• # question_answer If a variable takes values $0,\,\,1,\,\,2,...\,,n$ with frequencies ${{q}^{n}},\,\frac{n}{1}{{q}^{n-1}}p,\,\frac{n(n-1)}{1\cdot 2}{{q}^{n-2}}{{p}^{2}},...,\,{{p}^{n}}$ where $p+q=1,$ then the mean is A)  $np$                           B)  $nq$ C)  $n(p+q)$                   D)  None of these

$\therefore$ $\overline{x}\frac{...+n\cdot {{p}^{n}}]}{{{q}^{n}}+\frac{n}{1}{{q}^{n-1}}p+\frac{n(n-1)}{2}{{q}^{n-2}}{{p}^{2}}+...+{{p}^{n}}}$ $=\frac{{{0.}^{n}}{{C}_{0}}{{q}^{n}}{{p}^{0}}+1\cdot {{\,}^{n}}{{C}_{1}}{{q}^{n-1}}p+...+{{n}^{n}}{{C}_{n}}{{q}^{0}}{{p}^{n}}}{^{n}{{C}_{0}}{{q}^{n}}{{p}^{0}}{{+}^{n}}{{C}_{1}}{{q}^{n-1}}{{p}^{1}}+...{{+}^{n}}{{C}_{n}}{{q}^{n-n}}{{p}^{n}}}$ $=\frac{\sum\limits_{r\,=\,0}^{n}{r{{\cdot }^{n}}{{C}_{r}}{{q}^{n-r}}{{d}^{r}}}}{\sum\limits_{r\,=\,0}^{n}{^{n}{{C}_{r}}{{q}^{n-r}}{{p}^{r}}}}$ $=\frac{\sum\limits_{r\,=\,0}^{n}{r\cdot \frac{n}{r}{{\,}^{n-1}}{{C}_{r-1}}{{q}^{n-r}}\,p\cdot {{p}^{r-1}}}}{\sum\limits_{r\,=\,0}^{n}{^{n}{{C}_{r}}{{q}^{n-r}}{{p}^{r}}}}$ $=\frac{np{{(p+q)}^{n-1}}}{{{(q+p)}^{n}}}=np$    $(\because \,\,p+q=1)$