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The probability distribution of a discre...

The probability distribution of a discrete random variable x is given as under

Calculate
(i) the value of A, if E(X)=2.94.
(ii) variance of X.

Text Solution

Verified by Experts

(i) We have,`sumXP(X)=1/2+2/5+12/25+(2A)/10+(3A)/25+(5A)/25`
`=(25+20+24+10A+6A+10A)/50=(69+26A)/50`
Since, E(X)=`sumXP(X)`
`rArr2.94=(69+26A)/50`
`rArr 26A=50xx2.94-69`
`rArrA=(147-69)/26=78/26=3`
(ii) We know that,
Var(X)=`E(X^(2))-[E(X)]^(2)`
`=sumX^(2)P(X)-[sumXP(X)]^(2)`
`=1/2+4/5+48/25+(4A^(2))/10+(9A^(2))/25+(25A^(2))/25-[E(X)]^(2)`
`=(20+40+96+20A^(2)+18A^(2)+50A^(2))/50-[E(X)]^(2)`
`=(161+88A^(2))/50-[E(C)]^(2)=(161+88xx(3)^(2))/50-[E(X)]^(2)` `[becauseA=3]`
`=953/50-[2.94]^(2)` [`becauseE(X)=2.94`]
=19.0600-8.6436=10.4164
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