Let X be a random variable with distribution.

x	$-$2	$-$1	3	4	6
P(X = x)	${1 \over 5}$	a	${1 \over 3}$	${1 \over 5}$	b

If the mean of X is 2.3 and variance of X is $\sigma$², then 100 $\sigma$² is equal to :

<table class="tg"> <thead> <tr> <th class="tg-baqh">x</th> <th class="tg-baqh">$-$2</th> <th class="tg-baqh">$-$1</th> <th class="tg-baqh">3</th> <th class="tg-baqh">4</th> <th class="tg-baqh">6</th> </tr> </thead> <tbody> <tr> <td class="tg-baqh">P(X = x)</td> <td class="tg-baqh">${1 \over 5}$</td> <td class="tg-baqh">a</td> <td class="tg-baqh">${1 \over 3}$</td> <td class="tg-baqh">${1 \over 5}$</td> <td class="tg-baqh">b</td> </tr> </tbody> </table><br><br>$\overline X$ = 2.3<br><br>$-$a + 6b = ${9 \over {10}}$ ..... (1)<br><br>$\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1}$<br><br>$a + b = {4 \over {15}}$ .... (2)<br><br>From equation (1) and (2)<br><br>$a = {1 \over {10}},b = {1 \over 6}$<br><br>${\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}}$<br><br>${1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}$<br><br>$= {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}$<br><br>$= 4 + a + 3 + 36b - {(2.3)^2}$<br><br>$= 7 + a + 36b - {(2.3)^2}$<br><br>$= 7 + {1 \over {10}} + 6 - {(2.3)^2}$<br><br>$= 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$<br><br>$= {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$<br><br>$= {{1310 - {{(23)}^2}} \over {100}}$<br><br>$= {{1310 - 529} \over {100}}$<br><br>${\sigma ^2} = {{781} \over {100}}$<br><br>$100{\sigma ^2} = 781$