The weighted mean of the first n natural numb | vedclass

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Question

(B) The weighted mean is given by the formula: $\frac{\sum_{i=1}^{n} (i \cdot i^2)}{\sum_{i=1}^{n} i^2} = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2

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

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(B) The weighted mean is given by the formula: $\frac{\sum_{i=1}^{n} (i \cdot i^2)}{\sum_{i=1}^{n} i^2} = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2}$.We know that $\sum_{i=1}^{n} i^3 = \left[ \frac{n(n + 1)}{2} \right]^2$ and $\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}$.Substituting these values,we get:Weighted mean = $\frac{\left[ \frac{n(n + 1)}{2} \right]^2}{\frac{n(n + 1)(2n + 1)}{6}}$$= \frac{n^2(n + 1)^2}{4} \cdot \frac{6}{n(n + 1)(2n + 1)}$$= \frac{6n(n + 1)}{4(2n + 1)} = \frac{3n(n + 1)}{2(2n + 1)}$.

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

The weighted mean of the first $n$ natural numbers,where the weights are equal to the squares of the corresponding numbers,is:

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