What I gather from the question is that
has second moment
and variance
, and you're asked to find the expectation and variance of the random variable
.
From the given second moment and variance, we find the expectation of
:
![V(X) = E(X^2) - E(X)^2 \implies E(X) = √(E(X^2) - V(X)) = √(23)](https://img.qammunity.org/2023/formulas/mathematics/college/n53295k7u2cliqze99li2dxg3hbru0960y.png)
Expectation is linear, so
![E(Y) = E(2X+10) = 2 E(X) + 10 = \boxed{2√(23) + 10}](https://img.qammunity.org/2023/formulas/mathematics/college/qpmqhbhpxpzav16pvhdjb0v75qw0ua45mh.png)
Using the same variance identity, we have
![V(Y) = V(2X+10) = E((2X+10)^2) - E(2X+10)^2](https://img.qammunity.org/2023/formulas/mathematics/college/1ojh9kz2xlgp16rdto6a3ph5zu0oxtnbok.png)
and
![E((2X+10)^2) = E(4X^2 + 40X + 100) = 4E(X^2) + 40E(X) + 100 = 424 + 40√(23)](https://img.qammunity.org/2023/formulas/mathematics/college/lfek6atfjy294kgpkp76rz5s0yzizo2zm9.png)
so that
![V(Y) = V(2X+10) = (424 + 40√(23)) - (2√(23) + 10)^2 = \boxed{232}](https://img.qammunity.org/2023/formulas/mathematics/college/9fybu5j6yr16lht1pb3h7fdoyy0qqklbzm.png)
Alternatively, we can use the identity
![V(aX+b) = a^2 V(X) \implies V(2X+10) = 4V(X) = 232](https://img.qammunity.org/2023/formulas/mathematics/college/p6whdjwvzwqu855wgxkljax4aanzmktfkw.png)