If S_1 and S_2 are the variances of the first | vedclass

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(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2 - 1}{12}$.For the first $2k$ natural numbers,$S_1 = \frac{(2k)^2 -

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$[4, \infty)$

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(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2 - 1}{12}$.For the first $2k$ natural numbers,$S_1 = \frac{(2k)^2 - 1}{12} = \frac{4k^2 - 1}{12}$.For the first $k$ natural numbers,$S_2 = \frac{k^2 - 1}{12}$.Therefore,the ratio is $\frac{S_1}{S_2} = \frac{4k^2 - 1}{k^2 - 1} = \frac{4(k^2 - 1) + 3}{k^2 - 1} = 4 + \frac{3}{k^2 - 1}$.Since $k > 1$,$k^2 - 1 > 0$. As $k 	o 1^+$,$\frac{3}{k^2 - 1} 	o \infty$,and as $k 	o \infty$,$\frac{3}{k^2 - 1} 	o 0$.Thus,$4 + \frac{3}{k^2 - 1} \in (4, \infty)$.

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
$f_i$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

Size $(x_i)$	$3.5$	$4.5$	$5.5$	$6.5$	$7.5$	$8.5$	$9.5$
Frequency $(f_i)$	$3$	$7$	$22$	$60$	$85$	$32$	$8$

If $S_1$ and $S_2$ are the variances of the first $2k$ and $k$ $(k > 1)$ natural numbers respectively,then $(S_1 / S_2)$ lies in the interval

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