Question

If $a$ variable takes values $0,1,2,...,n$ with frequencies $q^n,\frac{n}{1}{q}^{n-1}p,\frac{n(n-1)}{1.2}{q}^{n-2}{p}^2,...,p^n$, where $p+q=1$, then the mean is

• A. $np$
• B. $nq$
• C. $n(p+q)$
• D. None of these

Answer 1

Answer: (A)

$np$

The required mean is
$\Rightarrow \overline{x}=\frac{0.q^n+1.\frac{n}{1}{q}^{n-1}p+2.\frac{n(n-1)}{2!}{q}^{n-2}{p}^2+...+n.p^n}{q^n+\frac{n}{1}{q}^{n-1}p+\frac{n(n-1)}{2!}{q}^{n-2}{p}^2+...+p^2}$
$\Rightarrow \overline{x}=\frac{{0.}^n{C_0{q}^n{p}^0+1.}^n{C}_1{q}^{n-1}p+...+n{.}^n{C}_n{p}^0{p}^n}{{}^n{C_0{q}^n{p}^0+}^n{C_1{q}^{n-1}{p}^1+...+}^n{C}_n{q}^{n-n}{p}^n}$
$\Rightarrow \overline{x}=\frac{{\sum }_{r=0}^{n}r{.}^n{C}_r{q}^{n-r}{p}^r}{{\sum }_{r=0}^n{}^n{C}_r{q}^{n-r}{p}^r}=\frac{{\sum }_{r=0}^n{r.\frac{n}{r}}^{n-1}{C}_{r-1}{q}^{n-r}p.p^{r-1}}{{\sum }_{r=0}^n{}^n{C}_r{q}^{n-r}{p}^r}$
$\Rightarrow \overline{x}=\frac{np\left({\sum }_{r=0}^n{}^{n-1}{C}_{r-1}{p}^{r-1}{q}^{(n-1)-(r-1)}\right)}{{\sum }_{r=0}^n{}^n{C}_r{q}^{n-r}{p}^r}\Rightarrow \overline{x}=\frac{np{(q+p)}^{n-1}}{{(q+p)}^n}$
$\therefore \overline{x}=np\{\because q+p=1\}$