/*! 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} Q. 8.6 8.6 . In Self-Test Problem 8.5, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

8.6 . In Self-Test Problem 8.5, how many components would one need to have on hand to be approximately 90percent certain that the stock would last at least 35days?

Short Answer

Expert verified

The components is n=56.

Step by step solution

01

Given information

The components need to have on hand, approximately 90percent. The stock would last at least 35days.

02

Explanation

Let Xidenote the lifetime of ith component. Understand that X1,X2,… is a sequence of independent and identically distributed random variables, each having a mean.
μ=EXi=∫01xf(x)dx=∫012x2dx=2x3301=23
Then the variance is,
σ2=VarXi=∫01x2f(x)dx-232=∫012x3dx-49=2x4401-49=118
LetSn=∑i=1nXiindicates the time of nth failure.

Verify that the value of nis satisfied.
PSn≥35≈.90?

03

Explanation

Using the central limit theorem:

.90≈PSn≥35=1−PSn≤35=1−PSn−nμσn≤35−nμσn≈≈1-Φ35-nμσn⇒Φ35-nμσn≈.10⇒35-nμσn≈Φ-1(.10)=-1.28

Then,

18(105−2n)3n≈−1.28n=56

Hence, the components n=56

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 91Ó°ÊÓ!

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

A.J. has 20jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of50minutes and a standard deviation of10minutes. M.J. has 20jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of52minutes and a standard deviation of 15minutes.

(a)Find the probability that A.J. finishes in less than 900minutes.

(b)Find the probability that M.J. finishes in less than900minutes.

(c)Find the probability that A.J. finishes before M.J.

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.

ItXhas, a meanμand standard deviationσ, the ratior=|μ|/σis called the measurement signal-to-noise ratioX. The idea is that Xcan be expressed asX=μ+(X−μ), μrepresenting the signal and X−μthe noise. If we define|(X−μ)/μ|=Dit as the relative deviation Xfrom its signal (or mean)μ, show that forα>0,

P{D≤α}≥1−1r2α2.

P{D≤α}≥1−1r2α2

A die is continually rolled until the total sum of all rolls exceeds 300. Approximate the probability that at least 80 rolls are necessary.

A tobacco company claims that the amount of nicotine in one of its cigarettes is a random variable with a mean of 2.2mg and a standard deviation of 0.3mg. However, the average nicotine content of 100randomly chosen cigarettes was 3.1mg. Approximate the probability that the average would have been as high as or higher than 3.1 if the company’s claims were true

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.