Question #167636

  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.

Expert's answer

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 0.68..\sqrt{0.68..} = 0.82


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