A discrete random variable has the PMF of x/15 for x = 1,2,3,4,5 and p(x) = 0 for all other values. $Z = 3X^2 +1$. what is Var(Z)?

Question

A discrete random variable has the PMF of $x/15$ for $x \in\{ 1,2,3,4,5\}$ and $p(x) = 0$ for all other values. $Z = 3X^2 +1$. what is $\mathsf {Var}(Z)$?

When I solve for $\mathsf E(Z)$ I get $3[1/225 + 4/225 + 9/225 + 16/225 + 25/225] + 1$.

Then when I solve for $\mathsf E(Z^2)$ I get $3^2[(1/225)^2 + (4/225)^2 + (9/225)^2 + (16/225)^2 + (25/225)^2] + 1^2$.

Then $\mathsf {Var}(Z) = \mathsf {Var}(aX+b) = 3^2(Var(Z))$.

Which $\mathsf {Var}(Z) = \mathsf E(Z^2) - (\mathsf E(Z))^2$.

When I plug in the value I get a negative value. HELP! I know variance is a non-negative number.

Answer 1

Edit: $\color{navy}{\text{The original post has been updated. Added new text to also answer the amended post.}}$

Why are you squaring the probability to find $\mathsf E(Z)$?   You just need: $$\mathsf E(3X+1) = 1 + 3 \sum_{x=1}^5 \frac {x^2}{15}$$

$$\color{navy}{\mathsf E(3X^2+1) = 1 + 3 \sum_{x=1}^5 \frac {x^3}{15}}$$

Your formula for $\mathsf E[(3X+1)^2]$ is further erroneous.   You have attempted to find $\mathsf E[(3X)^2+1]$ instead.   What you need is:

$$\mathsf E((3X+1)^2) = \sum_{x=1}^{5}\frac{(3x+1)^2x}{15}$$

$$\color{navy}{\mathsf E((3X^2+1)^2) = \sum_{x=1}^{5}\frac{(3x^2+1)^2x}{15}}$$

Answer 2

I do not undestand, where you get $\frac{1}{225}$. Lets count $EX$ $$EX=1\frac{1}{15}+2\frac{2}{15}+3\frac{3}{15}+4\frac{4}{15}+5\frac{5}{15}=\frac{55}{15},$$ and now count $$EX^2=\sum_{i=1}^{5}i^2\frac{i}{15}=\frac{225}{15},$$ therefore $$var(X)=\frac{225}{15}-\frac{55^2}{15^2}>1.5$$ Now can you do the same for $Z$.