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MathematicsStatisticsMeasures of DispersionEasy2 minPYQ_2020

MathematicsEasynumerical

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16,$ then the value of $m + n$ is equal to

Answer:

18.00

Solution:

$var (1, 2, \dots n) = 10 \Rightarrow \frac{1^{2} + 2^{2} + \dots + n^{2}}{n} - {(\frac{1 + 2 + \dots + n}{n})}^{2} = 10$

$\Rightarrow \frac{(n + 1) (2 n + 1)}{6} - {(\frac{n + 1}{2})}^{2} = 10$

$\Rightarrow n^{2} - 1 = 120 \Rightarrow n = 11$

$var (2, 4, 6 \dots ., 2 m) = 16 \Rightarrow var (1, 2 \dots ., m) = 4$

$\Rightarrow m^{2} - 1 = 48 \Rightarrow m = 7 \Rightarrow m + n = 18$

Stream:JEESubject:MathematicsTopic:StatisticsSubtopic:Measures of Dispersion

⏱ 2mℹ️ Source: PYQ_2020

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