Question

Find the mean and variance for the first n natural numbers

Answer 1

Answer:

$\boxed{\begin{aligned}& \qquad \:\sf \: Mean\:(\overline{x}) = \dfrac{n + 1}{2} \qquad \: \\ \\& \qquad \:\sf \:Variance = \: \dfrac{ {n}^{2} - 1}{12} \end{aligned}}$

Step-by-step explanation:

We know, first n natural numbers are 1, 2, 3, ..., n

So, distribution table is as follow:

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf {x}^{2} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf {1}^{2} \\ \\ \sf 2 & \sf {2}^{2} \\ \\ \sf 3 & \sf {3}^{2} \\ \\ \sf . & \sf .\\ \\ \sf . & \sf .\\ \\ \sf n & \sf {n}^{2} \end{array}} \\ \end{gathered}$

So,

$\sf \: \sum \: x = 1 + 2 + 3 + ... + n = \dfrac{n(n + 1)}{2}$

and

$\sf \: \sum \: {x}^{2} = {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$

Now,

$\sf \: Mean\:(\overline{x}) = \dfrac{\sum \: x}{n} = \dfrac{n(n + 1)}{2n} = \dfrac{n + 1}{2}$

Thus,

$\implies\sf \: \boxed{\sf \: Mean\:(\overline{x}) = \dfrac{n + 1}{2} \: }$

Now,

$\sf \: Variance$

$\sf \: = \: \dfrac{\sum \: {x}^{2} }{n} - {\:(\overline{x})}^{2}$

$\sf \: = \: \dfrac{n(n + 1)(2n + 1)}{6n} - {\bigg(\dfrac{n + 1}{2} \bigg) }^{2}$

$\sf \: = \: \dfrac{(n + 1)(2n + 1)}{6} - {\bigg(\dfrac{n + 1}{2} \bigg) }^{2}$

$\sf \: = \: \dfrac{n + 1}{2} \bigg( \dfrac{2n + 1}{3} - \dfrac{n + 1}{2} \bigg)$

$\sf \: = \: \dfrac{n + 1}{2} \bigg( \dfrac{2(2n + 1) - 3(n + 1)}{6} \bigg)$

$\sf \: = \: \dfrac{n + 1}{2} \bigg( \dfrac{4n + 2 - 3n - 3}{6} \bigg)$

$\sf \: = \: \dfrac{n + 1}{2} \bigg( \dfrac{n - 1}{6} \bigg)$

$\sf \: = \: \dfrac{ {n}^{2} - 1}{12}$

Thus,

$\implies\sf \: Variance = \: \dfrac{ {n}^{2} - 1}{12}$

Hence,

$\implies\boxed{\begin{aligned}& \qquad \:\sf \: Mean\:(\overline{x}) = \dfrac{n + 1}{2} \qquad \: \\ \\& \qquad \:\sf \:Variance = \: \dfrac{ {n}^{2} - 1}{12} \end{aligned}}$