Probability, linearity of expectation

Question

Hello, I would like to ask if part d) was solved right specifically the Var[Y] part.

Answer 1

(a)

We must have $\sum_xp_X(x)=1$ where $x$ ranges over the values with positive probability.

This leads to $p_X(4.1)=0.1$

(c)

$\mathbb EX=\sum_xxp_X(x)=-9.1\times0.1-6.5\times0.2-3.4\times0.2+1.2\times0.4+4.1\times0.1=-2$

$\mathsf{Var}(X)=\mathbb EX^2-(\mathbb EX)^2=\sum_xx^2p_X(x)-(-2)^2=21.3-4=17.3$

or alternatively:

$\mathsf{Var}(X)=\mathbb E(X-\mathbb EX)^2=\sum_x(x+2)^2p_X(x)=17.3$

(d)

$\mathbb{E}Y=\mathbb{E}\left(9-3X\right)=9-3\mathbb{E}X=9-3\times\left(-2\right)=15$

$\mathsf{Var}Y=\mathsf{Var}\left(9-3X\right)=\left(-3\right)^{2}\mathsf{Var}X=9\times17.3=155.7$

$\mathbb EY^2=\mathsf{Var}Y+(\mathbb EY)^2=155.7+225=380.7$

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Check yourself (and also check me).