1.2.1) The Median
The median is the middle of the data set. Therefore before selecting the median it is necessary to re-arrange the data set in an ascending order of the magnitude.
1.2.1.1) Median for set of raw data.
Lets get this by an simple example.
Ex. 2 9 7 5 3 4 5 4 7 3 1
We should re-arrange the data set in an ascending order.
1 2 3 3 4 4 5 5 7 7 9
Therefore the number in the middle is “4”
Definition Median:
The median is the middle value of an ordered set of data.
This ordered set could be either in ascending order or descending order.
There are two cases; where the data set does exactly have a middle value, and when it does not.
This is also recognized by the number of data elements in the data set. When the number of elements is odd and even. We normally denote median by
“Xm”
Thus we can get two different equations to derive the median.
When the number of elements are odd :-

When the number of elements are even :-

You can always get the median value by following the above 2 equations.
1.2.1.2) Median for a frequency distribution
I’ll get an example I used before
|
Number of students
|
30
|
45
|
78
|
69
|
48
|
25
|
5
|
|
GPA
|
1
|
1.5
|
2
|
2.5
|
3
|
3.5
|
4
|
|
Cumulative frequency
|
30
|
75
|
153
|
222
|
270
|
295
|
300
|
Total number of students =300
Sample size is even, therefore the median should be
Xm = [X(150)+X(151)]/2
Xm = 2
1.2.1.3) Median for a grouped frequency distribution
Here we encounter a problem. We should get the value before the summarization of the data.
For that purpose we use the method called interpolation. There is another method to do this estimation it is called extrapolation. In our guiding book it has used the extrapolation mechanism. So I’ll show how to use extrapolation.
I’m taking the example from our book.
|
Measurements
|
Cumulative frequency
|
Relative cumulative frequency
|
|
0.0 – 1.0
|
3
|
0.075
|
|
1.0 – 2.0
|
7
|
0.175
|
|
2.0 – 3.0
|
11
|
0.275
|
|
3.0 – 4.0
|
18
|
0.450
|
|
4.0 – 5.0
|
24
|
0.600
|
|
5.0 – 6.0
|
29
|
0.725
|
|
6.0 – 7.0
|
34
|
0.850
|
|
7.0 – 8.0
|
35
|
0.875
|
|
8.0 – 9.0
|
37
|
0.925
|
|
9.0 – 10.0
|
40
|
1.000
|
How did I arrive at the relative cumulative frequency…..
What it means is that the weight of each cumulative frequency value with respect to the total value.
So to get the relative frequency, get each value and divide it with the total value
3/40 = 0.075
——————————————————————————————————————–
#!/Note
if [ you do not understand what is going on in the later part of this chapter ]
then
please read the “percentiles” page.
fi
———————————————————————————————————————
P50 = 4.0 + (1/0.600-0.450)*0.05 = 4.33`7;
Here I looked at the 50th percentile. It should be on the range from 0.450 to 0.600.
I know the exact value for 0.450 that is 4
What I need to estimate is for the other 0.05
Also I know that from 0.450 to 0.600 represent 1
Then comes simple mathematics.
If 0.450 to 0.600 represent 1 what is for 0.05
That is 1 —> 0.600 – 0.450 = 0.15
For 0.05 —-> (1/0.15) * 0.05 = 1/3;
Ok now you know how to get the median lets move into next chapter………