
A large data set is represented in a graphical or tabular form denoting the frequency of occurrence of each set of data. This observation is known as frequency distribution. For example, a class has how many cricketers, tennis players, and badminton players can be grouped into a frequency table to find out other relevant data. This grouping is known as grouped data and can be used to find out mean deviation.
In a grouped data, the class intervals are arranged in such a way that they do not have any gaps, and each class has their own frequency.
Formula: - ∑ f | X-X| / ∑ f
where, f is the value of frequency
x is the mean, calculated as (sum of all the values/number of values) = ∑ f x / ∑ f
mid points are calculated as (lower limit + upper limit) / 2.
|
Class interval |
Players (f) |
Values (x) |
F(x) |
Mean dx – x-x |
Absolute deviation |
F |dx| = |fx-x| |
|
0-10 |
5 |
5 |
25 |
-22 |
22 |
110 |
|
10-20 |
8 |
15 |
120 |
-12 |
12 |
96 |
|
20-30 |
15 |
25 |
375 |
-2 |
2 |
30 |
|
30-40 |
16 |
35 |
560 |
8 |
8 |
128 |
|
40-50 |
6 |
45 |
270 |
18 |
18 |
108 |
|
|
∑ f =50 |
|
∑ f x = 1350 |
|
|
∑f |dx| = 472 |
X =∑ f x / ∑ f = 1350/50
X= 27
Mean Deviation=∑ f | X-X| / ∑ f
= 472/50
= 9.44
Hence, mean deviation is 9.44
In ungrouped data, some of the class intervals tend to be missing with irregular frequency distributions amongst them.
Example: - Find the mean deviation of the following ungrouped data.
|
X |
5 |
7 |
8 |
9 |
10 |
11 |
13 |
Solution
|
X |
|x-a| |
|
5 |
3 |
|
7 |
1 |
|
8 |
0 |
|
9 |
1 |
|
10 |
2 |
|
11 |
3 |
|
13 |
5 |
We need to find the median first. Here total entries are 7, which is an odd number.
Therefore, median = item corresponding to the value of (n+1) /2
= 8/2 = 4th item = 8
Mean deviation = ∑ f |X-Me| / N = ∑ f | D| / N
= 15/7 = 2.14
JEE Main marks vs rank vs percentile
JEE Advanced Eligibility Criteria
JEE Advanced Chemistry Syllabus
JEE Advanced Registration Dates