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Of all the weld failures in a certain assembly, 85% of

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

Solution for problem 7E Chapter 4.2

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 7E

Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal. A sample of 20 weld failures is examined.

a. What is the probability that exactly five of them are base metal failures?

b. What is the probability that fewer than four of them are base metal failures?

c. What is the probability that none of them are base metal failures?

d. Find the mean number of base metal failures.

e. Find the standard deviation of the number of base metal failures.

Step-by-Step Solution:

Answer :

Step 1 of 6:

Given, in a sample of 20 weld failures, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal.

Let X follows binomial distribution with probability mass function is

P(x: n,p) =  (1 - p , x = 0, 1, 2, ….., n.

Where,

n = sample size

   = 20

x = random variable

p = probability of success

   = 15%

   =  0.15

q = 1 - p (probability of failure)

Mean of the binomial distribution is

   = np

Variance of the binomial distribution is

   = npq

Step 2 of 6:

The claim is to find the probability that exactly five of them are base metal failures

We have to find  P(X=5)

P(x: n,p) =  (1 - p , x = 0, 1, 2, ….., n.

                  Where, n = 20, p = 0.15 and x = 5

P(x = 5) =  (1 - 0.15

                      = (15504) (0.0000759) (0.087354)

                                                   = 0.1028

Hence, the probability that exactly five of them are base metal failures is 0.1028.

Step 3 of 6:

The claim is to find the probability that fewer than four of them are base metal failures

We have to find  P(X < 4) = P(x = 0) + P( x = 1) + P(x = 2) + P(x = 3)

P(x: n,p) =  (1 - p , x = 0, 1, 2, ….., n.

                  Where, n = 20, p = 0.15 and x = 0, 1, 2 and 3

P(x < 4) =  (1 - 0.15+  (1 - 0.15+

             (1 - 0.15+  (1 - 0.15

                          = (1) (1) (0.03876)  + (20) (0.15)  (0.0456)  +

                             (190) (0.0225) (0.053646)  +  (1140)  ( 0.003375) (0.0631134)

                         = 0.03876  +  0.1368  +  0.2293  +  0.2428

                         = 0.6477

Hence, the probability that fewer than four of them are base metal failures 0.6477

Step 4 of 6

Chapter 4.2, Problem 7E is Solved
Step 5 of 6

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

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