Chapter 5.2, Problem 20EC

Rolling Dice Construct a probability distribution for the sum shown on the faces when two dice are rolled. Find the mean, variance, and standard deviation of the distribution.

To determine

To construct: The probability distribution for the sum of shown on the faces when two dice are rolled.

To find: The mean, variance and standard deviation of the distribution.

Answer to Problem 20EC

The probability distribution for the sum of shown on the faces when two dice are rolled is,

Sum of two dice X	Probability $P\left(X\right)$
2	$\frac{1}{36}$
3	$\frac{2}{36}$
4	$\frac{3}{36}$
5	$\frac{4}{36}$
6	$\frac{5}{36}$
7	$\frac{6}{36}$
8	$\frac{5}{36}$
9	$\frac{4}{36}$
10	$\frac{3}{36}$
11	$\frac{2}{36}$
12	$\frac{1}{36}$
Total	1

The mean, variance and standard deviation of the distribution are 7.000, 5.901 and 2.429.

Explanation of Solution

Given info:

The two dice are rolled.

Calculation:

The possibilities for rolling a pair of six-sided dice is,

$\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

Thus, the total number of outcomes is 36.

The possibilities of sum of two dice is,

$\left\{\begin{array}{l}2,3,4,5,6,7,\\ 3,4,5,6,7,8,\\ 4,5,6,7,8,9,\\ 5,6,7,8,9,10,\\ 6,7,8,9,10,11,\\ 7,8,9,10,11,12\end{array}\right\}$

The formula to find the probability of sum of two dice for 2 is,

$P\left(\text{Sum of two dice is 2}\right)=\frac{\text{Number of times having sum 2}}{\text{Total number of outcomes}}$

Substitute 1 for number of times having sum 2 and 36 for total number of outcomes,

$P\left(\text{Sum of two dice is 2}\right)=\frac{\text{1}}{36}$

Similarly, the probability distribution for the random variable x is,

Sum of two dice X	Probability $P\left(X\right)$
2	$\frac{1}{36}$
3	$\frac{1}{36}+\frac{1}{36}=\frac{2}{36}$
4	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{3}{36}$
5	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{4}{36}$
6	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{5}{36}$
7	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{6}{36}$
8	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{5}{36}$
9	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{4}{36}$
10	$\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{3}{36}$
11	$\frac{1}{36}+\frac{1}{36}=\frac{2}{36}$
12	$\frac{1}{36}$
Total	1

Requirements for a Probability Distribution:

• The sum of probabilities of all events in the sample space must be equal to 1. That is, ${\displaystyle \sum P\left(X\right)}=1$
• The probability values of each event in the sample space must be between or equal to 0 and 1. That is, $0\le P\left(X\right)\le 1$

The random variable X represents the sum of the two numbers and these values have their corresponding probabilities. Here, the sum of all probabilities equals to 1. Also, all the individual probabilities lie between 0 and 1.

Mean:

The mean for the probability distribution is calculated using the formula:

$\mu ={\displaystyle \sum X\cdot P\left(X\right)}$

Where each value of X is multiplied by its corresponding probability and all product terms are added to get the mean.

The mean for the distribution is as follows:

Sum of two dice X	Probability $P\left(X\right)$	$X\cdot P\left(X\right)$
2	0.028	0.056
3	0.056	0.168
4	0.0833	0.332
5	0.111	0.555
6	0.139	0.834
7	0.167	1.169
8	0.139	1.112
9	0.111	0.999
10	0.083	0.830
11	0.056	0.616
12	0.028	0.336
Total	1	${\displaystyle \sum X}\cdot P\left(X\right)=7.000$

Thus, the mean for the distribution is approximately 7.000.

Variance:

The formula to find the variance for the probability distribution is,

${\sigma }^2={\displaystyle \sum \left[X^2\cdot P\left(X\right)\right]}-{\mu }^2$

Here,

Sum of two dice X	Probability $P\left(X\right)$	$X^{2}P\left(X\right)$
2	0.028	0.112
3	0.056	0.504
4	0.833	1.328
5	0.111	2.775
6	0.139	5.004
7	0.167	8.183
8	0.139	8.896
9	0.111	8.991
10	0.833	8.3
11	0.056	6.776
12	0.028	4.032
Total	1	54.901

Substitute 54.901 for ${\displaystyle \sum \left[X^2\cdot P\left(X\right)\right]}$ and 7.000 for $\mu$ in variance formula

$\begin{array}{c}{\sigma }^2=54.901-{\left(7.000\right)}^2\\ =54.901-49.000\\ =5.901\end{array}$

Thus, the variance for the distribution is 5.901.

Standard deviation:

The formula to find the standard deviation for the probability distribution is,

$\sigma =\sqrt{{\sigma }^2}$

Substitute 5.901 for ${\sigma }^2$

$\begin{array}{c}\sigma =\sqrt{5.901}\\ =2.429\end{array}$

Thus, the standard deviation for the distribution is 2.429.