Calculate variance for a probability mass function

Question

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?

Answer 1

All the steps seem correct and you have expressed them clearly.

Answer 2

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}

$$Var(X)=E(X^2)-(E(X))^2=9515K-(1265K)^2$$