Question

For a Binomial distribution mean and variance is given by

• A. $np,npq$
• B. $n^{2}p,N^2{p}^2{q}^2$
• C. $n^2{p}^2,N^2{p}^2{q}^2$
• D. None of these

Answer 1

The correct option is A $np,npq$

The Probabilty function for Binomial Distribution

$B(n,p)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{(1-p)}^{n-k}E\left[X\right]={\sum }_{k=0}^{n}k\ast \left(\begin{array}{c}n\\ k\end{array}\right)p^k{(1-p)}^{n-k}E\left[X\right]={\sum }_{k=1}^n\frac{n!}{(k-1)!(n-k)!}{p}^k{(1-p)}^{n-k}$

Let $y=k_1$ and $m=n_1$. Subbing $k=y+1$ and $n=m+1$ into the last sum(and using the fact that the limits $x=1$ and $x=n$ correspond to $y=0$ and $y=n_1=m$, respectively)

$E\left[X\right]={\sum }_{y=0}^n\frac{(m+1)!}{\left(y\right)!(m-y)!}{p}^y{(1-p)}^{m-y}E\left[X\right]=p(m+1){\sum }_{y=0}^n\frac{\left(m\right)!}{\left(y\right)!(m-y)!}{p}^y{(1-p)}^{m-y}E\left[X\right]=np{\sum }_{y=0}^n\frac{\left(m\right)!}{\left(y\right)!(m-y)!}{p}^y{(1-p)}^{m-y}$

Acc to binomial theorem

${(a+b)}^m={\sum }_{y=0}^n\frac{\left(m\right)!}{\left(y\right)!(m-y)!}{a}^y{\left(b\right)}^{m-y}$

Setting $a=p$ and $b=p$

$E\left[X\right]=np{\sum }_{y=0}^n\frac{\left(m\right)!}{\left(y\right)!(m-y)!}{p}^y{(1-p)}^{m-y}=np{(p+1-p)}^m=np$

Similarly, but this time using $y=x_2$ and $m=n_2$

$E\left[X\right(X-1\left)\right]={\sum }_{k=0}^{n}k(k-1)\ast \left(\begin{array}{c}n\\ k\end{array}\right)p^k{(1-p)}^{n-k}$

On Solving with above procedure

$E\left[X\right(X-1\left)\right]=n(n-1)p^2$

So Variance $=E\left[X^2\right]-{E\left[X\right]}^2=EX\left(X1\right)+E\left(X\right)-{E\left[X\right]}^2=n\left(n1\right)p^2+np-(np{)}^2=np(1-p)$