Median is also a measure of central tendency. As the name itself suggests, it is the value of the middle item of a series arranged in ascending or descending order of magnitude. Thus if there are seven items in a series arranged in ascending or descending order of magnitude, then median will be the magnitude of the 3" item. This item would divide the series in two equal parts; one part containing values less than the median value and the other part containing values above the median value. Thus, median is the middle most element of a series when it is arranged in the order of the magnitude of the items. If however there are even number of items in a series, then we cannot find a central item which divides the series in to two equal parts. For example, if there are eight items in a series, then, there is no single item in the middle, but two items, namely, the 4" and the 5". Then we take the arithmetic mean of the two middle items as the median. We can see that as against arithmetic mean, which is based on all items of the distribution, the median is only a positional value depends on the position occupied by the item.

Thus we can say that,

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

INDIVIDUAL SERIES

STEPS

  1. Arrange the data in ascending or descending order
  2. Locate the \( {{{(\frac{N + 1}{2})}^{th} }} item \, if \, N \, is\, odd \)
  3. Find the mean of the \( {{{(\frac{N}{2})}^{th} }} \) item and the next item, if N is even
Median can be easily computed by sorting the data from smallest to largest and counting the middle value.

DATA OF ODD NUMBERS

  • Let us find the median of the following series.
  • 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 5th item = 12

  • Let us compute the median from Weekly wages of 7 workers.

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

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

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

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

    DATA OF EVEN NUMBERS

  • Let us find median of the following series
  • 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 5th item and the 6,th item

    = mean of 2250 and 2400

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

  • Let us find median of the following series using another method
  • 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 =5.5th item

    = 5th item + 0.5 (6th item - 5th item)

    = 2250 + 0.5 (2400 - 2250)

    = 2250 + 0.5 × 150 = 2325.

    DISCRETE SERIES

    In discrete series items are grouped and frequencies are given. There will not be any classes. For finding the median of a discrete series, first we arrange the items in ascending or descending order as we did in individual series. Then we find the less than cumulative frequencies. Now it is easy to locate the value of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item.

    STEPS

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

    = 60.5th item

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

    CONTINUOUS SERIES

    In continuous series data is given in frequency classes. For finding the median, first we identify the median class. Median class is the class in which the value corresponding to the frequency \( {{{(\frac{N}{2})} }} \) may lie. After identifying the median class, we use the following formula for finding median:

    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

    1. Take the cumulative frequencies
    2. Find \( {{{(\frac{N}{2})} }} \)
    3. Find the median class, which is the class corresponding to \( {{{(\frac{N}{2})} }} \)
    4. Find the values of L, cf, and h
    5. Use the formula Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)
    If classes are not in exclusive form, we need to convert them into exclusive form.

  • Let us find median of the following frequency distribution.
  • 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

  • Let us find median from inclusive type classes
  • 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

  • Let us find median from open end type classes
  • 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

  • Let us find median from the following distribution.
  • 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

  • Let us find median from the following distribution.
  • 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

  • Let us find median from the following distribution.
  • 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

  • Let us find median from the following distribution.
  • 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

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

    DEMERITS OF MEDIAN

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