Question

If a variate takes the values $0,1,4,9,...,n^2$ with frequencies $x^n$, ${}^n{\text{C}}_1{x}^{n-1}{y}^1$, ${}^n{\text{C}}_2{x}^{n-2}{y}^2$, ..., $y^n$ such that $x+y=1$. then mean equals

• A. $nx(nx+y)$
• B. $ny(ny+x)$
• C. $nx(ny+x)$
• D. $ny(nx+y)$

Answer 1

Answer: (B)

$ny(ny+x)$

Mean $=\frac{\sum f_i{x}_i}{\sum f_i}=\sum f_i{x}_i$

$(\because \sum f_i=1)$
$\begin{array}{ccc}{x}_i& f_i& f_i.x_i\\ 0& x^n& 0\\ 1& {}^n{\text{C}}_1{x}^{n-1}{y}^1& 1.{}^n{\text{C}}_1{x}^{n-1}{y}^1\\ 4& {}^n{\text{C}}_2{x}^{n-2}{y}^2& 4.{}^n{\text{C}}_2{x}^{n-2}{y}^2\\ ...& ...& ...\\ ...& ...& ...\\ ...& ...& ...\\ n^2& {}^n{\text{C}}_n{y}^n& n^2{}^n{\text{C}}_2{x}^{n-2}{y}^2\\ & \sum f_i=1& {\sum }_{i=1}^n{f}_i{x}_i=\sum r^2{}^n{\text{C}}_r{x}^{n-r}{y}^r\end{array}$

$\therefore$ ${\sum }_{i=1}^n{f}_i{x}_i={\sum }_{r=0}^n{r}^2{}^n{\text{C}}_r{x}^{n-r}{y}^r$
$={\sum }_{r=0}^n{(r(r-1)+r)}^n{\text{C}}_r{x}^{n-r}{y}^r$

$=n(n-1)y^2+ny$
$=ny[(n-1)y+1]$
$=ny[ny+(1-y)]=ny[ny+x]$

$\because$ $x+y=1$