# What is the standard deviation of 1, 2, 3, 4 and 5?

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De Rono Share
Nov 5, 2015

The standard deviation of $\left\{1 , 2 , 3 , 4 , 5\right\}$
$= {\left[\frac{{5}^{2} - 1}{12}\right]}^{\frac{1}{2}} = \sqrt{2}$

#### Explanation:

Let's develop a general formula then as a particular you get standard deviation of $1 , 2 , 3 , 4$ and $5$. If we have $\left\{1 , 2 , 3 , \ldots . , n\right\}$ and we need to find the standard deviation of this numbers.

Note that

$\text{Var} \left(X\right) = \frac{1}{n} {\sum}_{i = 1}^{n} {x}_{i}^{2} - {\left(\frac{1}{n} {\sum}_{i = 1}^{n} {x}_{i}\right)}^{2}$

$\implies \text{Var} \left(X\right) = \frac{1}{n} {\sum}_{i = 1}^{n} {i}^{2} - {\left(\frac{1}{n} {\sum}_{i = 1}^{n} i\right)}^{2}$

$\implies \text{Var} \left(X\right) = \frac{1}{n} \cdot \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{1}{n} \cdot \frac{n \left(n + 1\right)}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{\left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{n + 1}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \left[\frac{2 n + 1}{3} - \frac{n + 1}{2}\right]$

$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \cdot \frac{n - 1}{6}$
$\implies \text{Var} \left(X\right) = \frac{{n}^{2} - 1}{12}$

So, Standard deviation of $\left\{1 , 2 , 3 , \ldots . , n\right\}$ is ${\left[\text{Var} \left(X\right)\right]}^{\frac{1}{2}} = {\left[\frac{{n}^{2} - 1}{12}\right]}^{\frac{1}{2}}$

In particular, your case the standard deviation of $\left\{1 , 2 , 3 , 4 , 5\right\}$
$= {\left[\frac{{5}^{2} - 1}{12}\right]}^{\frac{1}{2}} = \sqrt{2}$.

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