( 1 y ) , 0 0 1 , 0 , elsewhere .

a Find the density function of U1 = 2Y − 1.

b Find the density function of U2 = 1 − 2Y.

c Find the density function of U3 = Y2.

d Find E(U1), E(U2), and E(U3) by using the derived density functions for these random variables.

e Find E(U1), E(U2), and E(U3) by the methods of Chapter 4.

a.

Expert Solution
Check Mark
To determine

Find the density function for U1=2Y1.

Answer to Problem 1E

The density function for U1=2Y1 is fU1(u)=1u21u1.

Explanation of Solution

Calculation:

From the given information, the probability density function for Y is f(y)=2(1y),0y1.

The distribution function for Y is,

FY(y)=0yf(t)dt=0y2(1t)dt=2[tt22]0y=2yy2

From the given information, the random variable U1 is defined as U1=2Y1.

Consider the distribution function for U1,

FU1(u1)=P(U1u)=P(2Y1u)=P(Yu+12)=2(u+12)(u+12)2

Limits for the random variable U1:

The range for the random variable Y is from  0 to 1 and U1=2Y1.

For Y=  0, the value of U1 is  ̶ 1.

For Y=1, the value of U1 is 1.

Hence, the range for the random variable U1 is from  ̶ 1 to 1.

The probability density function for U1 is,

fU1(u)=FU1(u)=ddu(2(u+12)(u+12)2)=(2×12)(2×u+12×12)=1u+12

             =1u2,    1u1

b.

Expert Solution
Check Mark
To determine

Find the density function for U2=12Y.

Answer to Problem 1E

The density function for U2=12Y is fU2(u)=1+u2,1u1.

Explanation of Solution

Calculation:

From the given information, U2=12Y.

Consider the distribution function for U2,

FU2(u)=P(U2u)=P(12Yu)=P(2Yu1)=P(Y>1u2)

             =1F(1u2)=1[2(1u2)(1u2)2]=1[1u(1u2)2]=u+(1u2)2

Limits for the random variable U2:

The range for the random variable Y is from  0 to 1 and U2=12Y.

For Y= 0, the value of U2 is 1.

For Y=1, the value of U2 is  ̶ 1.

Hence, the range for the random variable U2 is from  ̶ 1  to 1.

The probability density function for U2 is,

fU2(u)=FU2(u)=ddu(u+(1u2)2)=1+(21u2×12)=1+u2,1u1

c.

Expert Solution
Check Mark
To determine

Find the density function for U3=Y2.

Answer to Problem 1E

The density function for U3=Y2 is fU3(u)=1u1,0u1.

Explanation of Solution

Calculation:

From the given information, U3=Y2.

Consider the distribution function for U3,

FU3(u)=P(U3u)=P(Y2u)=P(Yu)=F(u)

             =2u(u)2=2uu

The probability density function for U3 is,

fU3(u)=FU3(u)=ddu(2uu)=1u1,0u1

d.

Expert Solution
Check Mark
To determine

Find the value of E(U1) by using the derived density function of U1.

Find the value of E(U2) by using the derived density function of U2.

Find the value of E(U3) by using the derived density function of U3.

Answer to Problem 1E

The value of E(U1) is 13.

The value of E(U2) is 13.

The value of E(U3)  is 16.

Explanation of Solution

Calculation:

The density function for U1 is fU1(u)=1u21u1

Consider,

E(U1)=11ufU1(u)du=11u×1u2du=12[u22u33]11=12×23

             =13

The density function for U2 is fU2(u)=1+u2,1u1.

Consider,

E(U2)=11ufU2(u)du=11u×1+u2du=12[u22+u33]11=12×23

             =13

The density function for U3 is fU3(u)=1u1,0u1.

Consider,

E(U3)=01ufU3(u)du=01u×(1u1)du=[u32(32)u22]01=2312

              =16

e.

Expert Solution
Check Mark
To determine

Find the value of E(U1) by the methods of Chapter 4.

Find the value of E(U2) by the methods of Chapter 4.

Find the value of E(U3) by the methods of Chapter 4.

Answer to Problem 1E

The value of E(U1) is 13.

The value of E(U2) is 13.

The value of E(U3)  is 16.

Explanation of Solution

Calculation:

Result:

Let X be the random variable, then E(aX+b)=aE(X)+b, where a and b are constants.

The density function for Y is f(y)=2(1y),0y1.

Consider,

E(y)=01yf(y)dy=01y×2(1y)dy=2[y22y33]01=13

E(U1)=E(2Y1)=2E(Y)1=2×131=13

E(U2)=E(12Y)=12E(Y)=12×13=13

Consider,

E(y2)=01y2f(y)dy=01y2×2(1y)dy=2[y33y44]01=16

E(U3)=E(Y2)=16

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO
Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?
A normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…

Chapter 6 Solutions

Mathematical Statistics with Applications

Ch. 6.3 - Suppose that two electronic components in the...Ch. 6.3 - Prob. 12ECh. 6.3 - If Y1 and Y2 are independent exponential random...Ch. 6.3 - Prob. 14ECh. 6.3 - Prob. 15ECh. 6.3 - Prob. 16ECh. 6.3 - Prob. 17ECh. 6.3 - A member of the Pareto family of distributions...Ch. 6.3 - Prob. 19ECh. 6.3 - Let the random variable Y possess a uniform...Ch. 6.3 - Prob. 21ECh. 6.4 - Prob. 23ECh. 6.4 - In Exercise 6.4, we considered a random variable Y...Ch. 6.4 - Prob. 25ECh. 6.4 - Prob. 26ECh. 6.4 - Prob. 27ECh. 6.4 - Let Y have a uniform (0, 1) distribution. Show...Ch. 6.4 - Prob. 29ECh. 6.4 - A fluctuating electric current I may be considered...Ch. 6.4 - The joint distribution for the length of life of...Ch. 6.4 - Prob. 32ECh. 6.4 - The proportion of impurities in certain ore...Ch. 6.4 - A density function sometimes used by engineers to...Ch. 6.4 - Prob. 35ECh. 6.4 - Refer to Exercise 6.34. Let Y1 and Y2 be...Ch. 6.5 - Let Y1, Y2,, Yn be independent and identically...Ch. 6.5 - Let Y1 and Y2 be independent random variables with...Ch. 6.5 - Prob. 39ECh. 6.5 - Prob. 40ECh. 6.5 - Prob. 41ECh. 6.5 - A type of elevator has a maximum weight capacity...Ch. 6.5 - Prob. 43ECh. 6.5 - Prob. 44ECh. 6.5 - The manager of a construction job needs to figure...Ch. 6.5 - Suppose that Y has a gamma distribution with =...Ch. 6.5 - A random variable Y has a gamma distribution with ...Ch. 6.5 - Prob. 48ECh. 6.5 - Let Y1 be a binomial random variable with n1...Ch. 6.5 - Let Y be a binomial random variable with n trials...Ch. 6.5 - Prob. 51ECh. 6.5 - Prob. 52ECh. 6.5 - Let Y1,Y2,,Yn be independent binomial random...Ch. 6.5 - Prob. 54ECh. 6.5 - Customers arrive at a department store checkout...Ch. 6.5 - The length of time necessary to tune up a car is...Ch. 6.5 - Prob. 57ECh. 6.5 - Prob. 58ECh. 6.5 - Prob. 59ECh. 6.5 - Prob. 60ECh. 6.5 - Prob. 61ECh. 6.5 - Prob. 62ECh. 6.6 - In Example 6.14, Y1 and Y2 were independent...Ch. 6.6 - Refer to Exercise 6.63 and Example 6.14. Suppose...Ch. 6.6 - Prob. 65ECh. 6.6 - Prob. 66ECh. 6.6 - Prob. 67ECh. 6.6 - Prob. 68ECh. 6.6 - Prob. 71ECh. 6 - Let Y1 and Y2 be independent and uniformly...Ch. 6 - As in Exercise 6.72, let Y1 and Y2 be independent...Ch. 6 - Let Y1, Y2,, Yn be independent, uniformly...Ch. 6 - Prob. 75SECh. 6 - Prob. 76SECh. 6 - Prob. 77SECh. 6 - Prob. 78SECh. 6 - Refer to Exercise 6.77. If Y1,Y2,,Yn are...Ch. 6 - Prob. 80SECh. 6 - Let Y1, Y2,, Yn be independent, exponentially...Ch. 6 - Prob. 82SECh. 6 - Prob. 83SECh. 6 - Prob. 84SECh. 6 - Let Y1 and Y2 be independent and uniformly...Ch. 6 - Prob. 86SECh. 6 - Prob. 87SECh. 6 - Prob. 88SECh. 6 - Let Y1, Y2, . . . , Yn denote a random sample from...Ch. 6 - Prob. 90SECh. 6 - Prob. 91SECh. 6 - Prob. 92SECh. 6 - Prob. 93SECh. 6 - Prob. 94SECh. 6 - Prob. 96SECh. 6 - Prob. 97SECh. 6 - Prob. 98SECh. 6 - Prob. 99SECh. 6 - The time until failure of an electronic device has...Ch. 6 - Prob. 101SECh. 6 - Prob. 103SECh. 6 - Prob. 104SECh. 6 - Prob. 105SECh. 6 - Prob. 106SECh. 6 - Prob. 107SECh. 6 - Prob. 108SECh. 6 - Prob. 109SECh. 6 - Prob. 110SECh. 6 - Prob. 111SECh. 6 - Prob. 112SECh. 6 - Prob. 113SECh. 6 - Prob. 114SECh. 6 - Prob. 115SECh. 6 - Prob. 116SE
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Linear Algebra: A Modern Introduction
Algebra
ISBN:9781285463247
Author:David Poole
Publisher:Cengage Learning
Continuous Probability Distributions - Basic Introduction; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=QxqxdQ_g2uw;License: Standard YouTube License, CC-BY
Probability Density Function (p.d.f.) Finding k (Part 1) | ExamSolutions; Author: ExamSolutions;https://www.youtube.com/watch?v=RsuS2ehsTDM;License: Standard YouTube License, CC-BY
Find the value of k so that the Function is a Probability Density Function; Author: The Math Sorcerer;https://www.youtube.com/watch?v=QqoCZWrVnbA;License: Standard Youtube License