1. Calculate Mode from the following mixed data
| Class | Frequency |
| 1 | 3 |
| 2 | 4 |
| 5 | 10 |
| 6 - 10 | 23 |
| 10 - 20 | 20 |
| 20 - 30 | 20 |
| 30 - 50 | 15 |
| 50 - 70 | 3 |
| 70 - 100 | 2 |
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `f*x`
`(4)=(2)xx(3)` | `cf`
`(6)` |
| 1 | 3 | 1 | 3 | 3 |
| 2 | 4 | 2 | 8 | 7 |
| 5 | 10 | 5 | 50 | 17 |
| 6 - 10 | 23 | 8 | 184 | 40 |
| 10 - 20 | 20 | 15 | 300 | 60 |
| 20 - 30 | 20 | 25 | 500 | 80 |
| 30 - 50 | 15 | 40 | 600 | 95 |
| 50 - 70 | 3 | 60 | 180 | 98 |
| 70 - 100 | 2 | 85 | 170 | 100 |
| --- | --- | --- | --- | --- |
| `n = 100` | ----- | `sum f*x=1995` | ----- |
Mean `bar x = (sum fx)/n`
`=1995/100`
`=19.95`
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(100/2)^(th)` observation
= value of `50^(th)` observation
From the column of cumulative frequency `cf`, we find that the `50^(th)` observation lies in the class `10 - 20`.
`:.` The median class is `10 - 20`.
Now,
`:. L = `lower boundary point of median class `=10`
`:. n = `Total frequency `=100`
`:. cf = `Cumulative frequency of the class preceding the median class `=40`
`:. f = `Frequency of the median class `=20`
`:. c = `class length of median class `=10`
Median `M = L + (n/2 - cf)/f * c`
`=10 + (50 - 40)/20 * 10`
`=10 + (10)/20 * 10`
`=10 + 5`
`=15`
Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`
`=3 * 15 - 2 * 19.95`
`=45 - 39.9`
`=5.1`
2. Calculate Mode from the following mixed data
| Class | Frequency |
| 2 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 - 9 | 8 |
| 10 - 14 | 15 |
| 15 - 19 | 8 |
| 20 - 29 | 4 |
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `f*x`
`(4)=(2)xx(3)` | `cf`
`(6)` |
| 2 | 1 | 2 | 2 | 1 |
| 3 | 2 | 3 | 6 | 3 |
| 4 | 2 | 4 | 8 | 5 |
| 5 - 9 | 8 | 7 | 56 | 13 |
| 10 - 14 | 15 | 12 | 180 | 28 |
| 15 - 19 | 8 | 17 | 136 | 36 |
| 20 - 29 | 4 | 24.5 | 98 | 40 |
| --- | --- | --- | --- | --- |
| `n = 40` | ----- | `sum f*x=486` | ----- |
Mean `bar x = (sum fx)/n`
`=486/40`
`=12.15`
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(40/2)^(th)` observation
= value of `20^(th)` observation
From the column of cumulative frequency `cf`, we find that the `20^(th)` observation lies in the class `10 - 14`.
`:.` The median class is `9.5 - 14.5`.
Now,
`:. L = `lower boundary point of median class `=9.5`
`:. n = `Total frequency `=40`
`:. cf = `Cumulative frequency of the class preceding the median class `=13`
`:. f = `Frequency of the median class `=15`
`:. c = `class length of median class `=5`
Median `M = L + (n/2 - cf)/f * c`
`=9.5 + (20 - 13)/15 * 5`
`=9.5 + (7)/15 * 5`
`=9.5 + 2.3333`
`=11.8333`
Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`
`=3 * 11.8333 - 2 * 12.15`
`=35.5 - 24.3`
`=11.2`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
