|
Mode using Grouping Method Example-6
( Enter your problem )
|
- Formula
- Example-1
- Example-2
- Example-3
- Example-4
- Example-5
- Example-6
|
Other related methods
- Mean, Median and Mode
- Quartile, Decile, Percentile, Octile, Quintile
- Population Variance, Standard deviation and coefficient of variation
- Sample Variance, Standard deviation and coefficient of variation
- Population Skewness, Kurtosis
- Sample Skewness, Kurtosis
- Geometric mean, Harmonic mean
- Mean deviation, Quartile deviation, Decile deviation, Percentile deviation
- Five number summary
- Box and Whisker Plots
- Mode using Grouping Method
- Less than type Cumulative frequency table
- More than type Cumulative frequency table
- Class and their frequency table
|
|
7. Example-6
Calculate Mode using Grouping Method from the following grouped data
| Class | Frequency | | 0 - 10 | 4 | | 10 - 20 | 16 | | 20 - 30 | 15 | | 30 - 40 | 20 | | 40 - 50 | 7 | | 50 - 60 | 5 |
Solution:
Mode using Grouping Method :
1) In column-I, write the original frequency
2) In column-II, combine the frequency two by two, starting from top
3) In column-III, combine the frequency two by two, starting from second
4) In column-IV, combine the frequency three by three, starting from top
5) In column-V, combine the frequency three by three, starting from second
6) In column-VI, combine the frequency three by three, starting from third
Grouping Table :
| Class | I
Frequency | II
(1+2) | III
(2+3) | IV
(1+2+3) | V
(2+3+4) | VI
(3+4+5) | | 0-10 | 4 | | | | | | | | 4+16=20 | | | | | | 10-20 | 16 | | | 4+16+15=35 | | | | | | 16+15=31 | | | | | 20-30 | 15 | | | | 16+15+20=51 | | | | 15+20=35 | | | | | | 30-40 | 20 | | | | | 15+20+7=42 | | | | 20+7=27 | | | | | 40-50 | 7 | | | 20+7+5=32 | | | | | 7+5=12 | | | | | | 50-60 | 5 | | | | | |
1) 20 is the maximum value in the column-I and it is an individual frequency of Class 30-40. Therefore, we have tick(✓) this Class.
2) 35 is the maximum value in the column-II and it is the sum of 15 and 20; i.e., of Class 20-30 and 30-40. Therefore, we have tick(✓) this both Class.
3) 31 is the maximum value in the column-III and it is the sum of 16 and 15; i.e., of Class 10-20 and 20-30. Therefore, we have tick(✓) this both Class.
4) 35 is the maximum value in the column-IV and it is the sum of 4, 16, and 15; i.e, of Class 0-10, 10-20 and 20-30. Therefore, we have tick(✓) this three Class.
5) 51 is the maximum value in the column-V and it is the sum of 16, 15, and 20; i.e, of Class 10-20, 20-30 and 30-40. Therefore, we have tick(✓) this three Class.
6) 42 is the maximum value in the column-VI and it is the sum of 15, 20, and 7; i.e, of Class 20-30, 30-40 and 40-50. Therefore, we have tick(✓) this three Class.
Analysis Table :
| Column | I | II | III | IV | V | VI | Total | | Max Frequency | 20 | 35 | 31 | 35 | 51 | 42 | | | 0-10 | | | | ✓ | | | 1 | | 10-20 | | | ✓ | ✓ | ✓ | | 3 | | 20-30 | | ✓ | ✓ | ✓ | ✓ | ✓ | 5 | | 30-40 | ✓ | ✓ | | | ✓ | ✓ | 4 | | 40-50 | | | | | | ✓ | 1 | | 50-60 | | | | | | | - |
Since 20-30 has the maximum ticks 5. So mode class is 20-30.
`:.` The mode class is `20 - 30`.
`:. L = `lower boundary point of mode class `=20`
`:. f_1 = ` frequency of the mode class `=15`
`:. f_0 = ` frequency of the preceding class `=16`
`:. f_2 = ` frequency of the succedding class `=20`
`:. c = ` class length of mode class `=10`
`Z=L+(f_1-f_0)/((f_1-f_0)+(f_1-f_2)) * c`
Here `f_1-f_0=-1 < 0` and `f_1-f_2=-5 < 0`
So we can not apply the above formula
Since, `f_2=20 > f_0=16`
Thus, Mode `Z=L+f_2/(f_2+f_0) * c`
`=20+20/(20+16) * 10`
`=20+20/36 * 10`
`=20+5.5556`
`=25.5556`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
|
|
