Chapter 6, Problem 3CRE

a.

To determine

To interpret: The values of ${\mu }_X$ and ${\sigma }_X$

Explanation of Solution

Given:

The table is:

Matches	0	1	2	3	4
Probability	0.308	0.433	0.213	0.043	0.003
Pay out ($)	0	0	1	3	120

The values are:

  $\begin{array}{l}{\mu }_X=\$0.70\\ {\sigma }_X=\$6.58\end{array}$

Formula used:

The properties of mean and standard deviation are:

  $\begin{array}{l}{\mu }_{aX+b}=a{\mu }_X+b\\ {\sigma }_{aX+b}=a{\sigma }_X\end{array}$

The mean of X shows that expected pay out on an average is $0.70 and payout will vary on an average of $6.58 from the payout of $ 0.70.

b.

To determine

To find: The expected value and standard deviation

Answer to Problem 3CRE

The mean and standard deviation are $ 3.50 and $ 32.90 respectively.

Explanation of Solution

Calculation:

The expected value and standard deviation could be calculated as:

  $\begin{array}{l}{\mu }_{5X}=5\times {\mu }_X=5\left(0.70\right)=\$3.50\\ {\sigma }_{5X}=5\times {\sigma }_X=5\left(6.58\right)=\$32.90\end{array}$

Thus, the mean and standard deviation are $ 3.50 and $ 32.90 respectively.

c.

To determine

To find: The expected value and standard deviation of all the winnings of Marla.

Answer to Problem 3CRE

The mean and standard deviation are $ 3.50 and $ 14.70 respectively.

Explanation of Solution

Calculation:

The expected value and standard deviation could be calculated as:

  $\begin{array}{l}{\mu }_{X_1+X_2+\mathrm{....}+X_5}=5\times {\mu }_X=5\left(0.70\right)=\$3.50\\ \sqrt{{\sigma }^2{}_{X_1+X_2+\mathrm{....}+X_5}}=\sqrt{5\times {\sigma }^2{}_X}=\sqrt{5{\left(6.58\right)}^2}=\$14.71\end{array}$

Thus, the mean and standard deviation are $ 3.50 and $ 14.71 respectively.

d.

To determine

To find: The player that is preferred by a Casino

Explanation of Solution

Calculation:

From the above two parts, it is clear than the mean of both the players are same. Thus, the decision would be made using the value of the standard deviation. Since, the standard deviation of Marla is lower. Thus, Marla is to be