Chapter 10.4, Problem 53E

Solution

a.

To determine: the expected value of the game.

The expected value of the game is 4.5

Given information:

The square root of the expected value of squared deviations from the mean, $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$

Concept Used:

The mean of a discrete probability distribution, where x is each value of random variables and $P\left(x\right)$ is the probability, $\mu ={\displaystyle \sum \left|x\cdot P\left(x\right)\right|}$

Calculation:

Calculate the mean using the formula for the mean of probability distribution.

  $\begin{array}{c}\mu =0\cdot 0.8+10\cdot 0.15+50\cdot 0.08+0.01\cdot 100\\ =4.5\end{array}$

b.

To determine: the standard deviation of the value of the prizes

The standard deviation of the value of the prizes is 13.96.

Given information:

The square root of the expected value of squared deviations from the mean, $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$

Calculation:

The expected value of the game is 4.5.

Substitute all the values in $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$ .

  $\begin{array}{c}\sigma =\sqrt{{\left(0-4.5\right)}^2\cdot 0.80+{\left(10-4.5\right)}^2\cdot 0.15+{\left(50-4.5\right)}^2\cdot 0.04+{\left(100-4.5\right)}^2\cdot 0.01}\\ =\sqrt{194.75}\\ \approx 13.96\end{array}$

c.

To explain: why the standard deviation is so large.

The standard deviation is large since the data points are spread out over a large range of values.

Given information:

The square root of the expected value of squared deviations from the mean, $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$

Calculation:

The expected value of the game is 4.5.

The standard deviation of the value of the prizes is 13.96. Notice that the standard deviation $\sigma =\$13.96$ is high from the mean $\mu =\$4.50$ , this only means that the data points are spread out over a large range.

d.

To explain: why are willing to pay $5 to play this game.

No, since the expected value is lower there is no high chance to win money in the long run.

Given information:

The square root of the expected value of squared deviations from the mean, $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$

Calculation:

The expected value of the game is 4.5.

The standard deviation of the value of the prizes is 13.96. Whereas, the fee is $5 which is higher than the expected value.

Since the expected value is lower there is no high chance to win money in the long run.