Chapter 18, Problem 41E

(a)

To determine

To find the expected value and the standard deviation of your prospective winnings.

Answer to Problem 41E

The mean is $\$2$ and the standard deviation is $\$3.61$ .

Explanation of Solution

It is given in the question that you roll a die, winning nothing if the number of spots is odd and $\$1\text{ for 2 or a 4, and \$10 for a six}$ . Thus, let X be the number of dollars won in one play. Then the mean and the standard deviation are calculated as:

  $\begin{array}{l}\mu =E(X)\\ =0\times \frac{3}{6}+1\times \frac{2}{6}+10\times \frac{1}{6}\\ =\$2\\ {\sigma }^2=Var(X)\\ $ ={(0-2)}^2\times \frac{3}{6}+{(1-2)}^2\times \frac{2}{6}+{(10-2)}^2\times \frac{1}{6}$=13\\ \sigma =SD(X)\\ =\sqrt{Var(X)}\\ =\sqrt{13}\\ =\$3.61\end{array}$

Thus, the mean is $\$2$ and the standard deviation is $\$3.61$ .

(b)

To determine

To find the mean and the standard deviation of your total winnings if you play twice.

Answer to Problem 41E

The mean is $\$4$ and the standard deviation is $\$5.10$ .

Explanation of Solution

It is given in the question that you roll a die, winning nothing if the number of spots is odd and $\$1\text{ for 2 or a 4, and \$10 for a six}$ . Thus, if we play twice then the mean and the standard deviation are calculated as:

  $\begin{array}{l}\mu =E(X+X)\\ =2E(X)\\ =2(2)\\ =\$4\\ \sigma =SD(X+X)\\ =\sqrt{2Var(X)}\\ =\sqrt{2(13)}\\ =\$5.10\end{array}$

Thus, the mean is

$\$4$ and the standard deviation is $\$5.10$ .

(c)

To determine

To find out what is the probability that you win at least $\$100$ if you play $40$ times.

Answer to Problem 41E

  $0.191$ .

Explanation of Solution

It is given in the question that you roll a die, winning nothing if the number of spots is odd and $\$1\text{ for 2 or a 4, and \$10 for a six}$ . Thus, in order to win at least $\$100$ if you play $40$ times, you must average at least

  $\frac{100}{40}=\$2.50$ per play.

And the mean is $\$2$ and the standard deviation is $\$3.61$ . Rolling a die is random and the outcomes are mutually independent so the Central Limit Theorem guarantees that the sampling distribution model is Normal with mean and the standard deviation as:

  $\begin{array}{l}{\mu }_{\overline{x}}=\$2\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{\$3.61}{\sqrt{40}}\\ =\$0.571\end{array}$

Now we will find the z -score is as:

  $\begin{array}{l}z=\frac{2.50-2}{0.571}\\ =0.876\end{array}$

Thus, we will use graphing utility to find the probability that you win at least $\$100$ if you play $40$ times as:

  $\begin{array}{l}P(z\ge 0.876)=normalcdf(0.876,E99,0,1)\\ =0.191\end{array}$