Answer to Question #167360 in Statistics and Probability for kevin

Question

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.

Answer 1

We know that,

$\text{Mean}(\mu) =\Sigma_{i=0} ^{5} x_iP(x_i)$

$Variance(\sigma^2)= \Sigma_{i=0}^5 x^2_{i}P(x_i)-\mu^2$

$\text{Mean}(\mu) =\Sigma_{i=0} ^{5} x_iP(x_i)= 0.39$

$Variance(\sigma^2)= \Sigma_{i=0}^5 x^2_{i}P(x_i)-\mu^2\\$

$\Rightarrow0.84-0.1521=0.6879 = 0.69$