Measures of dispersion - class-X

Measures of dispersion include:
1)Sample standard deviation
2)Interquartile range (IQR) or Interdecile range
3)Range
4)Mean difference
5)Median absolute deviation (MAD)
6)Average absolute deviation (or simply called average deviation)
7)Distance standard deviation

Mean deviation, standard deviation as well as quartile deviation is the measure of dispersion.
Hence, all of these are measure of dispersion.
(It is well known fact)

Statement 1: is correct because positional measure of dispersion describes about the position that a particular data value has within a data set.
Statement 2:is correct because quartiles and percentiles are positional measure of dispersion.

Since,

Percentile,Quartile are positional measure of dispersion because it tells about the position of a particular data value has within a data set.
Standard deviation, Variance are computational measure of dispersion.

Given $\sigma = 20$, coefficient of variation $=50$ %
We know coefficient of variation $=\cfrac{\sigma }{\bar{x}}\times 100=50$
$\Rightarrow \bar{x} = 2\times \sigma = 40$

The coefficient of variation (CV) is a standardized measure of dispersion 

If mean of the given dist. be $\bar{x}$ and S.D be $\sigma $
then given $\bar{x} = 40, \sigma^2 = 1486$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$

We know if a distribution having mean $\bar{x}$ and standard deviation $\sigma$
then coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$
$\therefore \cfrac{20}{\bar{x}}\times 100=50\Rightarrow \bar{x} = 40$
Hence required mean is $=40$

Given $\displaystyle \Sigma \left ( x _{i}-\overline{x} \right )^{2}=250$,$n=10,\overline{x}=50$

Now, $\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x _{i}-\overline{x} \right )^{2}}$

$= \sqrt{\dfrac{1}{10}\times 250}=5$ 
Hence coefficient of variation $\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$%

Given,   $\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50$
Thus standard deviation $ = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$ % $= 10$%

Given,  mean $\bar{x} = 4,$ and coefficient of variation $=58$ %
If S.D of the given distribution is $\sigma$ then we know that,
Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$ %
$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$

Coeffecient of variation $ = \frac {SD}{AM} \times 100 = \frac {10}{20} \times 100 = 50 $

Coefficient of variation is the ratio of standard deviation to the mean.

Coefficient of variation is given by $CV = \dfrac{SD}{Mean}\times 100 $
$\Rightarrow \dfrac{5}{14}\times 100 = 35.7%$

Given : standard deviation$(\sigma)=1.2,$ mean$(\overline {X})=10$.

$\sigma=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$