Suppose in Problem 8.14that the variance of the number of automobiles sold weekly is9.

(a)Give a lower bound to the probability that next week’s sales are between 10and22, inclusively.

(b)Give an upper bound to the probability that next week’s sales exceed18.

Short Answer

Expert verified

(a)the probability is at least .75

(b)P{X>18}913=.69.

Step by step solution

01

Given Information.

The variance of the number of automobiles sold weekly is9.

02

part (a) Explanation.

Let the random variable Xrepresents the number of automobiles sold (in a week). It is given that E[X]=16andσ2=Var(X)=9.

(a)By Chebyshev's inequality,

P{|X-16|6}σ262=936=14.

SinceP{10X22}=P{|X-16|6}we get:

P{10X22}=1-P{|X-16|6}34=.75.

03

Part (b) Explanation.

(b)Using Corollary 5.1.(textbook) fora=2¯we get:

P{X>16+2=18}σ2σ2+22=99+4=913=.69.

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

We have 100components that we will put to use in a sequential fashion. That is, the component 1is initially put in use, and upon failure, it is replaced by a component2, which is itself replaced upon failure by a componentlocalid="1649784865723" 3, and so on. If the lifetime of component i is exponentially distributed with a mean 10+i/10,i=1,...,100estimate the probability that the total life of all components will exceed1200. Now repeat when the life distribution of component i is uniformly distributed over(0,20+i/5),i=1,...,100.

Let Xbe a non-negative random variable. Prove that

E[X]EX21/2EX31/3

A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2000 hours is at least 0.95?

Let X1,X2,be a sequence of independent and identically distributed random variables with distributionF, having a finite mean and variance. Whereas the central limit theorem states that the distribution ofi=1nXiapproaches a normal distribution as ngoes to infinity, it gives us no information about how largenneed to be before the normal becomes a good approximation. Whereas in most applications, the approximation yields good results whenevern20, and oftentimes for much smaller values ofn, how large a value of nis needed depends on the distribution ofXi. Give an example of distribution Fsuch that the distributioni=1100Xiis not close to a normal distribution.

Hint: Think Poisson.

8.8. On each bet, a gambler loses 1 with probability.7, loses 2with probability .2, or wins 10with probability .1. Approximate the probability that the gambler will be losing after his first 100 bets.

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