Find the median and quartile deviation for the following data.
$x$ $3$ $4$ $5$ $6$ $7$
$f$ $12$ $35$ $52$ $41$ $18$

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Solution

Let us find the cumulative frequency for the given data.
$x$ $f$ Cumulative frequency $f_{c}$ Remarks
$3$ $12$ $12$ $12$ , the first $12$ scores correspond to $x = 3$
$4$ $35$ $47$ $(12 + 35)$ from $13^{t h}$ to $47^{t h}$ scores correspond to $x = 4$
$5$ $52$ $99$ $(47 + 52)$ from $48^{t h}$ to $99^{t h}$ scores correspond to $x = 5$
$6$ $41$ $140$ $(99 + 41)$ from $100^{t h}$ to $140^{t h}$ scores correspond to $x = 6$
$7$ $18$ $158$ $(140 + 18)$ from $141^{s t}$ to $158^{t h}$ scores correspond to $x = 7$ .
Observe that $n = 158$ , the last cumulative frequency.
$M e d i a n = \frac{n + 1}{2} t h s c o r e = \frac{158 + 1}{2} t h s c o r e = {79.5}^{t h} s c o r e \approx 80^{t h}$ score.
From the column $f_{c}$ , from $48^{t h}$ to $99^{t h}$ score corresponds to $x = 5$ . Hence, $80^{t h}$ position $= 5$ . Therefore, $m e d i a n = 5$ .

Suggest Corrections

Wages (₹)	0−5	5−10	10−15	15−20	20−25	25−30
Number of Workers	4	6	3	8	12	7

Variable( $x$ )	$10$	$15$	$20$	$25$	$30$	$35$	$40$	45
Frequency( $f$ )	$7$	$3$	$8$	$5$	$6$	$8$	$4$	9

$x$	$f$	Cumulative frequency $f_{c}$	Remarks
$3$	$12$	$12$	$12$ , the first $12$ scores correspond to $x = 3$
$4$	$35$	$47$	$(12 + 35)$ from $13^{t h}$ to $47^{t h}$ scores correspond to $x = 4$
$5$	$52$	$99$	$(47 + 52)$ from $48^{t h}$ to $99^{t h}$ scores correspond to $x = 5$
$6$	$41$	$140$	$(99 + 41)$ from $100^{t h}$ to $140^{t h}$ scores correspond to $x = 6$
$7$	$18$	$158$	$(140 + 18)$ from $141^{s t}$ to $158^{t h}$ scores correspond to $x = 7$ .