Question

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

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

Answer 1

Answer: (C)

Let $x$ be the variate which assumes the values
$0, 1,2,....,n$ with frequencies
$q^n, \,{}^nC_1 pq^{n - 1}, \,{}^nC_2 p^2q^{n - 2},..., p^n$
such that $p + q = 1$
and $\sigma^2($ variance) $= \frac{\sum f_{i}x^{2}_{i}}{\sum f_{i}} - \left(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\right)^{2}$
where $\Sigma f_i = q^n + \,{}^nC_1 pq^{n-1} + \,{}^nC_2p^2q^{n - 2} + ... +p^n = 1$
Now,

Now, $\sum f_i = 1, \frac{\sum f_ix_i}{\sum f_i}= np$ and $\sum f_ix_i^2 = \sum r^2 \,{}^nC_r p^r q^{n-r}$
Now, mean $= n p$
and $\sigma^2$ (variance) $ = \displaystyle\sum_{r = 0}^{n} r^2 \,{}^nC_r p^r q^{n-r} - (np)^2$
$ = \displaystyle\sum_{r = 0}^{n} [r(r - 1) + r]\,{}^nC_r p^r q^{n-r} - n^2 p^2$
? $ = \displaystyle\sum_{r = 0}^{n} r(r-1)\,{}^nC_r p^rq^{n-r} + = \displaystyle\sum_{r = 0}^{n} r^n\,{}^nC_r p^rq^{n-r} - n^2p^2$
$ = \displaystyle\sum_{r = 0}^{n} r(r-1) \frac{n(n-1)}{r(r-1)} \,{}^{n-2}C_{r-2} p^r q^{n-r}$
$+ = \displaystyle\sum_{r = 0}^{n} r \frac{n}{r} \,{}^{n-1}C_{r-1} p^r q^{n-r} - n^2p^2$
$= n(n - 1) p^2(p + q)^{n-2} + np(p + q)^{n- 1} - n^2p^2$
$\Rightarrow \sigma^2 = np(1 - p) = npq$