Chapter 6.2, Problem 42E

a.

To determine

To find: The mean and standard deviation.

Answer to Problem 42E

The mean and standard deviation are $0.50 and $15.80.

Explanation of Solution

Given:

The probability distribution is:

Payoff X	$0	$500
Probability	0.999	0.001

Calculation:

The mean and standard deviation can be calculated as:

  $\begin{array}{c}\mu ={\displaystyle \sum x\times P\left(x\right)}\\ =0\left(0.99\right)+500\left(0.001\right)\\ =0.50\end{array}$

Now,

  $\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum x^2\times P\left(x\right)-{\left({\displaystyle \sum x\times P\left(x\right)}\right)}^2}}\\ =\sqrt{0^2\times 0.999+{500}^2\times 0.001-{\left(0.50\right)}^2}\\ =15.80\end{array}$

Thus, the mean and standard deviation are $0.50 and $15.80 respectively.

b.

To determine

To find: The mean and standard deviation of W .

Answer to Problem 42E

The mean and standard deviation are $0.50 and $15.80.

Explanation of Solution

Calculation:

The mean and standard deviation of W can be calculated as:

  $\begin{array}{l}{\mu }_W={\mu }_X-1=0.50=\$0.50\\ {\sigma }_W={\sigma }_{X-1}={\sigma }_X=\$15.80\end{array}$

The average winnings are -0.50 per ticket.