Question

find mean and variance of first n natural numbers....

Answer 1

First n natural numbers are 1,2,3,…,n.

Mean=(1+2+3+…+n)÷n

=[n(n+1)/2]/n

(n+1)/2 …(i)

And, 1²+2²+3²+…+n²=n(n+1)(2n+1)/6

Or, (1²+2²+3²+…+n²)/n= {(n+1)(2n+1)/6} …(ii)

Variance of first n natural numbers={(1²+2²+3²+…+n²)/n}-(Mean)²

={(n+1)(2n+1)/6}-{(n+1)²/4}

=(n+1)/12{4n+2–3(n+1)}

=(n+1)(n-1)/12

=(n²-1)/12

hope it's very helpful for u dude....

Answer 2

Thus, Mean and variance of first $n$natural numbers is $\frac{(n^{2}-1) }{12}$

Step-by-step explanation:

Given,

Find mean and variance of first $n$ natural numbers.

First $n$ natural numbers are $1,2,3,.......,n$

∴ $Mean=\frac{(1+2+3+....+n)}{n}$

⇒$Mean=\frac{\frac{n(n+1)}{2} }{n}$

⇒$Mean=\frac{(n+1)}{2}$__1

And, $1^{2} +2^{2} +3^{2} +......+n^{2} =\frac{n(n+1)(2n+1)}{6}$

⇒$\frac{1^{2} +2^{2} +3^{2} +......+n^{2} }{n} =\frac{(n+1)(2n+1)}{6}$__2

∴Variance of first n natural numbers$=\frac{1^{2} +2^{2} +3^{2} +......+n^{2} }{n} -Mean^{2}$

$=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)}{4}^{2}$

$=\frac{n+1}{2}[\frac{2n+1}{3}-\frac{n+1}{2} ]$

$=\frac{n+1}{2}[\frac{4n+2-3n-3}{6}]$

$=\frac{(n+1)(n-1)}{12}$

$=\frac{(n^{2}-1) }{12}$

∴ Mean and variance of first $n$natural numbers is $\frac{(n^{2}-1) }{12}$