Tag: statistics and probability

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$(\overline {X})=10$.

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

It is a fundamental concept.
coefficient of deviation $=\cfrac{\sigma}{\bar{x}}\times 100$
where $\sigma$ and $\bar{x}$ are standard deviation and mean respectively.