Question

If 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. $pq$
• B. $np$
• C. $nq$
• D. $n{p}^2$

Answer 1

The correct option is B $np$

Given variable takes values $0,1,2,......n$

with frequencies $q^n,\frac{n}{1}{q}^{n-1}.p,\frac{n(n-1)}{1.2}{q}^{n-2}{p}^2,......p^n$

The frequencies are ${}^n{C}_0{q}^n{p}^0,{}^n{C}_1{q}^{n-1}{p}^1,{}^n{C}_2{q}^{n-2}{p}^2.......{}^n{C}_r{q}^{n-r}{p}^3,.......,{}^n{C}_n{q}^0{p}^n$

Sum of the frequencies are $(p+q{)}^n=1$

now the combined sum is

$0.\left({}^n{C}_0{q}^n{p}^0\right)+1.\left({}^n{C}_1{q}^{n-1}{p}^1\right)+2\left({}^n{C}_2{q}^{n-2}{p}^2\right)+.......+r\left({}^n{C}_r{q}^{n-r}{p}^r\right)+....+n\left({}^n{C}_r{q}^0{p}^n\right)$

$=\sum _{r=1}^{n}r\left({}^n{C}_r{q}^{n-r}{p}^r\right)$

$=\sum _{r=1}^n\frac{n(n-1)!}{(n-r)!(r-1)!}{q}^{(n-1.(r-1)}{p}^{r-1}p$

$=np\sum _n^{r-1=0}{}^{(n-1)}{C}_{(r-1)}{q}^{(n-1)(r-1)}{p}^{r-1}$

$=np\sum _{r_1=0}^n{}^{n_1}{C}_{r_1}{q}^{n_1-r_1},p^{r_1}$

$=np(p+q{)}^{n_1}$

$=np$