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Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.4 - Problem 64e
Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.4 - Problem 64e

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# Determine the mean and variance of the random | Ch 3.4 - 64E

ISBN: 9781118539712 55

## Solution for problem 64E Chapter 3.4

Applied Statistics and Probability for Engineers | 6th Edition

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Problem 64E Problem 64E

Determine the mean and variance of the random variable in Exercise 3-24.

3-24. An optical inspection system is used to distinguish among different part types. The probability of a correct classification of any part is 0.98. Suppose that three parts are inspected and that the classifications are independent. Let the random variable X denote the number of parts that are correctly classified. Determine the probability mass function of X.

Step-by-Step Solution:

Solution :

Step 1 of 1:

Let X denotes the number of parts that are correctly classified.

Then the range of X is 0, 1, 2, 3.

Our goal is:

We need to determine the mean and variance.

We know that X is a binomial random variable with n=3 and p=0.98.

Hence  is

P(correct classified) = 0.98 and P(incorrect classified) = 0.02.

P(X=0) =

P(X=0) = 0.020.020.02

P(X=0) =

Therefore, P(X=0) = .

P(X=1) = (3)(0.98)

P(X=1) = 0.001176

Therefore, P(X=1) = 0.001176.

P(X=2) = (3)(0.98)(0.98)

P(X=2) = 0.0576

Therefore, P(X=2) = 0.0576.

P(X=3) =

P(X=3) = 0.9412

Therefore, P(X=3) = 0.9412.

Now we need find mean.

=

= (0+10.001176+20.0576+30.9412)

= 2.94

Therefore, E(x) = 2.94.

=

= (+0.001176+0.0576+0.9412)

= 8.7023

Therefore, = 8.7023.

Then we need to find variance.

The formula for the variance is

Variance = []

Variance = 8.7023-

Variance = 8.7023-8.6436

Variance = 0.0587

Therefore, Variance = 0.0587.

Step 2 of 1

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