Question

For a $B.D.$, variance is given by the formula

• A. ${\sigma }^2=np{q}^2$
• B. ${\sigma }^2=\frac{\sqrt{npq}}{2}$
• C. ${\sigma }^2=npq$
• D. ${\sigma }^2=n^2{p}^2{q}^2$

Answer 1

Answer: (C)

Let $x$ be the variate which assumes the values
$0,1,2,....,n$ with frequencies
$q^n,{}^n{C}_{1}p{q}^{n-1},{}^n{C}_2{p}^2{q}^{n-2},...,p^n$
such that $p+q=1$
and ${\sigma }^2($ variance) $=\frac{\sum f_i{x}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{x}_i}{\sum f_i}\right)}^2$
where $\Sigma f_i=q^n+{}^n{C}_{1}p{q}^{n-1}+{}^n{C}_2{p}^2{q}^{n-2}+...+p^n=1$
Now,

Now, $\sum f_i=1,\frac{\sum f_i{x}_i}{\sum f_i}=np$ and $\sum f_i{x}_i^2=\sum r^2{}^n{C}_r{p}^r{q}^{n-r}$
Now, mean $=np$
and ${\sigma }^2$ (variance) $=\sum _{r=0}^n{r}^2{}^n{C}_r{p}^r{q}^{n-r}-(np{)}^2$
$=\sum _{r=0}^n[r(r-1)+r]{}^n{C}_r{p}^r{q}^{n-r}-n^2{p}^2$
? $=\sum _{r=0}^{n}r(r-1){}^n{C}_r{p}^r{q}^{n-r}+=\sum _{r=0}^n{r}^n{}^n{C}_r{p}^r{q}^{n-r}-n^2{p}^2$
$=\sum _{r=0}^{n}r(r-1)\frac{n(n-1)}{r(r-1)}{}^{n-2}{C}_{r-2}{p}^r{q}^{n-r}$
$+=\sum _{r=0}^{n}r\frac{n}{r}{}^{n-1}{C}_{r-1}{p}^r{q}^{n-r}-n^2{p}^2$
$=n(n-1)p^2(p+q{)}^{n-2}+np(p+q{)}^{n-1}-n^2{p}^2$
$\Rightarrow {\sigma }^2=np(1-p)=npq$