The mean of the values 0, 1, 2, dots, n havin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The weighted mean is given by $\bar{x} = \frac{\sum_{r=0}^{n} r \cdot {^nC_r}}{\sum_{r=0}^{n} {^nC_r}}$.
The numerator is $\sum_{r=0}^{n} r \cdot

Vedclass · Accepted Answer

$\frac{n}{2}$

Vedclass · Answer

(D) The weighted mean is given by $\bar{x} = \frac{\sum_{r=0}^{n} r \cdot {^nC_r}}{\sum_{r=0}^{n} {^nC_r}}$.
The numerator is $\sum_{r=0}^{n} r \cdot {^nC_r} = \sum_{r=1}^{n} r \cdot \frac{n}{r} \cdot {^{n-1}C_{r-1}} = n \sum_{r=1}^{n} {^{n-1}C_{r-1}} = n \cdot 2^{n-1}$.
The denominator is $\sum_{r=0}^{n} {^nC_r} = 2^n$.
Therefore, the mean is $\bar{x} = \frac{n \cdot 2^{n-1}}{2^n} = \frac{n}{2}$.

The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

Similar Questions

$\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5} = \dots$

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + \dots + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is $.............$.

If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$,then $C_0+2 C_1+3 C_2+\ldots+(n+1) C_n$ is equal to

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

The value of $\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \dots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module