Answer to Question #160608 in Statistics and Probability for Rosalyn

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 mean of the probability distribution. 

Answer 1

Solution:

First of all, let us restate the formula of the mean:

$E(\xi) = \Sigma_{i=1}^n p(x_i)x_i$

Mean of the given probability distribution:

$E(\xi) = p(0)*0 + p(1)*1 + p(2)*2 + p(3)*3 + p(4) *4 +p(5)*5 =$

$= 0.17 * 1 + 0.04*2+0.025*3+0.01*4+0.005*5 =$

$=0.17+0.08+0.075+0.04+0.025=0.17+0.12+0.1=$

$=0.39$

Answer:

$E(\xi) = 0.39$