Statistics and probability - 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$(\overline {X})=10$.

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

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.

First quartile also called the lower quartile or the 25th percentile(splits off the lowest 25% of data from the highest 75%)
Second quartile also called the median or the 50th percentile (cuts data set in half)
Third quartile  also called the upper quartile or the 75th percentile (splits off the highest 25% of data from the lowest 75%)
Since its a symmetrical distribution therefore the median will be 30

The range of the data=Highest vale-lowest value

Range =maximum value-minimum value

Range is the difference between the maximum value and the minimum value of the data set.

$\Rightarrow$  Here, $L=3.84$ and $R=0.46$

$Range$ $as$ $the$ $name$ $indicates$ $gives$ $us$ $all$ $the$ $area$ $available$ $under$ $light$ $and$ $hence$ $statement$ $is$ $true.$

The value of the mean deviation is minimum if the deviations are taken from the median. So, it is less than that measured from any other value.

In arithmetic, the range of a set of data is the difference between the largest and smallest values.

We have, $Z=3M-2\bar{x}$
$\therefore 2\bar{x}+Z=3M$
or $2\bar{x}-3M=-Z$

$\begin{matrix} 7,8,2,1,3,13,18, \ range, , of, , data=\left( { \max  imum-\min  imum } \right)  \ =\left( { 18-1 } \right) =17, , , , , Ans. \  \end{matrix}$

Let the highest score be $x _{m}$ and 

A decile is any of the nine values that divide the sorted data into ten equal parts, so that each part represents 1/10 of the sample or population.
For a continuous distribution, the formula for $r^{th}$ decile is given by $D _r = l _1 + \frac{\frac{rN}{10} - C}{f} \times (l _2 - l _1)$
Substituting r = 8, we have $D _8 = l _1 + \frac{\frac{8N}{10} - C}{f} \times (l _2 - l _1)$ 

$P(1)$ is not true 

Arrange the given observations in ascending order
$40,54,62,68,76,90$
Here, number of terms $n=6 (even) $
$\displaystyle \therefore $ Median (M) $\displaystyle =\frac{\left ( \frac{n}{2} \right )th:term+\left ( \frac{n}{2}+1 \right )th:term}{2}=\frac{62+68}{2}=65$

$\Sigma \left | x _{i}-M \right |=25+11+3+3+11+25=78$
Mean deviation from median $\displaystyle =\frac{\Sigma \left | x _{i}-M \right |}{n}=\frac{78}{6}=13 $
$\therefore $ Coefficient of M.D.=$\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2$

Arranging the given data in ascending order
40,54,62,68,76,90
Here, $n=6 (even)$
$M= \dfrac{\text{value of }3^{rd}\text{observation}+\text{value of }4^{th}\text{observation}}{2}$
Median $M=\dfrac{62+68}{2}=65$

Mean deviation about median $M.D=\dfrac{|40-65|+|54-65|+|62-65|+|68-65|+|76-65|+|90-65|}{65}$

$=\dfrac{25+11+3+3+11+25}{65}=1.2$

The difference between maximum and the minimum observation in the data is range.

Coefficient of range of a set of data is given by $\dfrac{max-min}{max+min}$
$\dfrac{max-min}{max+min}=\dfrac{1}{8}$
$8max-8min=max+min$
$7max=9min$
$\dfrac{max}{min}=\dfrac{9}{7}$

The largest value of data is $x _m=1575$

Coefficient of range$=\dfrac{x _m-x _0}{x _m+x _0}=\dfrac{7.44-x _0}{7.44+x _0}$

The largest value of data is $x _m=49.8$

Standard deviation is independent of the change of origin (adding or subtracting a constant) and is scaled by the absolute value of the change of scale (multiplying by a constant). If the original SD is 3.5, the new SD is | -2 | * 3.5 = 7.

Given $\displaystyle u =\frac{x}{h}-\frac{a}{h}$
Since,S.D. is not depend on change of origin but it is depend on change of scale.
$\displaystyle \therefore \sigma _{u}=\frac{\sigma _{x}}{h}$
$\Rightarrow h\sigma _{u}=\sigma _{x}$

If we change scale by using x + h then median increases by h.
so median is not independent of change of scale.
From histogram we can see highest frequency so made.
Hence, options 'A' and 'B' are correct.