Question

Find mean andsigma of n natural numbers

Answer 1

Answer:

Mean = $\frac{n+1}{2}$

Variance(σ²) = $\frac{n^{2} - 1 }{12}$

Standard deviation (σ) = $\sqrt{\frac{n^{2} -1}{12} }$

Step-by-step explanation:

Now,

We have to calculate mean of n natural number,

We know,

mean =  ∑x / n

         = 1 + 2 + 3 + .......... + n / n

and,

the sum of the first n natural numbers is = $\frac{n(n+1)}{2}$

So,

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

         = $\frac{n+1}{2}$

Now,

We cannot use the formula of (x-mean)^2 for variance as then the calculation will be very difficult.

So, we can use the formula :

σ² = ∑x^2 /  n - (∑x / n)^2

for this we want ∑x^2 which is :-

$1^{2}+ 2^{2} +3^{2}+.........+ n^{2}$ which is =

$\frac{n(n+1)(2n+1)}{6}$

So,

σ² = $\frac{n(n+1)(2n+1)}{6 n} - [\frac{n(n+1)}{2n} ]^{2}$

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

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

  = $\frac{(n+1)(n+5)}{12}$

  = $\frac{n^{2}-1 }{12}$

So,

σ = $\sqrt{\frac{n^{2}-1 }{12} }$

Thanks !

#BAL #answerwithquality