If the variance of the first n natural number | vedclass

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(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.Given $\frac{n^2 - 1}{12} = 10$,we have $n^2

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$11:7$

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(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.Given $\frac{n^2 - 1}{12} = 10$,we have $n^2 - 1 = 120$,so $n^2 = 121$,which gives $n = 11$.The first $m$ even natural numbers are $2, 4, 6, \dots, 2m$. This is equivalent to $2 	imes (1, 2, 3, \dots, m)$.The variance of a set of numbers multiplied by a constant $k$ is $k^2$ times the original variance.Thus,the variance of the first $m$ even natural numbers is $2^2 	imes \frac{m^2 - 1}{12} = 4 	imes \frac{m^2 - 1}{12} = \frac{m^2 - 1}{3}$.Given $\frac{m^2 - 1}{3} = 16$,we have $m^2 - 1 = 48$,so $m^2 = 49$,which gives $m = 7$.Therefore,$n: m = 11: 7$.

Class interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$1$	$3$	$4$	$2$

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $n: m=$

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