Chapter 13.4, Problem 4CYP

(a)

To determine

To find the expected value $E(X)$ .

Answer to Problem 4CYP

  $E(X)=4.05$ .

Explanation of Solution

Given:

Two dice are rolled and let the random variable X represents the sum of the potential outcomes.

Calculation:

The following is the table for the sum of outcomes of rolling two dice-

X-Value(Sum)	Frequency	Probability
2	1	$\frac{1}{36}$
3	2	$\frac{2}{36}$
4	3	$\frac{3}{36}$
5	4	$\frac{4}{36}$
6	5	$\frac{5}{36}$
7	6	$\frac{6}{36}$
8	5	$\frac{5}{36}$
9	4	$\frac{4}{36}$
10	3	$\frac{3}{36}$
11	2	$\frac{2}{36}$
12	1	$\frac{1}{36}$

The expected value

  $\begin{array}{l}E(X)=2\times \frac{1}{36}+4\times \frac{2}{36}+6\times \frac{3}{36}+8\times \frac{4}{36}+10\times \frac{5}{36}+6\times \frac{6}{36}\\ =\frac{146}{36}=4.05\end{array}$

(b)

To determine

To find the average value.

Answer to Problem 4CYP

  $7$ .

Explanation of Solution

Given:

Two dice are rolled and let the random variable X represents the sum of the potential outcomes.

Calculation:

The following is the table for the sum of outcomes of rolling two dice-

X-Value(Sum)	Frequency	Probability
2	1	$\frac{1}{36}$
3	2	$\frac{2}{36}$
4	3	$\frac{3}{36}$
5	4	$\frac{4}{36}$
6	5	$\frac{5}{36}$
7	6	$\frac{6}{36}$
8	5	$\frac{5}{36}$
9	4	$\frac{4}{36}$
10	3	$\frac{3}{36}$
11	2	$\frac{2}{36}$
12	1	$\frac{1}{36}$

The average value is

  $\begin{array}{l}\text{Average value}=2\times \frac{1}{36}+3\times \frac{2}{36}+4\times \frac{3}{36}+5\times \frac{4}{36}+6\times \frac{5}{36}+7\times \frac{6}{36}+12\times \frac{1}{36}+11\times \frac{2}{36}+10\times \frac{3}{36}+9\times \frac{4}{36}+8\times \frac{5}{36}\\ =\frac{252}{36}=7\end{array}$