First slide

Introduction to statistics

Question

If a variable takes values $0, 1, 2, \dots, n$ with frequencies $1,^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}$ then the AM is

Moderate

A

$n$
B

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

$n + 1$
D

$\frac{n}{2}$

Solution

The required mean is given by
$\begin{array}{l} \bar{X} = \frac{0 \cdot 1 + 1 \cdot^{n} C_{1} + 2 \cdot^{n} C_{2} + 3 \cdot^{n} C_{3} + \dots + n \cdot^{n} C_{n}}{1 +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}} \\ \bar{X} = \frac{\sum_{r = 0}^{n} r \cdot^{n} C_{r}}{\sum_{r = 0}^{n}^{n} C_{r}} = \frac{\sum_{r = 1}^{n} r \cdot {\frac{n}{r}}^{n - 1} C_{r - 1}}{\sum_{r = 0}^{n}^{n} C_{r}} = \frac{n \sum_{r = 1}^{n}^{n - 1} C_{r - 1}}{\sum_{r = 0}^{n}^{n} C_{r}} \\ \bar{X} = \frac{n \times 2^{n - 1}}{2^{n}} = \frac{n}{2} \end{array}$ `

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