First slide

Introduction to statistics

Question

The arithmetic mean of the following frequency distribution:
Variable (X) $\begin{array}{lllllll} : 0 1 2 3 \dots . n \end{array}$
Frequency (f) $\begin{array}{lllllll} :^{n} C_{0}^{n} C_{1}^{n} C_{2}^{n} C_{3} \dots .^{n} C_{n} \end{array}$
is

Moderate

A

$\frac{2^{n}}{n}$
B

$\frac{2^{n}}{n + 1}$
C

$\frac{n}{2}$
D

$\frac{2}{n}$

Solution

Let $\bar{X}$ denote the required mean. Then,
$\begin{array}{l} \bar{X} = \frac{\sum_{r = 0}^{n} r \cdot^{n} C_{r}}{\sum_{r = 0}^{n}^{n} C_{r}} = \frac{\sum_{r = 1}^{n} r \cdot {\frac{n}{r}}^{n - 1} C_{r - 1}}{\sum_{r = 0}^{n} r^{n} C_{r}} = \frac{n \sum_{r = 1}^{n}^{n - 1} C_{r - 1}}{2^{n}} \\ \Rightarrow \bar{X} = \frac{n \times 2^{n - 1}}{2^{n}} = \frac{n}{2} [∵ \sum_{r = 1}^{n}^{n - 1} C_{r - 1} = 2^{n - 1}] \end{array}$

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