Statistics is the branch of mathematics that collects, organizes and analyzes data scientifically and make meaningful interpretation of the data. Statistics helps to organize large volume of numbers for decision making. For this the numbers should be organized properly into meaningful groups that are convenient to handle. This is called a frequency table or frequency distribution.
Let us consider marks of 20 students in an examination are as below 45,25,20,12,18,38,27,6,48,32,12,9,27,41,15,36,45,28,19,29.
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This is raw data. It is difficult to make meaningful conclusion from this representation. To make analysis of this data easy, we need to have two kinds of measures and they are
(1) What is the central number in the data which is a reasonable representation of the data or which gives an idea about the figures?
(2) How far is the data points spread from this representative number? Do they all the data points stay around the central figure oar there many numbers far away from the representative number?
The measures which tell us about the representative figures are called measures of central tendencies and the second measures which tells us about the dispersion of data is called measures of dispersion.
There are many measures of central tendencies but there are three most commonly used measures they are called mean, median and mode. There are two commonly used measures of dispersion called range, standard deviation and mean deviation
In this discussion we will discuss in detail about the median
How to Calculate the Median of a Data?
The media of a data set is the single value from the data set that measures the central item of the data. To find the median of a data set, first arranges the values in ascending or descending order. If the data set has odd number of items, the middle term is the mean else if the data set has even number of items, the median is the average of the two middle terms.
Example 1: Calculate the median of the data set
4.2, 5.7, 8.9, 3.8, 1.6, 7.4, 6.8, 9.2, 6.9
Solution:
Let us arrange the data set in ascending order we get
1.6, 3.8, 4.2, 5.7, 6.8, 6.9 7.4, 8.9, 9.2
Median
The number at the middle is 6.8. The median of the data set is 6.8
Now let us describe how to calculate the median of a data that is arranged in a frequency table. Let us consider an example
Suppose the above data is grouped into frequency table as below
Data
Frequency
1-1.99
1
2-2.99
0
3-3.99
1
4-4.99
1
5-5.99
1
6-6.99
2
7-7.99
1
8-8.99
1
9-9.99
1
We will see how to calculate the median from this table
We know the median is the value, which is at the centre of the data points. So add up the figures in the frequency column. This comes to 9 in this example. This is the total number of data points. If we divide this by two we get 4.5. The median is the central data, which means we need to find the data point Commulative frequency 4.5 or just more.
Step 1: add the column in the frequency till you reach 4.5. If we add the first five rows we get 4. If we add the first six rows, we get 6. This means the median data is somewhere in the row 6-6.99.This class 6-6.99 contains the median and is called the median class.
Step 2: Now lets us find the fifth data point. Till 6-6.99, we had four data points. The range 6-6.99 has two data points
`(7-6)/(2)` = 0.5 where 6 is the first item of the median class and 7 is the first item of next class. This means that the range of 6 to 7 is split between two values. So we calculate the first value in this
We get 6 + 0.5 = 6.5 – This is the median
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This is different form the value in the raw data as we used frequency tables data and not raw data.
To simplify this we have a formula to calculate the median as below
Median =`((n+1)/(2)- F)/(f_(m))xx w + L_(m)` Where
n is the total number of items or the total frequency
F- sum of the class frequencies up to but NOT including the median class
fm – Frequency of the median class
w- class interval
Lm – Lower limit of the median class
If we apply the formula in the above example we get,
n= 9
w=1
F=4
fm=2
Lm=6
Median = `((n+1)/(2)- F)/(f_(m))xx w + L_(m)` =`((9+1)/(2)- 4)/(2)xx 1 + 6` = 0.5 + 6 = 6.5 – as before.
The biggest advantage of the median is that the extreme values doe not impact the median much unlike the mean. But it has certain disadvantages like we need to array the data before we can perform any calculation. Also, statistical calculations with median are more complex than the calculation with medians. Hence, mean is more widely used than the median
Exercises on Median
Prob 1 : Find the median of the data set
7, 6,14,12,8,9
Ans:
Step 1: Arrange the data in the ascending order. We get
6, 7, 8, 9, 12, 14
Step 2: There are two central values 8 and 9
The average of the two is `(8+9)/(2)` = 8.5 – This is the median
Prob 2: Find the median of the following data arrange din the frequency table
Class
Frequency
100-149.5
12
150-199.5
14
200-249.5
27
250-299.5
58
300-349.5
72
350-399.5
63
400-449.5
36
450-499.5
18
Ans:
Let us calculate the cumulative frequency of the classes
Class
Frequency
Cumulative Frequency
100-149.5
12
12
150-199.5
14
26
200-249.5
27
53
250-299.5
58
111
300-349.5
72
183
350-399.5
63
246
400-449.5
36
282
450-499.5
18
300
The total frequency n = 300
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Median class is that class which contains `(300)/(2)` = 150 or 150th and 151st terms. The required class is 300 – 349.5
Let us use the formula
n=300
w=50
F=111
fm=72
Lm=300
Median = `((n+1)/(2)- F)/(f_(m))xx w + L_(m)` = `((300+1)/(2)- 111)/(72)xx 50 + 300` = `((150.5-111)/72) xx 50 + 300` = 327.43
Prob 3: Suppose we made a frequency table with each class having a different width. Will the calculation of median be affected?
Ans:
No. As we saw, we were interested only in the class interval of the median class and the cumulative frequency up to median class. So the class intervals of other classes will not affect the calculation.
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