Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Q79E (page 247)

Flaws in the plastic-coated wire. The British Columbia Institute of Technology provides on its Web site (www.math.bcit.ca) practical applications of statistics at mechanical engineering firms. The following is a Poisson application. A roll of plastic-coated wire has an average of .8 flaws per 4-meter length of wire. Suppose a quality-control engineer will sample a 4-meter length of wire from a roll of wire 220 meters in length. If no flaws are found in the sample, the engineer will accept the entire roll of wire. What is the probability that the roll will be rejected? What assumption did you make to find this probability?

Short Answer

Expert verified

The probability that the roll will be rejected is 0.5507.

Step by step solution

01

Given information

There is a roll of plastic-coated wire that has an average of 0.8 flaws per 4-meter length of wire.

x be the number of flaws in a 4-meter length of wire

02

Finding the probability

x  ~  Poissonλwhereλ=0.8

The probability mass function of x is

Px=e-λλxx!

Roll will be rejected if there is at least one flaw in the sample of a 4-meter length of wire.

The probability is,

Px⩾1=1-Px=0=1-e-0.80.800!=1-0.4493=0.5507

Px⩾1=0.5507

Therefore, the probability that the roll will be rejected is 0.5507.

Since the process has a Poisson distribution, we have to assume that the flaws are randomly distributed, and the 4-meter length of sample wire represents the entire roll.

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.

Start your free trial

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

Blood diamonds. According to Global Research News (March 4, 2014), one-fourth of all rough diamonds produced in the world are blood diamonds, i.e., diamonds mined to finance war or an insurgency. (See Exercise 3.81, p. 200.) In a random sample of 700 rough diamonds purchased by a diamond buyer, let x be the number that are blood diamonds.

a. Find the mean of x.

b. Find the standard deviation of x.

c. Find the z-score for the value x = 200.

d. Find the approximate probability that the number of the 700 rough diamonds that are blood diamonds is less than or equal to 200.

4.113 Credit/debit card market shares. The following table reports the U.S. credit/debit card industry’s market share data for 2015. A random sample of 100 credit/debit card users is to be questioned regarding their satisfaction with

their card company. For simplification, assume that each card user carries just one card and that the market share percentages are the percentages of all card customers that carry each brand.

Credit debit Card

Market Share %

Visa

59

MasterCard

26

Discover

2

American Express

13

Source:Based on Nilson Reportdata, June 2015.

a. Propose a procedure for randomly selecting the 100 card users.

b. For random samples of 100 card users, what is the expected number of customers who carry Visa? Discover?

c. What is the approximate probability that half or more of the sample of card users carry Visa? American Express?

d. Justify the use of the normal approximation to the binomial in answering the question in part c.

Find the following probabilities for the standard normal

random variable z:

a.P(z≤2.1)

b.P(z≥2.1)

c.P(z≥-1.65)

d.P(-2.13≤z≤-.41)

e.P(-1.45≤z≤2.15)

f.P(z≤-1.43)

Assume that xis a binomial random variable with n = 100

and p = 5. Use the normal probability distribution to approximate

the following probabilities:

a.P(x≤48)

b.P(50≤x≤65)

c.P(x≥70)

d.P(55≤x≤58)

e.P(x=62)

f.P(x≤49orx≥72)

Buy-side vs. sell-side analysts’ earnings forecasts. Financial analysts who make forecasts of stock prices are categorized as either “buy-side” analysts or “sell-side” analysts. Refer to the Financial Analysts Journal (July/August 2008) comparison of earnings forecasts of buy-side and sell-side analysts, Exercise 2.86 (p. 112). The mean and standard deviation of forecast errors for both types of analysts are reproduced in the table. Assume that the distribution of forecast errors are approximately normally distributed.

a. Find the probability that a buy-side analyst has a forecast error of +2.00 or higher.

b. Find the probability that a sell-side analyst has a forecast error of +2.00 or higher


Buy-Side Analysts

Sell-Side Analysts

Mean

0.85

-0.05

Standard Deviation

1.93

0.85
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
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