Discussion Forum

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

Answer:

Explanation:

Ley 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

 $P(X=2)=P(1,1)=\frac{1}{36}$

 $P(X=3)=P{(1,2),(2,1)}=\frac{2}{36}$

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

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

  $P(X=8)=\frac{5}{36}$,

$P(X=9)=\frac{4}{36}$,$P(X=10)=\frac{3}{36}$,$P(X=11)=\frac{2}{36}$,

$P(X=12)=\frac{1}{36}$,

$\therefore$ Probability distribution table is given below

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

=$\frac{\left[2\times1+3\times2+4\times3+5\times4+6\times5+7\times6+8\times5+9\times4+10\times3+11\times2+12\times1\right]}{36}$

 $=\frac{252}{36}=7$

 Variance = $\sum X^{2} P(X)-\overline{X}^{2}$

$\frac{\left[2^{2}\times1+3^{2}\times2+4^{2}\times3+5^{2}\times4+6^{2}\times5+7^{2}\times6+ 8^{2}\times5+9^{2}\times4+10^{2}\times3+11^{2}\times2+12^{2}\times1\right]}{36}$-$7^{2}$

 =$\frac{1974}{36}-49$

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

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

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

 and SD=$\sqrt{\frac{35}{6}}$