Answer to Question #167636 in Statistics and Probability for rezalyn

Question

1. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, or 5 defective parts in one day are 0.75, 0.17, 0.04, 0.025, 0.01, and 0.005 respectively. Find the variance and standard deviation of the probability distribution.

Answer 1

Expected value of the number of defective parts is 0*0.75 + 1*0.17 + 2*0.04 + 3*0.025 + 4*0.01 + 5*0.005 = 0.39

Expected value of the square of number of defective parts is 0²*0.75 + 1²*0.17 + 2²*0.04 + 3²*0.025 + 4²*0.01 + 5²*0.005 = 0.84

Variance is 0.84 - 0.39 ² = 0.68...

Standard deviation is $\sqrt{0.68..}$ = 0.82