For a frequency distribution, standard deviation is computed by

English

Gujarati

Hindi

13.Statistics

normal

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}}} $

Solution

(c)It is obvious.

Standard 11

Mathematics

Share

Similar Questions

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 …………..

hard

(JEE MAIN-2024)

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

medium

The mean and standard deviation of $20$ observations are found to be $10$ and $2$, respectively. On respectively, it was found that an observation by mistake was taken $8$ instead of $12$ . The correct standard deviation is

medium

(JEE MAIN-2024)

The data is obtained in tabular form as follows.

${x_i}$ $60$ $61$ $62$ $63$ $64$ $65$ $66$ $67$ $68$

${f_i}$ $2$ $1$ $12$ $29$ $25$ $12$ $10$ $4$ $5$

hard

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2 \ldots$, $x _7$ is $4$ and the mean is $\overline{ x }$, then $\overline{ x }+ x _6$ is equal to:

hard

(JEE MAIN-2023)