/*! 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} Q143E Lead in metal shredder residue. ... [FREE SOLUTION] | 91影视

91影视

Lead in metal shredder residue. On the basis of data collectedfrom metal shredders across the nation, the amount xof extractable lead in metal shredder residue has an approximateexponential distribution with mean= 2.5 milligramsper liter (Florida Shredder鈥檚 Association).

a. Find the probability that xis greater than 2 milligramsper liter.

b. Find the probability that xis less than 5 milligrams perliter.

Short Answer

Expert verified

a.Px>2=.0068

b.Px<5=0.9999

Step by step solution

01

Given information

X is an exponential random variable and=2.5 .

02

Define the probability density function

The p.d.f of X is

fx=e-x;x0=2.5e-2.5x

03

Calculate P(x>2)

a

Px>2=1-Px2=1-022.5-2.5xdx=1-2.502e-2.5xdx=1-525-2e-552=1-e-5e-5-1=1-0.9932=0.0068

Hence, the probability of x that x is greater than 2 milligrams per liter is 0.0068.

04

Calculate P(x<5)

b

Px<5=052.5e-2.5xdx=2.505e-2.5xdx=525-2e-25252=e-252e-252-1=0.9999

Hence, the probability of x that x is less than 5 milligrams per liter is 0.9999.

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

Consider the probability distributions shown here:

  1. Use your intuition to find the mean for each distribution. How did you arrive at your choice?
  2. Which distribution appears to be more variable? Why?
  3. Calculate渭鈥塧苍诲鈥壪2 for each distribution. Compare these answers with your answers in parts a and b.

Suppose x is a binomial random variable with n = 3 and p = .3.

  1. Calculate the value of p(x),role="math" localid="1653657859012" x=0,1,2,3,using the formula for a binomial probability distribution.
  2. Using your answers to part a, give the probability distribution for x in tabular form.

Cell phone handoff behavior. Refer to the Journal of Engineering, Computing and Architecture (Vol. 3., 2009) study of cell phone handoff behavior, Exercise 3.47 (p. 183). Recall that a 鈥渉andoff鈥 describes the process of a cell phone moving from one base channel (identified by a color code) to another. During a particular driving trip, a cell phone changed channels (color codes) 85 times. Color code 鈥渂鈥 was accessed 40 times on the trip. You randomly select 7 of the 85 handoffs. How likely is it that the cell phone accessed color code 鈥渂鈥 only twice for these 7 handoffs?

LASIK surgery complications. According to studies, 1% of all patients who undergo laser surgery (i.e., LASIK) to correct their vision have serious post laser vision problems (All About Vision, 2012). In a sample of 100,000 patients, what is the approximate probability that fewer than 950 will experience serious post laser vision problems?

185 Software file updates. Software configuration management was used to monitor a software engineering team鈥檚 performance at Motorola, Inc. (Software Quality Professional, Nov. 2004). One of the variables of interest was the number of updates to a file that was changed because of a problem report. Summary statistics forn=421 n = 421 files yielded the following results: role="math" localid="1658219642985" x=4.71,s=6.09, QL=1, andQU=6 . Are these data approximately normally distributed? Explain.

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.