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Suppose the random variable X has the probability distribution given below:

 X -2 -1 0 1 2 P(X=X) 0.25 0.2 0.15 0.35 0.05

Let $Y=(2*(X^2))+6$.The expected value E(Y) is:

A) 9.5     B) 6.      C )15.5.        D )18

edited | 115 views

$Y = 2(X^2) + 6$

if X = 2 or X = -2 $\rightarrow$ Y = 14

X = 1 or X = -1 $\rightarrow$ Y = 8

X = 0 , Y$\rightarrow$ 6

$P(14) = 0.25 + 0.05 = 0.30$

$P(8) = 0.20+0.35 = 0.55$

$P(6) = 0.15$

$E(Y) = 14 \times 0.30 + 8 \times 0.55 + 6 \times 0.15 = 9.5$
by Boss (36.7k points)
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Thank you sir
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@ Utkarsh

Can you please explain how to attempt these type of problems  and why are putting  x = 0,1,2,-2?

How E(y) = 14 * 0.30 +8 *0.5 + 6?

Thanks
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Mayankprakash This question is just asking for basics of random variable.

Y is a dependent random variable which depends on value of X. So by substituting the value of X you can calculate value of Y and E(Y) is expectation(mean) of random variable Y.