Question

Find mean and standard deviation of first n natural numbers​

Answer 1

$\large\underline{\sf{Solution-}}$

We have to find mean and standard deviation of first n natural numbers .

So, We know first n natural numbers are

1, 2, 3, 4, --------, n

So, Mean of this series is given by

$\rm \: \overline{x} \: = \: \dfrac{1}{n} \displaystyle\sum_{k = 1}^n x_k$

$\rm \: \overline{x} \: = \: \dfrac{1 + 2 + 3 + - - - + n}{n}$

$\rm \: \overline{x} \: = \: \dfrac{\displaystyle\sum_{k = 1}^n k}{n}$

$\rm \: \overline{x} \: = \: \dfrac{\dfrac{n \: (n \: + \: 1)}{2} }{n}$

$\rm\implies \:\rm \: \overline{x} \: = \: \dfrac{n \: + \: 1}{2}$

Now, Standard Deviation is given by

$\rm \: \sigma \: = \: \sqrt{\dfrac{\displaystyle\sum_{k = 1}^n {x_k}^{2} }{n} - {( \overline{x})}^{2} }$

$\rm \: \sigma \: = \: \sqrt{\dfrac{ {1}^{2} + {2}^{2} + {3}^{2} + - - + {n}^{2} }{n} - {\bigg(\dfrac{n + 1}{2} \bigg) }^{2} }$

$\rm \: \sigma \: = \: \sqrt{\dfrac{\displaystyle\sum_{k = 1}^n {k}^{2} }{n} - {\bigg(\dfrac{n + 1}{2} \bigg) }^{2} }$

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

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

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

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

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

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

$\rm\implies \:\rm \: \sigma \: = \: \sqrt{\dfrac{ {n}^{2} - 1}{12} }$

Hence,

$\: \: \: \: \:\rm \: \overline{x} \: = \: \dfrac{n \: + \: 1}{2}$

and

$\: \: \: \: \: \:\rm \: \sigma \: = \: \sqrt{\dfrac{ {n}^{2} - 1}{12} }$

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Basic Result Used

$\rm \: \displaystyle\sum_{k = 1}^n k \: = \: \frac{n(n + 1)}{2}$

$\rm \: \displaystyle\sum_{k = 1}^n {k}^{2} \: = \: \frac{n(n + 1)(2n + 1)}{6}$

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ADDITIONAL INFORMATION

Standard Deviation for Continuous series using

1. Direct Method

$\rm \:\boxed{\tt{ \sigma \: = \frac{1}{N} \sqrt{N\displaystyle\sum \: f_i {x_i}^{2} \: - \: {(\displaystyle\sum f_ix_i)}^{2} } \: }}$

2. Short Cut Method

$\rm \:\boxed{\tt{ \sigma \: = \frac{1}{N} \sqrt{N\displaystyle\sum \: f_i {d_i}^{2} \: - \: {(\displaystyle\sum f_id_i)}^{2} } \: }}$

3. Step - Deviation Method

$\rm \:\boxed{\tt{ \sigma \: = \frac{h}{N} \sqrt{N\displaystyle\sum \: f_i {u_i}^{2} \: - \: {(\displaystyle\sum f_iu_i)}^{2} } \: }}$

Answer 2

$\underline{ \color{purple}{ \rm ANSWERS \: : }}$

$\boxed{ \blue{ \rm mean( x )= \frac{ n+ 1}{2} }}$

$\rm and \: \: \: standard \: \: \: \: deviation( \sigma) \: \: is \: \: \:$

$\rm \sigma = \sqrt{ \frac{ {n}^{2} - 1 }{12} }$