Median
Median of an observation is the value of the variable which divides it into two equal parts. It is the value which is
- If there are odd number of observations, the value in the middle of the ordered list.
- If there are even number of observations, it is the arithmetic mean of values in the middle of the ordered list.
The median is a positional average. Sometimes, in case of even number of observations, instead of arithmetic mean of values in the middle, one of the two is chosen randomly.
Steps to Calculate Median
- Find N/2, where N is the total number of observations.
- Sort the observations and tabulate it. Add a third column to calculate the cumulative (total) frequency.
- Spot the x with cumulative frequency just greater than N/2. That will be your median.
Example 7: Find the median of 2, 3, 3, 5, 6, 6, 6
Solution: Note that N = 7, hence, N/2 = 3.5. So, the 4th observation, x =5 is the median.
Example 8: Find the median of 1, 11, 11, 13, 16, 17.
Solution: Note that N = 6. The values in centre are, 11 and 13. Then by the rule, we can choose 11 or 13 as the median. Or, we can take their average, (11+13)/2 = 12 as their median as well.
Example 9: Pick any 5 numbers. Find its (arithmetic) mean and median.
Example 10: Obtain the median for the following frequency distribution:
x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
f: | 8 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |
Solution: Here, the value of x’s are already sorted.
x | f | Cumulative Frequency |
1 | 8 | 8 |
2 | 10 | 8+10 = 18 |
3 | 11 | 18+11= 29 |
4 | 16 | 29+16= 45 |
5 | 20 | 45+20= 65 |
6 | 25 | 65+25= 90 |
7 | 15 | 90+15= 105 |
8 | 9 | 105+9= 114 |
9 | 6 | 114+6= 120 |
N = 120 |
Hence, N = 120 and N/2 = 60.
Observe the cumulative frequency column. The value just greater than 60 in it is 65. Corresponding value to 65 of x is 5.
Hence, median is 5.
In the case of continuous frequency distribution, the class corresponding to the cf. just greater than NI2 is called the median class and the value of median is obtained by the following formula:
Median =
where l is the lower limit of the median class.
f is the frequency of the median class.
h is the magnitude of median class
‘c’ is the cumulative frequency of the class preceding the median class.
and N = ∑ f
Example 11: Find the median wage of the following distribution:
Wages (in Rs.): | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
No. of labourers: | 3 | 5 | 20 | 10 | 5 |
Solution:
Wages (x, in Rs.) | No. of labourers (f) | Cumulative Frequency |
20-30 | 3 | 3 |
30-40 | 5 | 3+5 = 8 |
40-50 | 20 | 8+20 = 28 |
50-60 | 10 | 28+10 = 38 |
60-70 | 5 | 38+5 = 43 |
N = 43 |
Now, N/2 = 43/2 = 21.5.
Cumulative frequency just greater than 21.5 is 28 and corresponding class is 40-50.
Thus, the median class is 40-50. Thus,
l = 40, f = 28, h = 10, c = 8, N = 43.
Using the formula,
Median =
Example 12: Find the median of the following distribution:
Wages (in Rs.): | 10-30 | 30-50 | 50-70 | 70-90 |
No. of labourers: | 4 | 6 | 18 | 12 |
(Try it yourself!)
Merits and Demerits of Median
Requisite | Merit | Demerit | Notes |
It is rigidly defined. | ✓ | ||
It is easy to understand and calculate. | ✓ | …and in some cases, it can be calculated by just looking at the distribution! | |
It is not affected much by extreme observations. | ✓ | ||
It is not suitable for further mathematical treatment. | ✓ | ||
It is not based on all observations | ✓ | Consider the distribution 10, 14, 20, 32, 35. Even if we replace 10 and 14 by two numbers less than 20, median will not change. | |
In comparison to mean, it is affected by fluctuation in sampling. | ✓ |
Mode
Mode is the value which occurs most frequently in a set of observations, and around which other values tend to be around. The mode of a distribution is the value of x corresponding to maximum frequency. A distribution may have two modes (bi-modal) or more than two (multimodal).
Example 13: Find the mode of the following distribution:
x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
f: | 4 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |
Solution: 25 has the highest frequency. Hence the mode is the corresponding x value, 6.
Example 14: Find the mode of the following distribution:
x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
f: | 4 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |
Solution: (Try it yourself!)
In case of class-intervals, suppose xk-xk+1 is the modal class with highest frequency fk, then
where l is the lower limit of the modal class.
Example 15: Find the mode of the following distribution:
Class Interval: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
Solution: Here, maximum frequency is 28. Thus the class 40-50 is the modal class.
Then, l = 40, h = 10, fk = 28, fk-1 = 12, fk+1 = 20.
Mode =
Example 16: Find the mode of the following distribution:
Class Interval: | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |
Frequency: | 5 | 8 | 17 | 12 | 8 |
Solution: (Try it yourself!)
Sometimes, mode can be approximated from mean, using the formula Mode = 3 Median – 2 Mean.
Requisite | Merit | Demerit | Notes |
It is ill-defined. | ✓ | Mode of a distribution need not be unique. | |
Easy to understand and calculate. | ✓ | …and in some cases, simply by inspection. | |
We can find mode of open-ended classes | ✓ | ||
It is not at all affected by extreme observations. | ✓ | ||
It is not suitable for further mathematical treatment. | ✓ | ||
It is not based on all observations | ✓ | ||
In comparison to mean, it is affected by fluctuation in sampling. | ✓ |
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