Chapter 6.8, Problem 8E

(a).

To determine

To Explain: A given graph corresponds to a probability density function.

Explanation of Solution

Given Information: The graph given is

  

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

• $f\left(x\right)\ge 0$ for all $x$ .
• $\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}f\left(x\right)dx=1$
• $\underset{b}{\overset{a}{{\displaystyle \int }}}f\left(x\right)dx$ is the area bound by the curve and the $x$ -axis between $a$ and $b$ .

The given graph is always on or above the $x$ -axis. So, the value of $f\left(x\right)$ is non-negative for all $x$ . Therefore, the graph satisfies the first condition mentioned above.

The integral of the function from $-\infty$ to $\infty$ is the area bounded by the given curve and the $x$ -axis. The area is in the shape of a triangle. So, the integral is given by

  $\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}f\left(x\right)dx=Areaoftriangle=\frac{1}{2}\times base\times height$

  $=\frac{1}{2}\times 10\times 0.2=1$

Therefore, the given function is a probability distribution function.

(b).

To determine

To Calculate: The probability of the value of a random variable being in the given interval

Answer to Problem 8E

The value of $P\left(X<3\right)$ is $0.15$ and $P\left(3<X<8\right)$ is 0.75

Explanation of Solution

Given: The graph of the probability density function is

  

The probabilities to be calculated are $P\left(X<3\right)$ and $P(3\le X\le 8)$ .

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 $P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$
• $\underset{b}{\overset{a}{{\displaystyle \int }}}f\left(x\right)dx$ is the area bound by the curve and the $x$ -axis between $a$ and $b$ .

(i).

  $P\left(X<3\right)$ is the area under the curve to the left of $x=3$ and the value of $f\left(3\right)$ is $0.1$

So

  $P\left(X<3\right)=P\left(-\infty <X<3\right)=\underset{0}{\overset{3}{{\displaystyle \int }}}f\left(x\right)dx$

  $=Areaoftriangle=\frac{1}{2}\times base\times height$

  $=\frac{1}{2}\times 3\times 0.1$

  $=0.15$

(ii).

  $P\left(3\le X\le 8\right)$ is the area under the curve between $x=3$ and $x=8$ .

From the graph , the value of $f\left(3\right)$ is $0.1$ and the value of $f\left(8\right)$ is also $0.1$

So

  $P\left(3\le X\le 8\right)=\underset{3}{\overset{8}{{\displaystyle \int }}}f\left(x\right)dx$

  $=Area=5\times 0.1+\frac{1}{2}\times 5\times \left(0.2-0.1\right)$

  $=0.5+0.25$

  $=0.75$

Therefore, $P\left(X<3\right)$ is $0.15$ and $P\left(3<X<8\right)$ is $0.75$

(c).

To determine

To Calculate: The mean of a probability distribution function.

Answer to Problem 8E

The mean of the random variable is $5.\overline{3}$

Explanation of Solution

Given: The graph of the probability density function is

  

Concept Used: If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the mean value of the random variable is

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

The given graph of $f\left(x\right)$ written in algebraic form is

  $f\left(x\right)=\frac{0.2}{6}x$ if $0\le x\le 6$ ,

  $f\left(x\right)=\frac{0.2}{4}\left(10-x\right)$ if $6\le x\le 10$ and

  $f\left(x\right)=0$ if $x<0$ or $x>10$

Now put this algebraic form of $f\left(x\right)$ in the integral for mean to write

  $\mu =\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}xf\left(x\right)dx=\underset{0}{\overset{6}{{\displaystyle \int }}}x\times \frac{0.2}{6}xdx+\underset{6}{\overset{10}{{\displaystyle \int }}}x\times \frac{0.2}{4}(10-x)dx$

  $=\frac{0.2}{6}\times {\frac{x^3}{3}]}_0^6+\frac{0.2}{4}\times {\left(10\cdot \frac{x^2}{2}-\frac{x^3}{3}\right)]}_6^{10}$

  $=\frac{0.2}{6}\left(\frac{216}{3}-0\right)+\frac{0.2}{4}\left(\left[10\cdot \frac{100}{2}-\frac{1000}{3}\right]-\left[10\cdot \frac{36}{2}-\frac{216}{3}\right]\right)$

  $=2.4+2.9\overline{3}$

  $=5.\overline{3}$

Therefore, the mean of the random variable is $5.\overline{3}$