**probability**of a

**function**for

**5 months**whose

**density**is given in

**units**of

**months.**The question depends on the concept of

**Probability**Â

**Density Function (PDF).**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: \[ P ( X \geq 1 ) = 1\ -\ 0.3608 \] \[ P ( X \geq 1 ) = 0.6392 \]

5/5 - (19 votes)