Answer to Question #296043 in Statistics and Probability for Subodh

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 < x < 1,

2 − x, 1 ≤ x < 2,

0, elsewhere.

Find the probability that over a period of one year, a

family runs their vacuum cleaner

(a) less than 120 hours;

(b) between 50 and 100 hours.

Answer 1

a)

$$x = 120/100 = 1.2$$

$$P (x < 1.2) = \int^1_0xdx+\int^{1.2}_1(2-x)dx$$

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

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

b)

$$x = 50/100 = 0.5,$$

$$x = 100/100 = 1$$

$$P (0.5<x < 1) = \int^{1}_{0.5}xdx=[\dfrac{x^2}{2}]\begin{matrix} 1 \\ 0.5 \end{matrix}$$

$$=0.5-0.125=0.375$$