S.D . of first n natural numbers is

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13.Statistics

easy

The $S.D$. of the first $n$ natural numbers is

$\frac{{n + 1}}{2}$

$\sqrt {\frac{{n(n + 1)}}{2}} $

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

None of these

Solution

(c) $S. D.$ of first $n$ natural numbers $ = \sqrt {\frac{1}{n}\Sigma {x^2} – {{\left( {\frac{{\Sigma x}}{n}} \right)}^2}} $,

$ = \sqrt {\frac{{n(n + 1)(2n + 1)}}{{6n}} – {{\left[ {\frac{{n(n + 1)}}{{2n}}} \right]}^2}} $

$ = \sqrt {\frac{{(n + 1)(2n + 1)}}{6} – {{\left( {\frac{{n + 1}}{2}} \right)}^2}} = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{2n + 1}}{3} – \frac{{n + 1}}{2}} \right)} $

$ = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{4n + 2 – 3n – 3}}{6}} \right)} $

$ = \sqrt {\frac{{{n^2} – 1}}{{12}}} $.

Standard 11

Mathematics

Find the mean and variance for the data

${x_i}$ $6$ $10$ $14$ $18$ $24$ $28$ $30$

${f_i}$ $2$ $4$ $7$ $12$ $8$ $4$ $3$

medium

Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :

hard

(JEE MAIN-2020)

If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 – 4)^2, (x_2 – 4)^2, …., (x_{50} – 4)^2$ is

hard

(JEE MAIN-2019)

If $\sum\limits_{i = 1}^{18} {({x_i} – 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} – 8)^2 = 45} $ then the standard deviation of $x_1, x_2, …… x_{18}$ is :-

normal

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

$X_i$ $0$ $1$ $2$ $3$ $4$ $5$

$f_i$ $k+2$ $2k$ $K^{2}-1$ $K^{2}-1$ $K^{2}-1$ $k-3$

where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.

hard

(JEE MAIN-2023)

${x_i}$	$6$	$10$	$14$	$18$	$24$	$28$	$30$
${f_i}$	$2$	$4$	$7$	$12$	$8$	$4$	$3$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$K^{2}-1$	$K^{2}-1$	$K^{2}-1$	$k-3$

The $S.D$. of the first $n$ natural numbers is

Solution

Similar Questions

Find the mean and variance for the data ${x_i}$ $6$ $10$ $14$ $18$ $24$ $28$ $30$ ${f_i}$ $2$ $4$ $7$ $12$ $8$ $4$ $3$

If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 – 4)^2, (x_2 – 4)^2, …., (x_{50} – 4)^2$ is

If $\sum\limits_{i = 1}^{18} {({x_i} – 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} – 8)^2 = 45} $ then the standard deviation of $x_1, x_2, …… x_{18}$ is :-

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution $X_i$ $0$ $1$ $2$ $3$ $4$ $5$ $f_i$ $k+2$ $2k$ $K^{2}-1$ $K^{2}-1$ $K^{2}-1$ $k-3$ where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.

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Find the mean and variance for the data

${x_i}$ $6$ $10$ $14$ $18$ $24$ $28$ $30$

${f_i}$ $2$ $4$ $7$ $12$ $8$ $4$ $3$

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

$X_i$ $0$ $1$ $2$ $3$ $4$ $5$

$f_i$ $k+2$ $2k$ $K^{2}-1$ $K^{2}-1$ $K^{2}-1$ $k-3$

where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.