First slide

Introduction to statistics

Question

The weighted A.M of first n natural numbers whose weights are equal to the corresponding numbers is equal to

Easy

A

2n+1
B

$\frac{1}{2} (2 n + 1)$
C

$\frac{1}{3} (2 n + 1)$
D

$\frac{2 n + 1}{6}$

Solution

The required mean is
$\begin{matrix} \bar{X} = \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + \dots + n \cdot n}{1 + 2 + 3 + \dots + n} \\ = \frac{{Σn}^{2}}{Σn} = \frac{n (n + 1) \frac{(2 n + 1)}{6}}{\frac{n (n + 1)}{2}} \\ = \frac{2 n + 1}{3} \end{matrix}$

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