Question

If the variable takes the values $0,1,2,\cdots ,n$ with frequencies proportional to $q^n{,}^n{C}_r{q}^{n-1}p{,}^n{C}_2{q}^{n-2}{p}^2,\cdots ,p^n$ where $p+q=1,$ then show that mean square deviation is $n^2{p}^2+npq$ and variance is $npq.$

Answer 1

${\mu }_1^{\prime }=E\left(x\right)=\sum _{x=0}^{0}x\left(\frac{n}{x}\right)p^x{q}^{n-x}$
$\left(\frac{n}{x}\right){=}^n{C}_x=\frac{n}{x}\left(\frac{n-1}{x-1}\right)$ $=np\sum _{x-1}^n\left(\frac{n-1}{x-1}\right)p^{x-1}{q}^{(n-1)-(x-1)}$
$=np{(p+q)}^{n-1}=np$ $\therefore p+q=1$
mean square deviation about $x=0$ is
${\mu }_2=E\left(x^2\right)=\sum _{x=0}^n{x}^2\left(\frac{n}{x}\right)p^x{q}^{n-x}$
$\sum _{x=0}^n[x(x-1)+1]\left(\frac{n}{x}\right)p^x{q}^{n-x}$
$\sum _{x=0}^{n}x(x-1)(\frac{n}{x}{p}^x{q}^{n-x}+\sum _{x=0}^n)x\left(\frac{n}{x}\right)p^x{q}^{n-x}$
$n(n-1)p^2\sum _{x=2}^n\left(\frac{n-2}{n-2}\right)p^{x-2}{q}^{(n-2)-(x-2)}+np$
$n(n-1)p^2{(q+p)}^{n-2}+np$ $n(n-1)p^2+np=n^2{p}^2+np(1-p)$
$=n^2{p}^2+npq$ $\therefore p+q=1$variance =${\mu }_2=\mu \frac{1}{2}-{\left({\mu }_1^{\prime }\right)}^2$
$=n^2{p}^2+npq-{\left(np\right)}^2=npq$