For a frequency distribution, standard deviation is computed by
- A$\sigma = \frac{{\sum f(x - \bar x)}}{{\sum f}}$
- B$\sigma = \frac{{\sqrt {\sum f{{(x - \bar x)}^2}} }}{{\sum f}}$
- C$\sigma = \sqrt {\frac{{\sum f{{(x - \bar x)}^2}}}{{\sum f}}} $
- D$\sigma = \sqrt {\frac{{\sum f(x - \bar x)}}{{\sum f}}} $
Similar Questions
The number of values of $a \in N$ such that the variance of $3,7,12 a, 43-a$ is a natural number is (Mean $=13$)
The number of values of $a \in N$ such that the variance of $3,7,12 a, 43-a$ is a natural number is (Mean $=13$)
- [JEE MAIN 2022]
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]
If the mean and variance of the following data:
$6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:
If the mean and variance of the following data:
$6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:
- [JEE MAIN 2021]
The $S.D$. of the first $n$ natural numbers is
The $S.D$. of the first $n$ natural numbers is
The mean and the variance of five observations are $4$ and $5.20,$ respectively. If three of the observations are $3, 4$ and $4;$ then the absolute value of the difference of the other two observations, is
The mean and the variance of five observations are $4$ and $5.20,$ respectively. If three of the observations are $3, 4$ and $4;$ then the absolute value of the difference of the other two observations, is
- [JEE MAIN 2019]