Answer to Question #323872 in Statistics and Probability for kathy

Question

The total number of hours, measured in units of

100 hours, that a family runs a vacuum cleaner over a

period of one year is a continuous random variable X

that has the density function

f(x) =

⎧

⎨

⎩

x, 0 2 − x, 1 ≤ x < 2,

0, elsewhere.

Find the variance of X.

Find the average number of hours per year that families run their vacuum cleaners.

Answer 1

$f\left( x \right) =\left\{ \begin{array}{c} x,0\leqslant x<1\\ 2-x,1\leqslant x<2\\ 0,elsewhere\\\end{array} \right. \\EX=\int{xf\left( x \right) dx}=\int_0^1{x\cdot xdx}+\int_1^2{x\left( 2-x \right) dx}=\\=\frac{1}{3}+x^2|_{1}^{2}-\frac{x^3}{3}|_{1}^{2}=1\\EX^2=\int{x^2f\left( x \right) dx}=\int_0^1{x^2\cdot xdx}+\int_1^2{x^2\left( 2-x \right) dx}=\\=\frac{1}{4}+\frac{2x^3}{3}|_{1}^{2}-\frac{x^4}{4}|_{1}^{2}=\frac{7}{6}\\VarX=\frac{7}{6}-1^2=\frac{1}{6}$