Expected value and standard deviation of a pmf function 
In a family, the probability mass function of the number of people $x$ who have contracted the flu is given by $$P(x) = Kx \qquad x\in\{0,1,\ldots,N\}$$ where $N$ is the number of people in the family.
(a) If nine people are expected to have flu in the family, calculate $K$ and $N$.
(b) Calculate the probability that the number of people in the family with flu is within one standard deviation of the mean.

My attempt:
This is a binomial distribution with probability parameter $p=0.5$ and mean $\mu=9$; hence $N = \mu/p=9/0.5 = 18$.
Now summing $Kx$ from $x = 0$ to $x=18$ gives $$\begin{align}
K + 2K + 3K+ \cdots + 18K &= 1 \\
K(1+ 2+ 3+ \cdots +18) &= 1 \\
171K &= 1 \end{align}$$
Hence $K = 1/171$.
The standard deviation for this distribution is $$Np(1-p) = 18\times 0.5\times0.5 = 4.5$$
Am I on the right path?

This is not a binomial function, more like a triangular one. We cannot immediately derive $N$ from $K$ or the other way round, but we can derive two equations first. For the expected value:
$$1\cdot K+2\cdot2K+\dots+N\cdot NK=9$$
$$K(1^2+2^2+\dots+N^2)=9$$
For the pmf summing to 1:
$$K(1+2+\dots+N)=1$$
Thus we have
$$\frac{1^2+2^2+\dots+N^2}{1+2+\dots+N}=9$$
$$\frac{2N(N+1)(2N+1)}{6N(N+1)}=\frac{2N+1}3=9\qquad N=13$$
Then $K(1+2+\dots+N)=91K=1$ and $K=\frac1{91}$.
The standard deviation is $\sqrt{\frac{1(1-9)^2+2(2-9)^2+\dots+13(13-9)^2}{91}}=\sqrt{10}=3.16\dots$ Thus one standard deviation is anywhere between 6 and 12 inclusive, and the probability of this happening is $\frac{6+7+\dots+12}{91}=\frac9{13}$.
