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NDA Math Statistics

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Question : 9 of 100

Marks: +1, -0

If V is the variance and M is the mean of first 15 natural numbers, then what is

V+M^2

$\frac{124}{3}$
$\frac{148}{3}$
$\frac{248}{3}$
$\frac{124}{9}$

Solution:

As, sum of first n natural number

=\frac{n(n+1)}{2}

So, sum of first 15 natural number

=\frac{15(15+1)}{2}=\frac{15\times 16}{2}=15\times 8

∴ M = Mean of first 15 natural number

=\frac{15\times 8}{18}=8

\therefore M=8

[first 15 natural numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]

Now, variance,

V=\frac{1}{n}\sum (x_i-M)^2

\therefore V=\frac{1}{15}[(1-8)^2+(2-8)^2

+(3-8)^2+(4-8)^2+(5-8)^2+(6-8)^2

+(7-8)^2+(8-8)^2+(9-8)^2+(10-8)^2

+(11-8)^2+(12-8)^2+(13-8)^2+(14-8)^2+(15)^2-8)^2+]

\Rightarrow V=\frac{1}{15}[49+36+25+16+9+4+1+0+1+4+9+16+25+36+49]

=\frac{1}{15}[280]=\frac{56}{3}

\therefore V+M^2=\frac{56}{3}+8^2=\frac{56}{3}+64

=\frac{56+192}{3}=\frac{248}{3}

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