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.
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) = 0.020.02
0.02
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
+1
0.001176+2
0.0576+3
0.9412)
= 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.
