In a lot of 10 components, 2 are sampled at random for Problem 10SE Chapter 2

Statistics for Engineers and Scientists | 4th Edition

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Problem 10SE

In a lot of 10 components, 2 are sampled at random for inspection. Assume that in fact exactly 2 of the 10 components in the lot are defective. Let X be the number of sampled components that are defective.

a. Find P(X = 0).

b. Find P(X = 1).

c. Find P(X = 2).

d. Find the probability mass function of X.

e. Find the mean of X.

f. Find the standard deviation of X.

Step-by-Step Solution:
Step 3 of 5

Solutions:

Step of 1 of 5:

In a lot of 10 components, 2 are sampled at random for inspection. Here exactly 2 of the 10 components in the lot are defective. Let X be the number of defectives. Then we have to find

P(X=0)P(X=1)P(X=2)Probability mass function of X.Mean of XStandard deviation of X

Step 2 of 5:

(a)

Here it is given that in a lot 10 components exactly two components are defectives. That means the remaining 8 are non-defectives. 2 are sampled at random for inspection. Since X is the number of defectives

P(X=0)= P(randomly selected two components are non - defectives)

The total ways to select two non- defective components are = 8C2

And the total ways to select two components out of 10 components are = 10C2

Therefore

P(X=0) =  = =  0.6222

Therefore the probability P(X=0) 0.6222 .

(b) We have to find the probability that P(X=1)

P(X=1) = P(that one of the two selected components is defective and the other one is non-defective.)

The  total number of ways to select one defective item and one non-defective item is

= 8C1 2C1 .

The total ways to select two components out of 10 components are =   10C2 .

Therefore ,

P(X=1) = =   16/45

=  0.3555.

Therefore the probability P(X=1) =  0.3555.

Step 3 of 5:

(c)  We have to find the probability that P(X=2)

P(X=2) = P(that two selected components are defective.)

The  total number of ways to select two defective components are =  2C2.

The total ways to select two components out of 10 components are =   10C2 .

Therefore ,

P(X=1) = =   1/45

=  0.022.

Therefore the probability P(X=2) = 0.022.

Step 4 of 5

Step 5 of 5

ISBN: 9780073401331

This full solution covers the following key subjects: Find, components, sampled, lot, defective. This expansive textbook survival guide covers 153 chapters, and 2440 solutions. This textbook survival guide was created for the textbook: Statistics for Engineers and Scientists , edition: 4th. The full step-by-step solution to problem: 10SE from chapter: 2 was answered by Patricia, our top Statistics solution expert on 06/28/17, 11:15AM. Statistics for Engineers and Scientists was written by Patricia and is associated to the ISBN: 9780073401331. Since the solution to 10SE from 2 chapter was answered, more than 305 students have viewed the full step-by-step answer. The answer to “In a lot of 10 components, 2 are sampled at random for inspection. Assume that in fact exactly 2 of the 10 components in the lot are defective. Let X be the number of sampled components that are defective.a. Find P(X = 0).________________b. Find P(X = 1).________________c. Find P(X = 2).________________d. Find the probability mass function of X.________________e. Find the mean of X.________________f. Find the standard deviation of X.” is broken down into a number of easy to follow steps, and 69 words.

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