# 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 the following function. What is the probability that the system functions for at least 5 months?

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

The question aims to find the probability of a function for 5 months whose density is given in units of months.

The question depends on the concept of Probability Density Function (PDF). The PDF is the probability function that represents the likelihood of all the values of the continuous random variable.

To calculate the probability of the given probability density function for 5 months, we must first calculate the value of the constant C. We can calculate the value of the constant C in the function by integrating the function to infinity. The value of any PDF, when integrated, equates to 1. The function is given as:

$\int_{-\infty}^{\infty} f(x) \, dx = 1$

$\int_{-\infty}^{0} 0 \, dx + \int_{0}^{\infty} Cx e^{-x/2} \, dx = 1$

$\int_{0}^{\infty} Cx e^{-x/2} \, dx = 1$

Integrating the above equation, we get:

$C \Bigg[ x \dfrac{ e^{-x/2} }{ -1/2 } + 2\dfrac{ e^{-x/2} }{ -1/2 } \Bigg]_{0}^{\infty} = 1$

$-2C \Bigg[ x e^ {-x/2} + 2 e^ {-x/2} \Bigg]_{0}^{\infty} = 1$

$-2C \Big[ 0 + 0\ -\ 0\ -\ 2(1) \Big] = 1$

$4C = 1$

$C = \dfrac{ 1 }{ 4 }$

The density of the function is now given as:

$f(x) = \left\{ \begin {array} ( \dfrac{ 1 }{ 4 } x e^{-x/2} & x \gt 0 \\ 0 & x\leq 0 \end {array} \right.$

To calculate the probability for the function that it will perform for 5 months is given as:

$P ( X \geq 5 ) = 1\ -\ \int_{0}^{5} f(x) \, dx$

$P ( X \geq 5 ) = 1\ -\ \int_{0}^{5} \dfrac{ 1 }{ 4 } x e^{-x/2} \, dx$

$P ( X \geq 5 ) = 1\ -\ \Bigg[ – \dfrac{ (x + 2) e^{-x/2} }{ 2 } \Bigg]_{0}^{5}$

Simplifying the values, we get:

$P ( X \geq 5 ) = 1\ -\ 0.7127$

$P ( X \geq 5 ) = 0.2873$

## Numerical Result

The probability that the system with the given function will run for 5 months is calculated to be:

$P ( X \geq 5 ) = 0.2873$

## Example

Find the probability of a system that will run for 1 month if its density function is given with units represented in months.

$f(x) = \left\{ \begin {array} ( x e^{-x/2} & x \gt 0 \\ 0 & x\leq 0 \end {array} \right.$

The probability of the density function for 1 month is given as:

$P ( X \geq 1 ) = 1\ -\ \int_{0}^{1} f(x) \, dx$

$P ( X \geq 1 ) = 1\ -\ \int_{0}^{1} x e^{-x/2} \, dx$

$P ( X \geq 1 ) = 1\ -\ \Bigg[ – (2x + 4) e^ {-x/2} \Bigg]_{0}^{1}$

Simplifying the values, we get:

$P ( X \geq 1 ) = 1\ -\ 0.3608$

$P ( X \geq 1 ) = 0.6392$