Question

A system consisting of one original unit plus a spare

can function for a random amount of time$X$. If the density

of$X$is given (in units of months) by

$f\left(x\right)=\left\{\begin{array}{ll}Cx{e}^{-x/2}& x>0\\ 0& x\le 0\end{array}\right.$

what is the probability that the system functions for at least $5$months?

Short answer

The probability that the system will function for at least$5$a month is$0.28729$

Step-by-step solution

Given information.

The density of a random amount of time $X$is given as

$f\left(x\right)=\left\{\begin{array}{ll}Cx{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& x>0\\ 0& x\le 0\end{array}\right.$

Explanation.

We first determine the value of the constant. We know that the probability density function of random variable integrates to$1$.

Therefore we have

$$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)dx=1 \\ {\int }_{-\infty }^{\infty }0dx+{\int }_0^{\infty }Cx{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx=1 \\ {\int }_0^{\infty }Cx{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx=1 \\ C{\left[\begin{array}{cc}x\int e^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx-\int & \frac{{\left(x\right)}^{\text{'}}{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \frac{-1}{2}}}dx\end{array}\right]}_0^{\infty }=1 \\ C{\left[\begin{array}{cc}x\frac{e^{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{-1}{2}}}& +2\frac{e^{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{-1}{2}}}\end{array}\right]}_0^{\infty }=1 \\ -2C{\left[\begin{array}{cc}x{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& +2e^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}\right]}_0^{\infty }=1 \\ -2C\left[0+0-0-2\left(1\right)\right]=1 \\ 4C=1 \\ C=\frac{1}{4} \end{gathered}$$

Explanation

Thus, the Density of a random amount of time$XX$is given as

$f\left(x\right)=\left\{\begin{array}{ll}\frac{1^{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{4}& x>0\\ 0& x\le 0\end{array}\right.$

Explanation

Now,

The probability that the system function for at least$5$months$=P\left[\begin{array}{cc}X& \ge 5\end{array}\right]$

$$\begin{gathered} P\left[\begin{array}{cc}X\ge & 5\end{array}\right]={\int }_5^{\infty }f\left(x\right)dx \\ ={\int }_5^{\infty }\frac{1}{4}x{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx \\ =\frac{1}{4}{\int }_5^{\infty }x{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx \\ =\frac{1}{4}{\left[\begin{array}{cc}x\int e^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx& -\frac{{\left(x\right)}^{\text{'}}{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \frac{-1}{2}}}dx\end{array}\right]}_5^{\infty } \\ =\frac{1}{4}{\left[\begin{array}{cc}x\frac{e^{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{-1}{2}}}& +2\frac{e^{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{-1}{2}}}\end{array}\right]}_5^{\infty } \\ =\frac{-2}{4}{\left[\begin{array}{cc}x{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& +2e^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}\right]}_5^{\infty } \\ =\frac{-1}{2}\left[\begin{array}{cc}0+0-& 5e^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2e^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}\right] \\ =\frac{7}{2}{e}^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =0.28729 \end{gathered}$$