Question

Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and the standard deviation of X.

Answer 1

Hint:For solving this problem we need to have a clear understanding of what variance and standard deviation signify. By calculating the probability distribution and using the standard formula we get that the variance and standard deviation of X are $\dfrac{35}{6}and\sqrt{\dfrac{35}{6}}$ respectively.

Complete step-by-step solution:
Unlike range and quartiles, the variance combines all the values in a data set to produce a measure of spread. The variance and standard deviation (the square root of the variance) are the most commonly used measures of spread. We know that variance is a measure of how spread out a data set is. It is calculated as the average squared deviation of each number from the mean of a data set. Standard deviation is the measure of spread most commonly used in statistical practice when the mean is used to calculate central tendency. Thus, it measures spread around the mean. Because of its close links with the mean, standard deviation can be greatly affected if the mean gives a poor measure of central tendency.
According to the given problem, X denotes the sum of the numbers obtained when two fair dice are rolled. So, X may have values \[2,\text{ }3,\text{ }4,\text{ }5,\text{ }6,\text{ }7,\text{ }8,\text{ }9,\text{ }10,\text{ }11\text{ }and\text{ }12\] . This is so because the maximum and minimum number obtained while rolling a single dice is one and six respectively. Hence the maximum and minimum number obtained while rolling a single dice is two and twelve respectively.

X	Outcomes	Number of outcomes	Probability P(X)
$2$	\[\left( 1,1 \right)\]	$1$	\[\dfrac{1}{36}\]
$3$	\[\left( 1,2 \right),\text{ }\left( 2,1 \right)\]	$2$	\[\dfrac{2}{36}\]
$4$	\[\left( 1,3 \right),\text{ }\left( 2,2 \right),\text{ }\left( 3,1 \right)\]	$3$	\[\dfrac{3}{36}\]
$5$	\[\left( 1,4 \right),\text{ }\left( 2,3 \right),\text{ }\left( 3,2 \right),\text{ }\left( 4,1 \right)\]	$4$	\[\dfrac{4}{36}\]
$6$	\[\left( 1,5 \right),\text{ }\left( 2,4 \right),\text{ }\left( 3,3 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 5,1 \right)\]	$5$	\[\dfrac{5}{36}\]
$7$	\[\left( 1,6 \right),\text{ }\left( 2,5 \right),\text{ }\left( 3,4 \right),\text{ }\left( 4,\text{ }3 \right),\text{ }\left( 5,2 \right),\text{ }\left( 6,1 \right)\]	$6$	\[\dfrac{6}{36}\]
$8$	\[\left( 2,6 \right),\text{ }\left( 3,5 \right),\text{ }\left( 4,4 \right),\text{ }\left( 5,\text{ }3 \right),\text{ }\left( 6,2 \right)\]	$5$	\[\dfrac{5}{36}\]
$9$	\[\left( 3,6 \right),\text{ }\left( 4,5 \right),\text{ }\left( 5,4 \right),\text{ }\left( 6,\text{ }3 \right)\]	$4$	\[\dfrac{4}{36}\]
$10$	\[\begin{array}{*{35}{l}} \left( 4,6 \right),\text{ }\left( 5,5 \right),\text{ }\left( 6,4 \right) \\\end{array}\]	$3$	\[\dfrac{3}{36}\]
$11$	\[\left( 5,6 \right),\text{ }\left( 6,5 \right)\]	$2$	\[\dfrac{2}{36}\]
$12$	\[\left( 6,6 \right)\]	$1$	\[\dfrac{1}{36}\]

\[\begin{align}
& Mean\,X=\sum XP\left( X \right) \\
& =\text{ }\dfrac{\left[ 2\times 1+3\times 2+4\times 3+5\times
4+\times 6\times 5+7\times 6+8\times 5+9\times 4+10\times 3+11\times 2+12\times 1 \right]}{36} \\
& \Rightarrow Mean \,X=\dfrac{252}{36}=7 \\
\end{align}\]
\[\begin{align}
& Variance=\sum {{X}^{2}}P\left( X \right)-{{{\bar{X}}}^{2}} \\
& =\text{ }\dfrac{\left[ {{2}^{2}}\times 1+{{3}^{2}}\times 2+{{4}^{2}}\times 3+{{5}^{2}}\times
4+{{6}^{2}}\times 5+{{7}^{2}}\times 6+{{8}^{2}}\times 5+{{9}^{2}}\times 4+{{10}^{2}}\times
3+{{11}^{2}}\times 2+{{12}^{2}}\times 1 \right]}{36}-72 \\
& \Rightarrow Variance=\dfrac{1974}{36}-49=\dfrac{210}{36} \\
& \therefore Variance=\dfrac{35}{6} \\
& Standard\text{ }Deviation=\sqrt{Variance}=\sqrt{\dfrac{35}{6}} \\
\end{align}\]

Note: These types of problems may seem simple but there are high chances of miscalculations due to the lengthy calculations for variance and standard deviation. We need to carefully perform the calculations after using the correct formula.