Find mean, median and mode of the given data
$18, 21, 18, 3, 12, 15, 18, 42, 6, 3, 24, 12$

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Solution

Step 1: Finding the mean
We know that,
$M e a n = \frac{S u m o f a l l t h e o b s e r v a t i o n s}{T o t a l n u m b e r o f o b s e r v a t i o n}$
$= \frac{18 + 21 + 18 + 3 + 12 + 15 + 18 + 42 + 6 + 3 + 24 + 12}{12}$
$= \frac{192}{12} = 16$
Step 2: Finding the median
Arranging the given data in ascending order
$3, 3, 6, 12, 12, 15, 18, 18, 18, 21, 24, 42$
We can observe that total number of observation are even i.e.; $12$ .
So, median $=$ Mean of $(\frac{N}{2}) t h$ and $(\frac{N}{2} + 1) t h$ observation.
$=$ Mean of $(\frac{12}{2}) t h$ and $(\frac{12}{2} + 1) t h$ observation.
$=$ Mean of $6 t h$ and $7 t h$ observation.
$= \frac{15 + 18}{2} = \frac{33}{2} = 16.5$
Step 3 : Finding the mode
By observing the data, we can see, $18$ is having the heighest frequency i.e.; $3$
So, the mode of given data is $18$ .
Hence, the mean, median and mode are $16, 16.5$ and $18$ respectively.

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Calculate mean deviation from the median of the following data :

$\begin{matrix} Class Interval : & 0 - 6 & 6 - 12 & 12 - 18 & 18 - 24 & 24 - 30 Frequency : & 4 & 5 & 3 & 6 & 2 \end{matrix}$

Class interval:	0–6	6–12	12–18	18–24	24–30
Frequency	4	5	3	6	2

Class:	$3 - 6$	$6 - 9$	$9 - 12$	$12 - 15$	$15 - 18$	$18 - 21$	$21 - 24$
Frequency:	$2$	$5$	$10$	$23$	$21$	$12$	$3$

Number	$0 - 3$	$3 - 6$	$6 - 9$	$9 - 12$	$12 -$	$15 - 18$	$18 - 19$	$21 -$
Frequency	$4$		$9$	$7$	$6$	$3$	$1$

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