Find missing frequency for grouped data Find 1 Missing frequency when Median is given example

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Home > Statistical Methods calculators > Find missing frequency for grouped data example

Find missing frequency for grouped data ( Enter your problem )

( Enter your problem )

Find 1 Missing frequency when Mean is given example
Find 1 Missing frequency when Median is given example
Find 2 Missing frequencies when Mean is given example
Find 2 Missing frequencies when Median is given example
Find 2 Missing frequencies when Mode is given example
Find 2 Missing frequencies when Quartile is given example
Find 3 Missing frequencies when Mean or Median or mode are given example

1. Find 1 Missing frequency when Mean is given example
(Previous example)

3. Find 2 Missing frequencies when Mean is given example
(Next example)

2. Find 1 Missing frequency when Median is given example

1. Find missing frequency from the following data
Class Frequency
50 - 100 35
100 - 150 60
150 - 200 ?
200 - 250 40
250 - 300 20
median = 168

Solution:

Class (1) $(1)$	Frequency (f) $(f)$ (2) $(2)$	cf $cf$ (3) $(3)$
50-100	35	35 35=0+35 $35=0+35$ (3)= $(3)=$ Previous (3)+(2) $(3)+(2)$
100-150	60	95 95=35+60 $95=35+60$ (3)= $(3)=$ Previous (3)+(2) $(3)+(2)$
150-200	a	95 + a 95+a=95+a $95 + a=95+a$ (3)= $(3)=$ Previous (3)+(2) $(3)+(2)$
200-250	40	135 + a 135+a=95+a+40 $135 + a=95 + a+40$ (3)= $(3)=$ Previous (3)+(2) $(3)+(2)$
250-300	20	155 + a 155+a=135+a+20 $155 + a=135 + a+20$ (3)= $(3)=$ Previous (3)+(2) $(3)+(2)$
---	---	---
--	n=a+155 $n=a + 155$

To find median class
Here, median is

168 $168$ .

$:.$ The median class is

$150 - 200$ .

Now,

$:. L =$ lower boundary point of median class

$=150$

$:. n =$ Total frequency

$=a + 155$

$:. cf =$ Cumulative frequency of the class preceding the median class

$=95$

$:. f =$ Frequency of the median class

$=a$

$:. c =$ class length of median class

$=50$

Median

$M = L + (( n)/2 - cf)/f * c$

$168=150 + (((a + 155))/2 - (95))/a * 50$

$168 - 150 = (((a + 155))/2 - (95))/a * 50$

$18 = ((a + 155) - 2(95))/(2 * a) * 50$

$18=((a+155) - (190))/(2a) * 50$

$18*2a=((a+155)-(190)) * 50$

$36a=(a-35) * 50$

$36a=50a-1750$

$-14a=-1750$

$14a=1750$

$a=1750/14$

$a=125$

Thus, the missing frequency is

$125$ .

2. Find missing frequency from the following data
Class Frequency
10 - 20 15
20 - 30 25
30 - 40 ?
40 - 50 12
50 - 60 8
60 - 70 5
70 - 80 3
median = 32

Solution:

Class $(1)$	Frequency $(f)$ $(2)$	$cf$ $(6)$
10 - 20	15	15
20 - 30	25	40
30 - 40	a	40+a
40 - 50	12	52+a
50 - 60	8	60+a
60 - 70	5	65+a
70 - 80	3	68+a
---	---
	$n=68+a$

To find median class
Here, median is

$32$ .

$:.$ The median class is

$30 - 40$ .

Now,

$:. L =$ lower boundary point of median class

$=30$

$:. n =$ Total frequency

$=0$

$:. cf =$ Cumulative frequency of the class preceding the median class

$=40$

$:. f =$ Frequency of the median class

$=a$

$:. c =$ class length of median class

$=10$

Median

$M=L+(( n)/2-cf)/f * c$

$32=30 + (((68+a))/2 - (40))/a * 10$

$32 - 30 = (((68+a))/2 - (40))/a * 10$

$2 = ((68+a) - 2(40))/(2 * a) * 10$

$2=((a+68) - (80))/(2a) * 10$

$2*2a=((a+68)-(80)) * 10$

$4a=(a-12) * 10$

$4a=10a-120$

$-6a+120 = 0$

$6a = 120$

$a = 120/6$

$a = 20 ->(1)$

Thus, the missing frequency is

$20$ .

This material is intended as a summary. Use your textbook for detail explanation.
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1. Find 1 Missing frequency when Mean is given example
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