1. Find missing frequency from the following data
| Class | Frequency |
| 0 - 5 | 1 |
| 5 - 10 | 7 |
| 10 - 15 | 11 |
| 15 - 20 | ? |
| 20 - 25 | ? |
| 25 - 30 | 4 |
| 30 - 35 | 2 |
Total Frequency (N) = 40 and mean = 16.5
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `d=(x-A)/h=(x-17.5)/5`
`A=17.5,h=5`
`(4)` | `f*d`
`(5)=(2)xx(4)` |
| 0 - 5 | 1 | 2.5 `2.5=(0+5)/2` | -3 `d=(2.5-17.5)/5=(-15)/5=-3`
`d=(x-17.5)/5` | `-3` `-3=1xx-3`
`(5)=(2)xx(4)` |
| 5 - 10 | 7 | 7.5 `7.5=(5+10)/2` | -2 `d=(7.5-17.5)/5=(-10)/5=-2`
`d=(x-17.5)/5` | `-14` `-14=7xx-2`
`(5)=(2)xx(4)` |
| 10 - 15 | 11 | 12.5 `12.5=(10+15)/2` | -1 `d=(12.5-17.5)/5=(-5)/5=-1`
`d=(x-17.5)/5` | `-11` `-11=11xx-1`
`(5)=(2)xx(4)` |
| 15 - 20 | a | 17.5 `17.5=(15+20)/2` | 0 `d=(17.5-17.5)/5=(0)/5=0`
`d=(x-17.5)/5` | `0` `0=axx0`
`(5)=(2)xx(4)` |
| 20 - 25 | b | 22.5 `22.5=(20+25)/2` | 1 `d=(22.5-17.5)/5=(5)/5=1`
`d=(x-17.5)/5` | `b` `b=bxx1`
`(5)=(2)xx(4)` |
| 25 - 30 | 4 | 27.5 `27.5=(25+30)/2` | 2 `d=(27.5-17.5)/5=(10)/5=2`
`d=(x-17.5)/5` | `8` `8=4xx2`
`(5)=(2)xx(4)` |
| 30 - 35 | 2 | 32.5 `32.5=(30+35)/2` | 3 `d=(32.5-17.5)/5=(15)/5=3`
`d=(x-17.5)/5` | `6` `6=2xx3`
`(5)=(2)xx(4)` |
| --- | --- | --- | --- | --- |
| `n=25+a+b` | ----- | ----- | `sum f*d=-14+b` |
`n = 40`
`a + b + 25 = 40`
`a + b = 15 ->(1)`
Mean `bar x = A + (sum fd)/n * h`
`16.5 = 17.5 + (-14+b)/(40) * 5`
`-1 = (-14+b)/(40) * 5`
`(-1 * 40) / 5 = -14+b`
`-8 = -14+b`
`-8 + 14 = b`
`b = 6`
Substituting in `(1)`
`a + 6 = 15`
`a = 9`
Thus, the missing frequencies are `9` and `6` respectively.
2. Find missing frequency from the following data
| Class | Frequency |
| 0 - 20 | 5 |
| 20 - 40 | 13 |
| 40 - 60 | ? |
| 60 - 80 | 15 |
| 80 - 100 | ? |
| 100 - 120 | 20 |
| 120 - 140 | 7 |
Total Frequency (N) = 100 and mean = 74
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `d=(x-A)/h=(x-50)/20`
`A=50,h=20`
`(4)` | `f*d`
`(5)=(2)xx(4)` |
| 0 - 20 | 5 | 10 | -2 | `-10` |
| 20 - 40 | 13 | 30 | -1 | `-13` |
| 40 - 60 | a | 50 | 0 | `0` |
| 60 - 80 | 15 | 70 | 1 | `15` |
| 80 - 100 | b | 90 | 2 | `2b` |
| 100 - 120 | 20 | 110 | 3 | `60` |
| 120 - 140 | 7 | 130 | 4 | `28` |
| --- | --- | --- | --- | --- |
| `n=60+a+b` | ----- | ----- | `sum f*d=80+2b` |
`n = 100`
`60 + a+b= 100`
`a+b=40 ->(1)`
Mean `bar x = A + (sum fd)/n * h`
`74=50 + (80+2b)/(100) * 20`
`24 = (80+2b)/(100) * 20`
`(24 * 100) / 20=80+2b`
`120=80+2b`
`120 - 80=2b`
`2b=40 ->(2)`
`b=20`
Substituting in `(1)`
`a+20=40`
`a=20`
Thus, the missing frequencies are `20 and 20` respectively.
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
