Answer to Question #350443 in Statistics and Probability for zqvr1r3b

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

Answer 1

$$f(x) = \begin{cases} x, &0<x<1 \\ 2-x, &1\le x<2 \\ 0, & elsewhere \end{cases}$$

100 hours is taken as one unit of $X.$

a) Find the probability that over a period of one year, a family runs their vacuum cleaner less than 120 hours

$$P(X<1.2)=\displaystyle\int_{0}^{1}xdx+\displaystyle\int_{1}^{1.2}(2-x)dx$$

$$=[\dfrac{x^2}{2}]\begin{matrix} 1\\ 0 \end{matrix}+[2x-\dfrac{x^2}{2}]\begin{matrix} 1.2\\ 1 \end{matrix}$$

$$=0.5-0+2.4-0.72-2+0.5=0.68$$

b) Find the probability that over a period of one year, a family runs their vacuum cleaner between 50 and 100 hours.

$$P(0.5<X<1)=\displaystyle\int_{0.5}^{1}xdx$$

$$=[\dfrac{x^2}{2}]\begin{matrix} 1\\ 0.5 \end{matrix}=0.5-0.125=0.375$$