Introduction to statistics

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If the variance of the first ‘n’ natural numbers is 10 and the variance of the first ‘m’ even natural numbers is 16, then m + n is equal to ____.

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Solution

$V a r (1, 2, ........, n) = 10 \Rightarrow \frac{1^{2} + 2^{2} + ....... + n^{2}}{n} - {(\frac{1 + 2 + ...... + n}{n})}^{2} = 10$
$\Rightarrow \frac{(n + 1) (2 n + 1)}{6} - {(\frac{n + 1}{2})}^{2} = 10$
$\Rightarrow n^{2} - 1 = 120 \Rightarrow n = 11$
$V a r (2, 4, 6, ......., 2 m) = 16 \Rightarrow var (1, 2, ....., m) = 4$
$\Rightarrow m^{2} - 1 = 48 \Rightarrow m = 7 \Rightarrow m + n = 18$

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Introduction to statistics

Remember concepts with our Masterclasses.

If the variance of the first ‘n’ natural numbers is 10 and the variance of the first ‘m’ even natural numbers is 16, then m + n is equal to ____.

Var1,2,........,n=10⇒12+22+.......+n2n−1+2+......+nn2=10⇒n+12n+16−n+122=10⇒n2−1=120⇒n=11Var2,4,6,.......,2m=16⇒var1,2,.....,m=4⇒m2−1=48⇒m=7⇒m+n=18

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