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)` | Frequency `(f)`
`(2)` | `cf`
`(3)` |
| 50-100 | 35 | 35 `35=0+35`
`(3)=`Previous `(3)+(2)` |
| 100-150 | 60 | 95 `95=35+60`
`(3)=`Previous `(3)+(2)` |
| 150-200 | a | 95 + a `95 + a=95+a`
`(3)=`Previous `(3)+(2)` |
| 200-250 | 40 | 135 + a `135 + a=95 + a+40`
`(3)=`Previous `(3)+(2)` |
| 250-300 | 20 | 155 + a `155 + a=135 + a+20`
`(3)=`Previous `(3)+(2)` |
| --- | --- | --- |
| -- | `n=a + 155` | |
To find median class
Here, median is `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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