Calculate variance for a probability mass function In a family of ten people, the probability mass function of the number of people
who contracted the flu is given by
$$P_X(x) = K(2x + 9); x = 0,1,..., 10$$
Calculate the variance of the number of people with flu in the family.
My attempt:
The first thing I needed to find was "K". Thus $$\sum_{x=0}^{10}{P_X(x)=1}$$
$$\sum_{x=0}^{10}{K(2x+9)=1}$$
$$\frac{2k.10(10+1)}{2} + 90K = 1$$
$$K=\frac{1}{200}$$
And then use $V(X)=E(X^2)-{E(X)}^2=47.575-(6.325)^2=7.5693$
$$E(X)=\sum_{x=0}^{10}{\frac{x(2x+9)}{200}}=6.325$$
$$E(X^2)=\sum_{x=0}^{10}{\frac{x^2(2x+9)}{200}}=47.575$$
For y second attempt, I have $K=\frac{1}{209}$. What am I missing? Am I on the right path?

Note that$$\sum_{x=0}^{10}9=9(11)=99$$
$$10K(11)+99K=1$$
$$110K+99K=1$$
$$209K=1$$
$$K=\frac1{209}$$
\begin{align}
E(X^2) &= K \sum_{x=0}^{10} x^2(2x+9)\\
&=K\left( 2\left(\frac{(10)(11)}{2}\right)^2+9\cdot\frac{(10)(11)(21)}{6}\right) \\
&= 10(11)K\left(\frac{10(11)}{2} + \frac{3(21)}{2} \right)\\
&=\frac{10(11)}{2}K\left(110+63 \right)\\
&=9515K
\end{align}
\begin{align}
E(X) &= K \sum_{x=0}^{10} x(2x+9)\\
&=K\left( 2\cdot\frac{(10)(11)(21)}{6}+9\left(\frac{(10)(11)}{2}\right)\right) \\
&= 10(11)K\left(7 + \frac{9}{2} \right)\\
&=\frac{10(11)}{2}K\left(23 \right)\\
&=1265K
\end{align}
$$Var(X)=E(X^2)-(E(X))^2=9515K-(1265K)^2$$
