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Find the standard deviation of the first n natural numbers.
Solution
$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x_{i} & 1 & 2 & 3 & 4 & 5 & \ldots & \ldots & n \\ \hline x_{i}^{2} & 1 & 4 & 9 & 16 & 25 & \ldots & \ldots & n^{2} \\ \hline \end{array}$
Now, $\quad \Sigma x_{i}=1+2+3+4+\ldots+n=\frac{n(n+1)}{2}$
and $\Sigma x_{i}^{2}=1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$
$\therefore \quad \alpha=\sqrt{\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}}=\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^{2}(n+1)^{2}}{4 n^{2}}}$
$=\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{4}}=\sqrt{\frac{2\left(2 n^{2}+3 n+1\right)-3\left(n^{2}+2 n+1\right)}{12}}$
$=\sqrt{\frac{4 n^{2}+6 n+2-3 n^{2}-6 n-3}{12}}=\sqrt{\frac{n^{2}-1}{12}}$
Similar Questions
Let $\mathrm{X}$ be a random variable with distribution.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
