Calculate Mode using Grouping Method from the following grouped data
X | Frequency |
10 | 8 |
11 | 15 |
12 | 20 |
13 | 100 |
14 | 98 |
15 | 95 |
16 | 90 |
17 | 75 |
18 | 50 |
19 | 30 |
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 :x | I Frequency | II (1+2) | III (2+3) | IV (1+2+3) | V (2+3+4) | VI (3+4+5) |
10 | 8 | | | | | |
| | 8+15=23 | | | | |
11 | 15 | | | 8+15+20=43 | | |
| | | 15+20=35 | | | |
12 | 20 | | | | 15+20+100=135 | |
| | 20+100=120 | | | | |
13 | 100 | | | | | 20+100+98=218 |
| | | 100+98=198 | | | |
14 | 98 | | | 100+98+95=293 | | |
| | 98+95=193 | | | | |
15 | 95 | | | | 98+95+90=283 | |
| | | 95+90=185 | | | |
16 | 90 | | | | | 95+90+75=260 |
| | 90+75=165 | | | | |
17 | 75 | | | 90+75+50=215 | | |
| | | 75+50=125 | | | |
18 | 50 | | | | 75+50+30=155 | |
| | 50+30=80 | | | | |
19 | 30 | | | | | |
1) 100 is the maximum value in the column-I and it is an individual frequency of x 13. Therefore, we have tick(✓) this x.
2) 193 is the maximum value in the column-II and it is the sum of 98 and 95; i.e., of x 14 and 15. Therefore, we have tick(✓) this both x.
3) 198 is the maximum value in the column-III and it is the sum of 100 and 98; i.e., of x 13 and 14. Therefore, we have tick(✓) this both x.
4) 293 is the maximum value in the column-IV and it is the sum of 100, 98, and 95; i.e, of x 13, 14 and 15. Therefore, we have tick(✓) this three x.
5) 283 is the maximum value in the column-V and it is the sum of 98, 95, and 90; i.e, of x 14, 15 and 16. Therefore, we have tick(✓) this three x.
6) 260 is the maximum value in the column-VI and it is the sum of 95, 90, and 75; i.e, of x 15, 16 and 17. Therefore, we have tick(✓) this three x.
Analysis Table :Column | I | II | III | IV | V | VI | Total |
Max Frequency | 100 | 193 | 198 | 293 | 283 | 260 | |
10 | | | | | | | - |
11 | | | | | | | - |
12 | | | | | | | - |
13 | ✓ | | ✓ | ✓ | | | 3 |
14 | | ✓ | ✓ | ✓ | ✓ | | 4 |
15 | | ✓ | | ✓ | ✓ | ✓ | 4 |
16 | | | | | ✓ | ✓ | 2 |
17 | | | | | | ✓ | 1 |
18 | | | | | | | - |
19 | | | | | | | - |
Since 4 is the maximum ticks repeated 2 times, So grouping method fails to give the modal class.
We use formula Mode = 3 Median - 2 Mean
Find Mean, Median
`x` `(1)` | Frequency `(f)` `(2)` | `f*x` `(3)=(2)xx(1)` | `cf` `(5)` |
10 | 8 | 80 | 8 |
11 | 15 | 165 | 23 |
12 | 20 | 240 | 43 |
13 | 100 | 1300 | 143 |
14 | 98 | 1372 | 241 |
15 | 95 | 1425 | 336 |
16 | 90 | 1440 | 426 |
17 | 75 | 1275 | 501 |
18 | 50 | 900 | 551 |
19 | 30 | 570 | 581 |
--- | --- | --- | --- |
| `n=581` | `sum f*x=8767` | -- |
Mean `bar x = (sum fx)/n`
`=8767/581`
`=15.0895`
Median :M = value of `((n+1)/2)^(th)` observation
= value of `(582/2)^(th)` observation
= value of `291^(st)` observation
From the column of cumulative frequency `cf`, we find that the `291^(st)` observation is `15`.
Hence, the median of the data is `15`.
We have given Mean (`bar X`) `=15.0895`, Median(`M`) `=15`, Mode(`Z`) `=?`
`Z=3 M - 2 bar X`
`Z=3*15-2*15.0895`
`Z=45-30.179`
`Z=14.821`
This material is intended as a summary. Use your textbook for detail explanation.
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