Thus we can say that,

- Median refers to the middle value in a distribution
- It has a middle position in a series
- It is also called positional average
- It will not be affected by extreme items
- It splits the observations into two halves

**INDIVIDUAL SERIES**

### STEPS

- Arrange the data in ascending or descending order
- Locate the \( {{{(\frac{N + 1}{2})}^{th} }} item \, if \, N \, is\, odd \)
- Find the mean of the \( {{{(\frac{N}{2})}^{th} }} \) item and the next item, if N is even

**DATA OF ODD NUMBERS**

8, 21, 12, 5, 32, 9, 25, 23, 5

First we arrange the items in ascending order.

5, 5, 8, 9, 12, 21, 23, 25, 32

Here there are 9 items in this series. That is N=9, which is odd.

Hence, \( Median \, =\,value \, of\, {{{(\frac{N + 1}{2})}^{th} }} item \)

\( =\,value \, of\, {{{(\frac{9 + 1}{2})}^{th} }} item \)

\( =\,value \, of\, {{{(\frac{10}{2})}^{th} }} item \)

= value of 5^{th} item = 12

Wages | 100 | 148 | 80 | 9 | 155 | 200 | 145 |
---|

Now we can create a table showing wages arranged in ascending order

Sl.No | Wages arranged in ascending order |
---|---|

1 | 80 |

2 | 90 |

3 | 100 |

4 | 145 |

5 | 148 |

6 | 155 |

7 | 200 |

Median = \( {{{(\frac{N + 1}{2})}^{th} }} item \)

\( ={{{(\frac{7 + 1}{2})}^{th} }} item \) = 4^{th} item.

$$ {\biggl[\frac{1\,2\,3\,\,\,4\,\,\,5\,6\,7}{|\,|\,|\,\,\,|\,\,\,|\,|\,|\,}\biggl]} $$

Size of 4^{th} item = 145, hence we can say that median wage is 145

**DATA OF EVEN NUMBERS**

2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800.

First we arrange the items in ascending order.

1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600.

Here there are 10 items. That is N = 10, which is even.

Since N is even, median = mean of the \( {{{(\frac{N}{2})}^{th} }} \) item and the next item

= mean of the \( {{{(\frac{10}{2})}^{th} }} \) item and the next item

= mean of the 5^{th} item and the 6,^{th} item

= mean of 2250 and 2400

= \( {{{(\frac{2250 + 2400}{2})} }} \) = 2325.

2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800.

First we arrange the items in ascending order.

1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600.

Median = mean of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item

= \( {{{(\frac{10 + 1}{2})}^{th} }} \) item

=
\( {{{(\frac{11}{2})}^{th} }} \) item =5.5^{th} item

= 5^{th} item + 0.5 (6^{th} item - 5^{th} item)

= 2250 + 0.5 (2400 - 2250)

= 2250 + 0.5 × 150 = 2325.

**DISCRETE SERIES**

STEPS

- Arrange the data in ascending or descending order of magnitude
- Take the cumulative frequencies
- Median = the value of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item
- Look at the cumulative frequency column
- Locate the value corresponding to \( {{{(\frac{N + 1}{2})}^{th} }} \) if it is a whole number, otherwise corresponding to higher integer.

Income | Number of persons |
---|---|

100 | 24 |

140 | 25 |

75 | 15 |

200 | 20 |

260 | 6 |

190 | 30 |

We need to arrange the data in ascending order and also need to find cumulative frequency (cf). It is shown in the below given table.

Ascending order | Number of persons (f) | cf |
---|---|---|

75 | 15 | 15 |

100 | 24 | 39 |

140 | 25 | 64 |

190 | 30 | 94 |

200 | 20 | 114 |

260 | 6 | 120 |

Median = size of \( {{{(\frac{N + 1}{2})}^{th} }} \) item

= \( {{{(\frac{120 + 1}{2})}^{th} }} \) item

= 60.5^{th} item

60.5 is included in cumulative frequency 64 and therefore median = 140.

**CONTINUOUS SERIES**

Where,

L = Lower limit of the median class

N = Total frequency

cf = Cumulative frequency of the class preceding the median class

f = Frequency of the median class

h = Class width of the median class

STEPS

- Take the cumulative frequencies
- Find \( {{{(\frac{N}{2})} }} \)
- Find the median class, which is the class corresponding to \( {{{(\frac{N}{2})} }} \)
- Find the values of L, cf, and h
- Use the formula Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Marks | Number of students |
---|---|

0-10 | 1 |

10-20 | 3 |

20-30 | 7 |

30-40 | 12 |

40-50 | 21 |

50-60 | 16 |

60-70 | 9 |

70-80 | 4 |

80-90 | 2 |

First we constructs the cumulative frequency table. Here N is 75 and \( {{{(\frac{N}{2})} }} \) = 37.5. Mark the median class. That is the class corresponding to the cumulative frequency 38. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class.

Marks | Number of students | cf |
---|---|---|

0-10 | 1 | 1 |

10-20 | 3 | 4 |

20-30 | 7 | 11 |

30-40 | 12 | 23 |

40-50 (median class) |
21 |
44 |

50-60 | 16 | 60 |

60-70 | 9 | 69 |

70-80 | 4 | 73 |

80-90 | 2 | 75 |

N = 75 |

L = 40, \( {{{(\frac{N}{2})} }} \)= 37.5, cf = 23, f = 21, h = 10

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 40 + \frac{{37.5} - {23}}{21} × 10} \) = 46.9

**INCLUSIVE CLASS**

X | Frequency |
---|---|

1-5 | 22 |

6-10 | 34 |

11-15 | 53 |

16-20 | 60 |

21-25 | 48 |

26-30 | 26 |

31-35 | 18 |

36-40 | 14 |

In order to find median from distribution with inclusive classes, we have to convert the classes into the exclusive form along with the construction of the cumulative frequency table. In this distribution N = 275 and \( {{{(\frac{N}{2})} }} \) = 137.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 138. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class.Let us create a table with exclusive classes.

Inclusive Class | Converted to Exclusive | f | cf |
---|---|---|---|

1-5 | 0.5-5.5 | 22 | 22 |

6-10 | 5.5-10.5 | 34 | 56 |

11-15 | 10.5-15.5 | 53 | 109 |

16-20 | 15.5-20.5 |
60 |
169 |

21-25 | 20.5-25.5 | 48 | 217 |

26-30 | 25.5-30.5 | 26 | 243 |

31-35 | 30.5-35.5 | 18 | 261 |

36-40 | 35.5-40.5 | 14 | 275 |

L = 15.5, \( {{{(\frac{N}{2})} }} \) = 137.5, cf = 109, f = 60, h = 5

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 15.5 + \frac{{137.5} - {109}}{60} × 5} \) = 17.83

**OPEN END CLASS**

X | Frequency |
---|---|

Less than 100 | 40 |

100-200 | 89 |

200-300 | 148 |

300-400 | 64 |

400 and above | 39 |

In the above distribution, the first and last classes are open end classes. But as in the calculation of AM, it is not needed to make assumption about their class intervals. We can just leave them as they are. In this distribution N = 380 and \( {{{(\frac{N}{2})} }} \) = 190. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 138. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with cumulative frequencies.

X | Frequency | cf |
---|---|---|

Less than 100 | 40 | 40 |

100-200 | 89 | 129 |

200-300 |
148 |
277 |

300-400 | 64 | 341 |

400 and above | 39 | 380 |

L = 200, \( {{{(\frac{N}{2})} }} \) = 190, cf = 129, f = 148, h = 100

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 200 + \frac{{190} - {129}}{148} × 100} \) = 241.21

**LESS THAN CUMULATIVE FREQUENCIES**

Value | Frequency |
---|---|

Less than 10 | 4 |

" " 20 | 16 |

" " - 30 | 40 |

" " 40 | 76 |

" " 50 | 96 |

" " 60 | 112 |

" " 70 | 120 |

Less than 80 | 125 |

In the above given distribution cumulative frequencies are given. We have to find the simple frequency of each class, along with the construction of the cumulative frequency table. The classes are 0-10, 10-20, etc. Simple frequency of first class is 4 itself. Simple fréquency of second class is 16 — 4 = 12 and simple frequency of third class is 40 — 16 = 24: and so on. In this distribution N = 125 and \( {{{(\frac{N}{2})} }} \) = 62.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 63. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Now we can create a table with exclusive class showing normal frequency.

Value | Converted to Exclusive | f | cf |
---|---|---|---|

Less than 10 | 0-10 | 4 | 4 |

" " 20 | 10 - 20 | 12 | 16 |

" " 30 | 20 - 30 | 24 | 40 |

" " - 40 | 30 40 |
36 |
76 |

" " 50 | 40 - 50 | 20 | 96 |

" " 60 | 50 - 60 | 16 | 112 |

" " 70 | 60 - 70 | 8 | 120 |

Less than 80 | 70 - 80 | 5 | 125 |

L = 30, \( {{{(\frac{N}{2})} }} \) = 62.5, cf = 40, f = 36, h = 10

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 30 + \frac{{62.5} - {40}}{36} × 10} \) = 36.25

**MORE THAN CUMULATIVE FREQUENCIES**

Value | Frequency |
---|---|

More than 10 | 50 |

" " 20 | 43 |

" " 30 | 28 |

" " 40 | 13 |

More than 50 | 4 |

In the above given distribution, more than cumulative frequencies are given. We have to find the simple frequency of each class, and less than cumulative frequencies. These classes are 10-20, 20-30, etc. Simple frequency of first class is 50-43 = 7. Simple frequency of second class is 43-28 = 15 and simple frequency of third class is 28-13 = 15, and so on. In this distribution N = 50 and \( {{{(\frac{N}{2})} }} \) = 25. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 25. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with simple frequencies and cumulative frequencies.

Value | Converted to Exclusive | f | cf |
---|---|---|---|

More than 10 | 10-20 | 7 | 7 |

" " 20 | 20-30 | 15 | 22 |

" " 30 | 30-40 |
15 |
37 |

" " 40 | 40-50 | 9 | 46 |

More than 50 | 50-60 | 4 | 50 |

L = 30, \( {{{(\frac{N}{2})} }} \) = 25, cf = 22, f = 15, h = 10

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 30 + \frac{{25} - {22}}{15} × 10} \) = 32

**WITH MIDPOINTS OF CLASSES**

Mid values | Frequencies |
---|---|

2.5 | 4 |

7.5 | 12 |

12.5 | 18 |

17.5 | 10 |

22.5 | 7 |

27.5 | 4 |

In the above given distribution, only mid values are given. We have to find corresponding classes and cumulative frequencies for determining the median. These classes are 0-5, 5-10, etc. In this distribution N = 55 and \( {{{(\frac{N}{2})} }} \) = 27.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 28. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with classes and cumulative frequencies.

X | f | cf |
---|---|---|

0 - 5 | 4 | 4 |

5 - 10 | 12 | 16 |

10 - 15 |
18 |
34 |

15 - 20 | 10 | 44 |

20 - 25 | 7 | 51 |

25 - 30 | 4 | 55 |

L = 10, \( {{{(\frac{N}{2})} }} \) = 27.5, cf = 16, f = 18, h = 5

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 10 + \frac{{27.5} - {16}}{18} × 5} \) = 13.2

**WITH UNEQUAL CLASS INTERVALS**

Values | Frequencies |
---|---|

0 - 10 | 7 |

10 - 20 | 13 |

20 - 50 | 24 |

50 - 70 | 48 |

70 - 80 | 26 |

80 - 100 | 12 |

In the above given distribution, class intervals are unequal. But the frequencies need not to be adjusted to make the class intervals equal. Bu it should be remember that the value of *h* in the formula is the class interval of the median class. In this distribution N = 130 and
\( {{{(\frac{N}{2})} }} \) = 65. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 65. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with cumulative frequencies.

Values | Frequencies | cf |
---|---|---|

0 - 10 | 7 | 7 |

10 - 20 | 13 | 20 |

20 - 50 | 24 | 44 |

50 - 70 |
48 |
92 |

70 - 80 | 26 | 118 |

80 - 100 | 12 | 130 |

L = 50, \( {{{(\frac{N}{2})} }} \) = 65, cf = 44, f = 48, h = 20

Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

Median = \( { 50 + \frac{{65} - {44}}{48} × 20} \) = 58.75

**LOCATING MEDIAN GRAPHICALLY**

We had studied the method of finding median graphically. In order to find the median graphically, we draw the less than ogive and greater than ogive in the same graph and from the point of intersection of the ogives we draw a line perpendicular to the x-axis. Then, the point where the perpendicular touches the x-axis will give the value of the median. We can also very well locate median by drawing a single ogive. For example, if we draw the less than ogive, then take \( {{{\frac{N}{2}} }} \) on the y-axis and draw a perpendicular from y-axis to meet the ogive. From the point where it meets the ogive, draw another perpendicular on the x-axis. That point on the x-axis will give the value of the median.

Let us find median of the following distribution using less than ogive.

Values | Frequencies |
---|---|

0 - 2 | 2 |

2 - 4 | 4 |

4 - 6 | 5 |

6 - 8 | 8 |

8 - 10 | 7 |

10 - 12 | 4 |

We can create a table showing cumulative frequencies.

Values | Frequencies | cf |
---|---|---|

0 - 2 | 2 | 2 |

2 - 4 | 4 | 6 |

4 - 6 | 5 | 11 |

6 - 8 | 8 | 19 |

8 - 10 | 7 | 26 |

10 - 12 | 4 | 30 |

N = 30 |

Using above given cumulative frequency table we loacate median. You can see how median is located using less than ogive from the below given graph.

### MERITS OF MEDIAN

- It is easy to compute and understand
- It gives best results with open-end classes
- It is not influenced by the magnitude of extreme deviations
- It is the most appropriate average in dealing with qualitative data
- The value of median can be determined graphically

### DEMERITS OF MEDIAN

- It is only a positional average
- It may not be the true representative of the series in many cases.
- It is not based on all items
- The value of median is affected by sampling fluctuations
- It is not capable of algebraic treatment