Question #350443

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

1
Expert's answer
2022-06-14T11:17:43-0400
f(x)={x,0<x<12x,1x<20,elsewheref(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.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)=01xdx+11.2(2x)dxP(X<1.2)=\displaystyle\int_{0}^{1}xdx+\displaystyle\int_{1}^{1.2}(2-x)dx

=[x22]10+[2xx22]1.21=[\dfrac{x^2}{2}]\begin{matrix} 1\\ 0 \end{matrix}+[2x-\dfrac{x^2}{2}]\begin{matrix} 1.2\\ 1 \end{matrix}

=0.50+2.40.722+0.5=0.68=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)=0.51xdxP(0.5<X<1)=\displaystyle\int_{0.5}^{1}xdx

=[x22]10.5=0.50.125=0.375=[\dfrac{x^2}{2}]\begin{matrix} 1\\ 0.5 \end{matrix}=0.5-0.125=0.375


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