Question

A manufacturer produces widgets that are packaged in boxes of 150. The probability of a widget being defective is 0.023. Find the mean and standard deviation for the number of defective widgets in a box.

The mean, mu , is ___. (Round to the nearest hundredth as needed.)

The standard deviation, sigma , is ___. (Round to the nearest thousandth as needed.)

Answer 1

n 150,p 0.023 Expected value of X, Mean-μ-np-150 * 0.023-3.45 Expectation or mean is: μ = 3.45 Std Dev: σ-Vnp(1-p)-V150 * (0.023) (1-0.023) - V3.37065 ~ 1.835933 ~N 1.836 Standard deviation: σ 1.836 Thank yous