Range and mean deviation - class-X

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 $D _8 = l _1 + \frac{\frac{8N}{10} - C}{f} \times (l _2 - l _1)$ 

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 $x _m=49.8$