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Home > Statistical Methods calculators > Mean, Median and Mode for grouped data example
Mode Example for grouped data ( Enter your problem )
( Enter your problem )
  1. Formula & Example
  2. Mean Example
  3. Median Example
  4. Mode Example
 		Other related methods
  1. Mean, Median and Mode
  2. Quartile, Decile, Percentile, Octile, Quintile
  3. Population Variance, Standard deviation and coefficient of variation
  4. Sample Variance, Standard deviation and coefficient of variation
  5. Population Skewness, Kurtosis
  6. Sample Skewness, Kurtosis
  7. Geometric mean, Harmonic mean
  8. Mean deviation, Quartile deviation, Decile deviation, Percentile deviation
  9. Five number summary
  10. Box and Whisker Plots
  11. Mode using Grouping Method
  12. Less than type Cumulative frequency table
  13. More than type Cumulative frequency table
  14. Class and their frequency table
3. Median Example
(Previous example)		2. Quartile, Decile, Percentile, Octile, Quintile
(Next method)

4. Mode Example


Mode of grouped data
Mode of discrete frequency distribution
1. Calculate Mode from the following grouped data
XFrequency
01
15
210
36
43


Solution:
`x`
`(1)`		Frequency `(f)`
`(2)`
01
15
210
36
43
------
`n=25`


Mode :
the frequency of observation `2` is maximum (`10`)

`:. Z = 2`

2. Calculate Mode from the following grouped data
XFrequency
103
1112
1218
1312
143


Solution:
`x`
`(1)`		Frequency `(f)`
`(2)`
103
1112
1218
1312
143
------
`n=48`


Mode :
the frequency of observation `12` is maximum (`18`)

`:. Z = 12`

Mode of continuous frequency distribution
In grouped frequency distribution, we have to find a class with the maximum frequency, and this class is called modal class.
Mode is a value inside modal class and it is found by formula

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
where
`:. L = `lower boundary point of mode class

`:. f_1 = ` frequency of the mode class

`:. f_0 = ` frequency of the preceding class

`:. f_2 = ` frequency of the succedding class

`:. c = ` class length of mode class


3. Calculate Mode from the following grouped data
ClassFrequency
2 - 43
4 - 64
6 - 82
8 - 101


Solution:
Class
`(1)`		Frequency `(f)`
`(2)`
2-43
4-64
6-82
8-101
------
--`n = 10`


To find Mode Class
Here, maximum frequency is `4`.

`:.` The mode class is `4 - 6`.

`:. L = `lower boundary point of mode class `=4`

`:. f_1 = ` frequency of the mode class `=4`

`:. f_0 = ` frequency of the preceding class `=3`

`:. f_2 = ` frequency of the succedding class `=2`

`:. c = ` class length of mode class `=2`

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

`=4 + ((4 - 3)/(2*4 - 3 - 2)) * 2`

`=4 + (1/3) * 2`

`=4 + 0.6667`

`=4.6667`

4. Calculate Mode from the following grouped data
ClassFrequency
0 - 25
2 - 416
4 - 613
6 - 87
8 - 105
10 - 124


Solution:
Class
`(1)`		Frequency `(f)`
`(2)`
0-25
2-416
4-613
6-87
8-105
10-124
------
--`n = 50`


To find Mode Class
Here, maximum frequency is `16`.

`:.` The mode class is `2 - 4`.

`:. L = `lower boundary point of mode class `=2`

`:. f_1 = ` frequency of the mode class `=16`

`:. f_0 = ` frequency of the preceding class `=5`

`:. f_2 = ` frequency of the succedding class `=13`

`:. c = ` class length of mode class `=2`

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

`=2 + ((16 - 5)/(2*16 - 5 - 13)) * 2`

`=2 + (11/14) * 2`

`=2 + 1.5714`

`=3.5714`

5. Calculate Mode from the following grouped data
ClassFrequency
10 - 2015
20 - 3025
30 - 4020
40 - 5012
50 - 608
60 - 705
70 - 803


Solution:
Class
`(1)`		Frequency `(f)`
`(2)`
10 - 2015
20 - 3025
30 - 4020
40 - 5012
50 - 608
60 - 705
70 - 803
------
`n = 88`


To find Mode Class
Here, maximum frequency is `25`.

`:.` The mode class is `20 - 30`.

`:. L = `lower boundary point of mode class `=20`

`:. f_1 = ` frequency of the mode class `=25`

`:. f_0 = ` frequency of the preceding class `=15`

`:. f_2 = ` frequency of the succedding class `=20`

`:. c = ` class length of mode class `=10`

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

`=20 + ((25 - 15)/(2*25 - 15 - 20)) * 10`

`=20 + (10/15) * 10`

`=20 + 6.6667`

`=26.6667`

6. Calculate Mode from the following grouped data
ClassFrequency
20 - 25110
25 - 30170
30 - 3580
35 - 4045
40 - 4540
45 - 5035


Solution:
Class
`(1)`		Frequency `(f)`
`(2)`
20 - 25110
25 - 30170
30 - 3580
35 - 4045
40 - 4540
45 - 5035
------
`n = 480`


To find Mode Class
Here, maximum frequency is `170`.

`:.` The mode class is `25 - 30`.

`:. L = `lower boundary point of mode class `=25`

`:. f_1 = ` frequency of the mode class `=170`

`:. f_0 = ` frequency of the preceding class `=110`

`:. f_2 = ` frequency of the succedding class `=80`

`:. c = ` class length of mode class `=5`

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

`=25 + ((170 - 110)/(2*170 - 110 - 80)) * 5`

`=25 + (60/150) * 5`

`=25 + 2`

`=27`



This material is intended as a summary. Use your textbook for detail explanation.
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3. Median Example
(Previous example)		2. Quartile, Decile, Percentile, Octile, Quintile
(Next method)


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