Chapter 13.4, Problem 3CYU

a.

To determine

To calculate:

The expected value from each throw.

Answer to Problem 3CYU

The expected value from each throw is $36$ .

Explanation of Solution

Given information:

  

Calculation:

A game consists of trying to pop a balloon with a dart. There are $1625$ -point balloons around the perimeter of the square board, inside of the $25-$ point balloon perimeter , there are $850$ - point balloons, and within these balloons there is a single $100$ - point balloon.

Outcome	Frequency	Probability
$25$ points	$16$	$\frac{16}{25}=0.64$
$50$ points	$8$	$\frac{8}{25}=0.32$
$100$ points	$1$	$\frac{1}{25}=0.04$
Total	$25$

To get the expected value from each dart throw, take the sum of the outcomes amd their corresponding probabilities.

  $\begin{array}{l}ExpectedValue=\left(25\right)\left(0.64\right)+\left(50\right)\left(0.32\right)+\left(100\right)\left(0.4\right)\\ =16+16+4\\ =36\end{array}$

Therefore, the expected value from each throw is $36$ .

b.

To determine

To calculate:

The average value of the game.

Answer to Problem 3CYU

The average value of the game $35$ .

Explanation of Solution

Given information:

  

Calculation:

To simulate the balloon popping game, use a random number generator to generate integers. Use the integers $1$ through $25$ .

So, let $1-16$ represent $25$ points, let $17-24$ represent $50$ points and let $25$ represents $100$ points. Run $50$ trials of the game and record the frequency of observed outcomes.

The frequency of outcomes will vary between sets of $50$ trials. One such set of trials can be recorded in a frequency table as follows:

Outcome	Frequency
$25$ points	$30$
$50$ points	$20$
$100$ points	$0$
Total	$50$

To get the frequency in the frequency table to get the observed probabilities for each outcome.

Outcome	Frequency	Probability
$25$ points	$30$	$\frac{30}{50}=0.60$
$50$ points	$20$	$\frac{20}{50}=0.40$
$100$ points	$0$	$\frac{0}{50}=0$
Total	$50$	$\frac{50}{50}=1$

To get the average value of the game, take the sum of the products of simulated outcomes and their corresponding probabilities.

  $\begin{array}{l}AverageValue=\left(25\right)\left(0.60\right)+\left(50\right)\left(0.40\right)+\left(100\right)\left(0\right)\\ =15+20\\ =35\end{array}$

Therefore, the average value of the game $35$ .

c.

To determine

To explain:

The expected value and average value are compared.

Answer to Problem 3CYU

The expected value and average value are very close.

Explanation of Solution

The expected value was found to be $36$ and the average value was found to be $35$ . Therefore. The expected value and average value are very close.