Question

The mean of the values $0,1,2,3,....,n$ with the corresponding weights ${}^{n}C_0,{}^{n}C_1,{}^{n}C_2,....,{}^{n}C_n$ respectively is

• A. $\frac{n+1}{2}$
• B. $\frac{n-1}{2}$
• C. $\frac{2^n-1}{2}$
• D. $\frac{2^n+1}{2}$
• E. $\frac{n}{2}$

Answer 1

The correct option is E

$\frac{n}{2}$

Explanation for the correct option:

Find the weighted mean of the given series.

A series $0,1,2,3,....,n$ is given with corresponding weights ${}^{n}C_0,{}^{n}C_1,{}^{n}C_2,....,{}^{n}C_n$.

We know that, the weighted mean $\overline{x}$ is given by $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n}\left(x_i{w}_1\right)}{\sum _{i=1}^n\left(w_1\right)}$.

Therefore, the weighted mean of the given series is as follows:

$$\begin{gathered} \overline{x}=\frac{\left(0\right){}^{n}C_0+\left(1\right){}^{n}C_1+\left(2\right){}^{n}C_2+....+\left(n\right){}^{n}C_n}{{}^{n}C_0+{}^{n}C_1+{}^{n}C_2+....+{}^{n}C_n} \\ \Rightarrow \overline{x}=\frac{0+{}^{n}C_1+\left(2\right){}^{n}C_2+....+\left(n\right){}^{n}C_n}{2^n} \\ \Rightarrow \overline{x}=\frac{n{\left(2\right)}^{n-1}}{2^n} \\ \Rightarrow \overline{x}=\frac{n}{2} \end{gathered}$${Since, ${}^{n}C_0+{}^{n}C_1+{}^{n}C_2+....+{}^{n}C_n=2^n$ and $0·{}^{n}C_0+1·{}^{n}C_1+2·{}^{n}C_2+....+n·{}^{n}C_n=n\left(2^{n-1}\right)$}

Therefore, the mean of the given series is $\frac{n}{2}$.

Hence, option $E$ is the correct option.