For a given distribution of marks mean is 35.16 and its standard deviation is 19.76 . co-efficient o

English

Gujarati

Hindi

13.Statistics

easy

For a given distribution of marks mean is $35.16$ and its standard deviation is $19.76$. The co-efficient of variation is..

A

$\frac{{35.16}}{{19.76}}$

B

$\frac{{19.76}}{{35.16}}$

C

$\frac{{35.16}}{{19.76}} \times 100$

D

$\frac{{19.76}}{{35.16}} \times 100$

Solution

(d) Coefficient of variation $ = \frac{{{\rm{S}}{\rm{.D}}{\rm{.}}}}{{{\rm{Mean}}}} \times 100$

$ = \frac{{19.76}}{{35.16}} \times 100$.

Standard 11

Mathematics

Share

Similar Questions

If the variance of the frequency distribution is $3$ then $\alpha$ is ……

$X_i$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

Frequency $f_i$ $3$ $6$ $16$ $\alpha$ $9$ $5$ $6$

medium

(JEE MAIN-2023)

If the mean and variance of eight numbers $3,7,9,12,13,20, x$ and $y$ be $10$ and $25$ respectively, then $\mathrm{x} \cdot \mathrm{y}$ is equal to

hard

(JEE MAIN-2020)

The mean and standard deviation of $40$ observations are $30$ and $5$ respectively. It was noticed that two of these observations $12$ and $10$ were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^{2}$ is equal to$………$

hard

(JEE MAIN-2022)

The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2.$ then their new variance is equal to :

hard

(JEE MAIN-2023)

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)