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_____.
Answer (integer)
18
Solution
Variance ${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$
<br><br>variance of (1, 2, ….. n)
<br><br>10 = $${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$$
<br><br>on solving we get n = 11
<br><br>variance of 2, 4, 6…….2m = 16
<br><br>$\Rightarrow$ $${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$$ = 16
<br><br>$\Rightarrow$ m<sup>2</sup> = 49
<br><br>$\Rightarrow$ m = 7
<br><br>$\therefore$ m + n = 18
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Dispersion
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