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The number of flaws in a given area of aluminum foil

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 2E Chapter 4.3

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 2E

The number of flaws in a given area of aluminum foil follows a Poisson distribution with a mean of 3 per nr. Let X represent the number of flaws in a 1 m2sample of foil.

a. P(X = 5)

b. P(X = 0)

c. P(X<2)

d. P(X> 1)

e. μX

f. σX

Step-by-Step Solution:

Step 1 of 7</p>

Let X represents the no.of flaws in a sample

Here Xpoisson()

The pmf of Poisson distribution is p(x)=,x=0,1,2,............

Given mean =3

The mean of poisson distribution is

So =3

Step 2 of 7</p>

a) P(X=5)=

          =0.101

 Hence P(X=5)=0.101

Step 3 of 7</p>

b)P(X=0)=

             =0.0498

 Hence P(X=5)=0.0498

Step 4 of 7</p>

c) P(X<2)=P(0)+P(1)

              P(1) =

                      =0.1494

Hence P(X<2)=0.0498+0.1494

                       =0.1992

Step 5 of 7

Chapter 4.3, Problem 2E is Solved
Step 6 of 7

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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