Expected value and standard deviation of a pmf function

Question

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?

Answer 1

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 $\frac{6+7+\dots+12}{91}=\frac9{13}$.

Answer 2

This is not a binomial distribution.

Guide:

First we have $$\sum_{x=0}^N (Kx)=1$$

$$\frac{KN(N+1)}{2}=1$$

Also, the mean can be computed as follows:

$$E[X] = \sum_{x=0}^N x(Kx)= K \sum_{x=0}^Nx^2=9$$

$$\frac{N(N+1)(2N+1)}{6}=\frac9{K}$$ Hence $$\frac{2N+1}{3}=9$$

I will leave solving of $K$ and $N$ as an exercise.

After that, you can compute the variance using $E[X^2]-E[X]^2$.