Introduction to statistics

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

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Solution

We have,
$\frac{n^{2} - 1}{12} = 10 \Rightarrow n^{2} - 1 = 120 \Rightarrow n = 11$
It is given that
$\begin{array}{l} Var (2, 4, 6, \dots, 2 m) = 16 \\ \Rightarrow 2^{2} Var (1, 2, 3, \dots, m) = 16 \end{array}$
$\begin{array}{l} \Rightarrow 2^{2} (\frac{m^{2} - 1}{12}) = 16 \\ \Rightarrow m^{2} - 1 = 48 \Rightarrow m = 7 . \end{array}$
Hence, $m + n = 7 + 11 = 18$ .

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Introduction to statistics

Remember concepts with our Masterclasses.

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

We have,n2−112=10⇒n2−1=120⇒n=11It is given thatVar⁡(2,4,6,…,2m)=16⇒ 22Var⁡(1,2,3,…,m)=16⇒ 22m2−112=16⇒ m2−1=48⇒m=7.Hence, m+n=7+11=18.

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