Question

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

• A. $\frac{31}{6}$ and $\sqrt{\frac{31}{6}}$
• B. $\frac{35}{6}$ and $\sqrt{\frac{35}{6}}$
• C. $\frac{17}{6}$ and $\sqrt{\frac{17}{6}}$
• D. $\frac{31}{6}$ and $\sqrt{\frac{35}{6}}$

Answer 1

Answer: (B)

$\frac{35}{6}$ and $\sqrt{\frac{35}{6}}$

Let x denote the sum of the numbers obtained when two fair dice are rolled.
So, X may have values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
$P(X=2)=P(1,1)=\frac{1}{36}$
$P(X=3)=P\left\{\right(1,2),(2,1\left)\right\}=\frac{2}{36}$

$P(X=4)=\frac{3}{36};P(X=5)=\frac{4}{36}$

$P(X=6)=\frac{5}{36};P(X=7)=\frac{6}{36};P(X=8)=\frac{5}{36}$

$P(X=9)=\frac{4}{36};P(X=10)=\frac{3}{36};P(X=11)=\frac{2}{36};$
$P(X=12)=\frac{1}{36}$
$\therefore$ Probability distribution table is given below

X $23456789101112$

P(X) $\frac{1}{36}\frac{2}{36}\frac{3}{36}\frac{4}{36}\frac{5}{36}\frac{6}{36}\frac{5}{36}\frac{4}{36}\frac{3}{36}\frac{2}{36}\frac{1}{36}$

Mean $\overline{X}=\sum XP\left(X\right)$

$=\frac{[2\times 1+3\times 2+4\times 3+5\times 4+\times 6\times 5+7\times 6+8\times 5+9\times 4+10\times 3+11\times 2+12\times 1]}{36}$
$=\frac{252}{36}=7$
Variance $=\sum X^{2}P\left(X\right)-\overline{X^2}$

$=\frac{[2^2\times 1+3^2\times 2+4^2\times 3+5^2\times 4+6^2\times 5+7^2\times 6+8^2\times 5+9^2\times 4+{10}^2\times 3+{11}^2\times 2+{12}^2\times 1]}{36}-7^2$
$=\frac{1974}{36}-49$

$=\frac{1974-1764}{36}$

$=\frac{210}{36}=\frac{35}{6}$
$\therefore$ Variance $=\frac{35}{6}$

And, $SD=\sqrt{\frac{35}{6}}$