If V is the variance and M is the mean of first 15 natural numbers then what is $V+M^{2}$ equal to ?

Given :

 V is the variance and M is the mean of first 15 natural numbers.

To find :

We have to find $V+M^{2}$.

Solution :

The first 15 natural numbers are 1,2,3,....14,15.

$ Mean = \frac{1}{N}\sum x_{i}$

$ =\frac{1}{15}( 1+2+3+....+15)$

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

$ = 8$

$Variance = \frac{1}{N}\sum ( x_{i} -\mu )^{2} $

$ =\frac{1}{15} \times \left[( 1-8)^{2} +( 2-8)^{2} +....+( 15-8)^{2}\right]$

$=\frac{1}{15} \times \left[ 7^{2} +6^{2} +5^{2} +4^{2} +3^{2} +2^{2} +1^{2} +0^{2} +1^{2} +...+7^{2}\right]$

$ =\frac{1}{15} \times 2[ 49+36+25+16+9+4+1]$

$=\frac{1}{15} \times 2\times 140$

$ = \frac{56}{3}$

Therefore,

$V+M^{2}  =\frac{56}{3} +8^{2}$

$ = \frac{56}{3} +64$

$=\frac{56+64\times 3}{3}$

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

$ =\frac{248}{3}$ .

The value of $V+M^{2}$ is $\frac{248}{3}$
  

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Updated on: 10-Oct-2022

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