Chapter 6.8, Problem 7E

(a).

To determine

To Prove: That a certain probability distribution function appropriately depicts the given situation.

Explanation of Solution

Given Information:

• A spinner randomly indicates a real number between $0$ and $10$ .
• The probability of a spinner randomly indicating a number from any interval is the same.

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

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

Let the probability distribution function be $f\left(x\right)$ .

The value of $f\left(x\right)$ is $0$ when $x<0$ or $x>10$ because the spinner does not choose numbers other than those between $0$ and $10$ , that is the probability of choosing other numbers is $0$ .

Between $x=0$ and $x=10$ , the probability distribution function has a constant value $k$ because the spinner is fair and selects numbers from any interval with equal probability.

Therefore

  $f\left(x\right)=k$ when $0\le x\le 10$

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

Now, to find the value of $k$ , apply the condition that the value of the integral of a probability distribution from $-\infty$ to $\infty$ is 1. So

  $\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}f\left(x\right)dx=\underset{0}{\overset{10}{{\displaystyle \int }}}kdx=1$

  $\Rightarrow 10\times k=1$

  $\Rightarrow k=0.1$

Therefore, the probability distribution function for the given scenario is

  $f\left(x\right)=0.1$ when $0\le x\le 10$

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

(b).

To determine

To Calculate: The mean of a probability distribution function.

Answer to Problem 7E

The mean of the random variable is $5$ .

Explanation of Solution

Given: From the previous sub-part, the probability density function is $f\left(x\right)=0.1$ if $0\le x\le 10$ and $f\left(x\right)=0$ for all other values of $x$ .

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

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

Since the probability distribution function is constant in the interval of interest, intuition tells that the mean value is the median value of the interval, that is $5$ .

Now, to actually calculate the mean, put the expression for $f\left(x\right)$ in the equation for mean to write

  $\mu =\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}xf\left(x\right)dx=\underset{0}{\overset{10}{{\displaystyle \int }}}x\times 0.1dx$

  $={0.1\times \frac{x^2}{2}]}_0^{10}$

  $=0.1\times 50$

  $=5$

Therefore, the mean of the random variable is $5$