Chapter 16, Problem 12E

(a)

To determine

To construct: the probability model for amount win.

Explanation of Solution

Given:

A dice is rolled.

Calculation:

When a dice is rolled there is 6 possible outcomes {1, 2, 3, 4, 5, 6}.

The probability of rolling a 6 = $\frac{1}{6}$ and

the probability of not getting a 6 = $\frac{5}{6}\left(1-\frac{1}{6}\right)$ ..

The random variable X is defined as the amount won when a die is rolled once.

The probability distribution for the amount won is,

X	$0	$50	$100
$P\left(x\right)$	$\frac{25}{36}$	$\frac{5}{36}$	$\frac{1}{36}$

(b)

To determine

To find: the expected amount that win.

Answer to Problem 12E

23.61$

Explanation of Solution

Formula used:

  $E\left(X\right)={\displaystyle \sum xP\left(x\right)}$

Calculation:

The expected amount of win is,

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =\left(0\times \frac{25}{36}\right)+\left(50\times \frac{5}{36}\right)+\left(100\times \frac{6}{36}\right)\\ =0+6.94+16.67\\ =23.61\end{array}$

(c)

To determine

To find: the researcher would be willing to pay to play this online.

Answer to Problem 12E

38.16$

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(X^2\right)={\displaystyle \sum x^{2}P\left(x\right)}\\ SD=\sqrt{E\left(X^2\right)-\left[E{\left(X\right)}^2\right]}\end{array}$

Calculation:

The expected of $X^2$ is,

  $\begin{array}{c}E\left(X^2\right)={\displaystyle \sum {}_i{x}^{2}P\left(x\right)}\\ =\left(0^2\times \frac{25}{36}\right)+\left({50}^2\times \frac{5}{36}\right)+\left({100}^2\times \frac{6}{36}\right)\\ =0+347.22+1666.67\\ =2,013.89\end{array}$

The standard deviation of the amount might win is,

  $\begin{array}{c}SD=\sqrt{E\left(X^2\right)-\left[E{\left(X\right)}^2\right]}\\ =\sqrt{2013.89-{\left(23.61\right)}^2}\\ =\sqrt{2013.89-557.4321}\\ =\sqrt{1456.4579}\\ =38.16\end{array}$

Therefore, the standard deviation of the amount may win is $38.16.