Question
CO 4) Consider the following table. Defects in batch Probability 0 0.30 1 0.28 2 0.21...
CO 4) Consider the following table.
Defects in batch Probability
0 0.30
1 0.28
2 0.21
3 0.09
4 0.08
5 0.04
Find the variance of this variable.
0.67
1.49
1.99
1.41
Homework Answers
Answer #1
Solution :
From given probability distribution table,
= 
X * P(X)
= 0 * 0.30 + 1 * 0.28 + 2 * 0.21 + 3 * 0.09 + 4 * 0.08 + 5 * 0.04
= 0 + 0.28 + 0.42 + 0.27 + 0.32 + 0.20
= 1.49
variance =
X2 * P(X) -
2
= [02 * 0.30 + 12 * 0.28 + 22 * 0.21 + 32 * 0.09 + 42 * 0.08 + 52 * 0.04] - 1.492
= 0 + 0.28 + 0.84 + 0.81 + 1.28 + 1 - 2.2201
= 1.99
variance = 1.99
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