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

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.**

## Expert Answer

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:

\[ P ( X \geq 1 ) = 1\ -\ 0.3608 \]

\[ P ( X \geq 1 ) = 0.6392 \]