Find the mean and variance for the first $n$ natural numbers
Find the mean and variance for the first $n$ natural numbers
The mean of first $n$ natural numbers is calculated as follows.
Mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$
$\therefore$ Mean $=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$
Varianvce $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} $
$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{x_i} - \left( {\frac{{n + 1}}{2}} \right)} \right]}^2}} $
$ = \frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - \frac{1}{n}\sum\limits_{i = 1}^n {2\left( {\frac{{n + 1}}{n}} \right)} } {x_i} + \frac{1}{n}{\sum\limits_{i = 1}^n {\left( {\frac{{n + 1}}{2}} \right)} ^2}$
$=\frac{1}{n} \frac{n(n+1)(2 n+1)}{6}-\left(\frac{n+1}{n}\right)\left[\frac{n(n+1)}{2}\right]+\frac{(n+1)^{2}}{4 n} \times n$
$=\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{2}+\frac{(n+1)^{2}}{4}$
$=\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^{2}}{4}$
$=(n+1)\left[\frac{4 n+2-3 n-3}{12}\right]$
$=\frac{(n+1)(n-1)}{12}$
$=\frac{n^{2}-1}{12}$
Similar Questions
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$
$0$
$1$
$5$
$6$
$10$
$12$
$17$
$f_i$
$3$
$2$
$3$
$2$
$6$
$3$
$3$
The variance $\sigma^2$ of the data is $ . . . . . .$
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
- [JEE MAIN 2024]
Let $ \bar x , M$ and $\sigma^2$ be respectively the mean, mode and variance of $n$ observations $x_1 , x_2,...,x_n$ and $d_i\, = - x_i - a, i\, = 1, 2, .... , n$, where $a$ is any number.
Statement $I$: Variance of $d_1, d_2,.....d_n$ is $\sigma^2$.
Statement $II$ : Mean and mode of $d_1 , d_2, .... d_n$ are $-\bar x -a$ and $- M - a$, respectively
Let $ \bar x , M$ and $\sigma^2$ be respectively the mean, mode and variance of $n$ observations $x_1 , x_2,...,x_n$ and $d_i\, = - x_i - a, i\, = 1, 2, .... , n$, where $a$ is any number.
Statement $I$: Variance of $d_1, d_2,.....d_n$ is $\sigma^2$.
Statement $II$ : Mean and mode of $d_1 , d_2, .... d_n$ are $-\bar x -a$ and $- M - a$, respectively
- [JEE MAIN 2014]
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
- [JEE MAIN 2024]
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in R$ $\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is. . . . .
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in R$ $\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is. . . . .
- [IIT 2024]
The variance of $10$ observations is $16$. If each observation is doubled, then standard deviation of new data will be -
The variance of $10$ observations is $16$. If each observation is doubled, then standard deviation of new data will be -