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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