IfXis uniformly distributed over (-1,1),find

(a)P{X>12}

(b)the density function of the random variableX.

Short Answer

Expert verified

Therefore,

(a)PX>12=12

(b)We have found thatY=X~UNIF(0,1)

Step by step solution

01

Given information:

If Xis uniformly distributed over(-1,1)

02

Part (a) step 2 Explanation:

We have that X~UNIF(-1,1)and wish to calculate

PX>12=P-12>X>12=PX<-12X>12=PX<-12+PX>12=-11212dx+12112dx=14+14=12

03

Part (b) step 3 Explanation:

The CDF ofY=X, where

FxX=PXx=P-xXx=-xx12dt=12t-xx=2x2=x

for allx(0,1). We then differentiate the CDF to obtain the PDF asfx(x)=ddxF(X)=ddx(x)=1for x(0,1). Therefore, we have found that Y=X~UNIF(0,1).

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 Vaia!

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

Let X and Y be independent random variables that are both equally likely to be either 1, 2, . . . ,(10)N, where N is very large. Let D denote the greatest common divisor of X and Y, and let Q k = P{D = k}.

(a) Give a heuristic argument that Q k = 1 k2 Q1. Hint: Note that in order for D to equal k, k must divide both X and Y and also X/k, and Y/k must be relatively prime. (That is, X/k, and Y/k must have a greatest common divisor equal to 1.) (b) Use part (a) to show that Q1 = P{X and Y are relatively prime} = 1 q k=1 1/k2 It is a well-known identity that !q 1 1/k2 = π2/6, so Q1 = 6/π2. (In number theory, this is known as the Legendre theorem.) (c) Now argue that Q1 = "q i=1  P2 i − 1 P2 i  where Pi is the smallest prime greater than 1. Hint: X and Y will be relatively prime if they have no common prime factors. Hence, from part (b), we see that Problem 11 of Chapter 4 is that X and Y are relatively prime if XY has no multiple prime factors.)

Every day Jo practices her tennis serve by continually serving until she has had a total of 50successful serves. If each of her serves is, independently of previous ones,

successful with probability .4, approximately what is the probability that she will need more than 100serves to accomplish her goal?

Hint: Imagine even if Jo is successful that she continues to serve until she has served exactly 100 times. What must be true about her first 100 serves if she is to reach her goal?

The life of a certain type of automobile tire is normally distributed with mean 34,000miles and standard deviation 4,000miles.

(a) What is the probability that such a tire lasts more than 40,000miles?

(b) What is the probability that it lasts between 30,000and35,000miles?

(c) Given that it has survived 30,000miles, what is the conditional probability that the tire survives another 10,000miles?

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. If you continually make such bets, approximate the probability that

(a) you are winning after 34 bets;

(b) you are winning after 1000 bets;

(c) you are winning after 100,000 bets

Assume that each roll of the roulette ball is equally likely to land on any of the 38 numbers

You arrive at a bus stop at 10a.m., knowing that the bus will arrive at some time uniformly distributed between 10and 10:30.

(a) What is the probability that you will have to wait longer than 10minutes?

(b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free