Find the Median from the following table-
Class Interval $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$
Frequency $10$ $17$ $26$ $22$ $15$

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Solution

Class Interval Frequency
$(f)$ Cumulative
Frequency $(C . F .)$
$0 - 20$ $10$ $10$
$20 - 40$ $17$ $27$
$40 - 60$ $26$ $53$
$60 - 80$ $22$ $75$
$80 - 100$ $15$ $90$
$\sum f = N = 90$
Here $N = 90$
$M i d - p o i n t = \frac{N}{2} = \frac{90}{2} = 45$
$45 t h$ term is in the cumulative frequency $53$ . Median class is $40 - 60$ .
here $l_{1} = 40$ ,
$l_{2} = 60$ ,
$f = 26$ ,
$C = 27$
$M d = l_{1} + \frac{(\frac{N}{2} - C)}{f} (l_{2} - l_{1})$
$= 40 + ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{90}{2} - 27}{26} ⎞ ⎟ ⎟ ⎟ ⎠ (60 - 40)$
$= 40 + \frac{(45 - 27)}{26} \times 20$
$= 40 + \frac{18}{26} \times 20$
$= 40 + \frac{180}{26}$
$= 40 + 13.84$
$= 53.84$
Thus, the median $= 53.84$ .

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Class interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Frequency	$8$	$10$	$12$	$9$	$9$

50

x

y

Class-interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	Total
Frequency	$17$	$x$	$32$	$y$	$19$	$120$

Class Interval	0-10	10-20	20-40	40-60	60-80	80-100
Frequency	10	8	22	10	20	10

53

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$C l a s s$	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
$F r e q u e n c y$	$15$	$18$	$21$	$p$	$17$

Class interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$	$120 - 140$
Frequency	$4$	$15$	$23$	$12$	$16$	$8$	$2$

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