Find the standard deviation of the first n natural numbers.
Find the standard deviation of the first n natural numbers.
$ \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 ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
- [AIEEE 2012]
The $S.D$. of the first $n$ natural numbers is
The $S.D$. of the first $n$ natural numbers is
Let $r$ be the range and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ be the $S.D.$ of a set of observations ${x_1},\,{x_2},\,.....{x_n}$, then
Let $r$ be the range and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ be the $S.D.$ of a set of observations ${x_1},\,{x_2},\,.....{x_n}$, then
Let the mean and variance of the frequency distribution
$\mathrm{x}$
$\mathrm{x}_{1}=2$
$\mathrm{x}_{2}=6$
$\mathrm{x}_{3}=8$
$\mathrm{x}_{4}=9$
$\mathrm{f}$
$4$
$4$
$\alpha$
$\beta$
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:
Let the mean and variance of the frequency distribution
| $\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
| $\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:
- [JEE MAIN 2021]
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............
- [JEE MAIN 2024]