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Sagot :
What I gather from the question is that [tex]X[/tex] has second moment [tex]E(X^2)=81[/tex] and variance [tex]V(X) = 58[/tex], and you're asked to find the expectation and variance of the random variable [tex]Y=2X+10[/tex].
From the given second moment and variance, we find the expectation of [tex]X[/tex] :
[tex]V(X) = E(X^2) - E(X)^2 \implies E(X) = \sqrt{E(X^2) - V(X)} = \sqrt{23}[/tex]
Expectation is linear, so
[tex]E(Y) = E(2X+10) = 2 E(X) + 10 = \boxed{2\sqrt{23} + 10}[/tex]
Using the same variance identity, we have
[tex]V(Y) = V(2X+10) = E((2X+10)^2) - E(2X+10)^2[/tex]
and
[tex]E((2X+10)^2) = E(4X^2 + 40X + 100) = 4E(X^2) + 40E(X) + 100 = 424 + 40\sqrt{23}[/tex]
so that
[tex]V(Y) = V(2X+10) = (424 + 40\sqrt{23}) - (2\sqrt{23} + 10)^2 = \boxed{232}[/tex]
Alternatively, we can use the identity
[tex]V(aX+b) = a^2 V(X) \implies V(2X+10) = 4V(X) = 232[/tex]
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