QJEEM21Ma3913: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3913 Solution: Let X be a random variable with distribution. .tg…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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

\overline X

= 2.3

-

a + 6b =

{9 \over {10}}

..... (1)

\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1}

a + b = {4 \over {15}}

.... (2)

From equation (1) and (2)

a = {1 \over {10}},b = {1 \over 6}

{\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}}

{1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}

= {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}

= 4 + a + 3 + 36b - {(2.3)^2}

= 7 + a + 36b - {(2.3)^2}

= 7 + {1 \over {10}} + 6 - {(2.3)^2}

= 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}

= {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}

= {{1310 - {{(23)}^2}} \over {100}}

= {{1310 - 529} \over {100}}

{\sigma ^2} = {{781} \over {100}}

100{\sigma ^2} = 781

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