Answer to Question #325158 in Statistics and Probability for Jam

Question

From past experience, a company has found that in a box of four triple-A batteries, 74% contain no defective battery, 7% contain one defective battery, 4% contain two defective batteries, 6% contain three defective batteries, and 9% contain all defective batteries. Calculate the mean, variance, and standard deviation for the defective batteries

Answer 1

Let X - the random variable of the number of defective batteries in a box.

The mean:

$\mu=\sum x_i\cdot P(x_i)=\\ =0\cdot0.74+1\cdot0.07+2\cdot0.04+\\ +3\cdot0.06+4\cdot0.09=0.69.$

The variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\\ =\{ 0-0.69, 1-0.69, 2-0.69, 3-0.69,4-0.69\}=\\ =\{-0.69,0.31,1.31,2.31,3.31\},$

$\sigma^2=(-0.69)^2\cdot0.74+0.31^2\cdot0.07+1.31^2\cdot0.04+\\ +2.31^2\cdot0.06+3.31^2\cdot0.09=1.734.$

The standard deviation:

$\sigma=\sqrt{1.734}=1.317.$