Question

Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 0 3 Probability 0.268 0.305 0.240 0.140 2 O 4 0.036 5 0.011 (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is (Round to one decimal place as needed.) The variance is (Round to one decimal place as needed.) The standard deviation is (Round to one decimal place as needed.) (b) Interpret the results. The standard deviation is The mean is , so the average batch of 1000 machine parts has , so the typical number of defects in a batch of 1000 (Round to one decimal place as needed.)

Answer 1

let'x' be a Handom vanliable denotes the number of defects per 1000 machine parts The probability distribution of x is as forrow - Defects o 1 2 3 4 5 Probability 0.268 0.305 0.240 0.140 0.036 0.011 mean of x, E(X) = {X.P(x) = (0X0-268) +(1X0.305) +(2x0-210) +3(0.140) + (4x0.036) + 15X0:011) (a) EEN = 1.404 ) E(X)=1.4 my (d) Variance al X = ? F(x) = {X?P(x) -DX0:268)+(130:30s) + (90.240) +(9X0.140) +(16X0.036) +125X0:011) E(X) = 3.4 m

Variance u(x) = E(x² - (Ex)]² 3.4-(10432 Standard deviation, (3.0(X) = variance (2) = 1.2 my The mean is 1.4, so the average batch of 1000 machine has parts has liu defective parts. The Standard deviation is 1.2, so the typical number of defects in a batch of 1000 is les me