Question

If a variable takes the values $0,1,2,...,n$ with corresponding frequencies as binomial coefficients ${}^n{C}_0,{}^n{C}_1,...,{}^n{C}_n$, then the mean of distribution is

• A. $\frac{\left(n\right)(n+1)}{4}$
• B. $\frac{n}{2}$
• C. $\frac{\left(n\right)(n-1)}{2}$
• D. $\frac{\left(n\right)(n+1)}{2}$

Answer 1

The correct option is B $\frac{n}{2}$

We know that,
$\text{mean}=\left(\frac{\sum f_i{x}_i}{\sum f_i}\right)\Rightarrow \text{Mean}=\frac{0\cdot {}^n{C}_0+1\cdot {}^n{C}_1+2\cdot {}^n{C}_2+3\cdot {}^n{C}_3+\cdots +n\cdot {}^n{C}_n}{{}^n{C}_0+{}^n{C}_1+\cdots +{}^n{C}_n}\Rightarrow \text{Mean}=\frac{0\cdot {}^n{C}_0+1\cdot {}^n{C}_1+2\cdot {}^n{C}_2+3\cdot {}^n{C}_3+\cdots +n\cdot {}^n{C}_n}{2^n}$
The numerator is of form
$\sum _{r=0}^{r=n}r\cdot {}^n{C}_r=n\sum _{r=1}^{r=n}\cdot {}^{n-1}{C}_{r-1}=n\cdot 2^{n-1}$
Therefore,
$\text{Mean}=\frac{n\cdot 2^{(n-1)}}{2^n}\Rightarrow \text{Mean}=\frac{n}{2}$