JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 9)
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
18
Explanation
Variance $${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$$
variance of (1, 2, ….. n)
10 = $${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$$
on solving we get n = 11
variance of 2, 4, 6…….2m = 16
$$ \Rightarrow $$ $${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$$ = 16
$$ \Rightarrow $$ m2 = 49
$$ \Rightarrow $$ m = 7
$$ \therefore $$ m + n = 18
variance of (1, 2, ….. n)
10 = $${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$$
on solving we get n = 11
variance of 2, 4, 6…….2m = 16
$$ \Rightarrow $$ $${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$$ = 16
$$ \Rightarrow $$ m2 = 49
$$ \Rightarrow $$ m = 7
$$ \therefore $$ m + n = 18
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