Chapter 6.8, Problem 4E

(a)

To determine

To Verify: The given function is a probability density function.

Answer to Problem 4E

  $f\left(x\right)$ is indeed a probability density function.

Explanation of Solution

Given Information: The probability density function is $f\left(x\right)=x{e}^{-x}$ if $x\ge 0$ and $f\left(x\right)=0$ if $x<0$ .

Concept Used: If $f\left(x\right)$ is a probability density function, then

  $\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}f\left(x\right)dx=1$

Integrate the probability density function from $-\infty$ to $\infty$ and check if the value is one. So, write

  $\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}f\left(x\right)dx=\underset{0}{\overset{\infty }{{\displaystyle \int }}}x{e}^{-x}dx$

  $={-x{e}^{-x}-e^{-x}]}_0^{\infty }$

  $=-(0-1)$

  $=1$

Therefore, $f\left(x\right)$ is indeed a probability density function.

(b)

To determine

To Calculate: The probability that the value of a given random variable is within a certain range for a given probability distribution function.

Answer to Problem 4E

The value of $P\left(1\le X\le 2\right)$ is $0.329$

Explanation of Solution

Given:

The probability density function is $f\left(x\right)=x{e}^{-x}$ if $x\ge 0$ and $f\left(x\right)=0$ if $x<0$ .

The probability to be calculated is $P\left(X<2\right)$ .

Concept Used: If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$ , is given by

  $P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

From the given information, $a$ is $1$ and $b$ is $2$ .

So, the required probability is

  $\text{P}\left(1\le X\le 2\right)=\underset{1}{\overset{2}{{\displaystyle \int }}}x{e}^{-x}dx$

  $=-{\left(x{e}^{-x}+e^{-x}\right)]}_1^2$

  $=-\left(2e^{-2}+e^{-2}-e^{-1}-e^{-1}\right)$

  $=0.329$

Therefore, the value of $P\left(1\le X\le 2\right)$ is $0.329$