Median

Median

Median of a set of observations is the middle-most value when observations are arranged in order of magnitude. The number of observations smaller than median is same as the number greater than it. Thus median divided the observations into two equal parts. Median is, in certain sense, the real measure of central tendency, as gives the it gives the value of most central observation. It is unaffected by the presence of the extremely large and small observations. It can be calculated from frequency distribution with open-end class. Median finds largest application in psychological and achievement test,e.g. to find boy of average intelligence, the candidates may be ranked in order of intelligence and median is employed.

Calculation of Median The median is calculated as follows:1)From simple series. 2)From grouped frequency distribution.

1) From simple series First, The given data are arranged in order of magnitude a)If the number of observations be odd, the value of middle most item is the median i.e. N +1 Median = value of ----------- th observation. 2

Problem - 1 Q) Find median of 32,22,,29,17,40,26,21. Solution:Arranged the data in order of magnitude i.e. 17,21,26,26,29,32,40 Median is the middle most item i.e. N +1 7+1 the ----------- = -----------= 4 th observation. 2 2 Therefore the median of the above series is 26.(Ans)

b) If the number of observations be even, then the arithmetic mean of two middle most items is taken as median. For example consider the following problem:Q) Find the median of 3.1,2.6,5.0,4.7,2.4,3.9,5.1,3.6 Solution:First arrange the data in order of magnitude i.e. 2.4,2.6,3.1,3.6,3.9,4.7,5.0,5.1 N +1 8+1 Median= ----------- = -----------= 4.5 th observation. 2 2 i.e. arithmetic mean of 4 th and 5 th observation 3.6 + 3.9 Or Median = ----------------------= 3.75 (Ans)

2

2) From grouped frequency distribution. Median from a grouped frequency distribution is that value which corresponds to cumulative frequency N/2.Median from a grouped frequency distribution can be calculated by the application of simple interpolation in a cumulative frequency distribution.

If F1 and F2 be the cumulative frequencies shown in the table which are just smaller than and just larger than N/2 and they corresponds to the class boundaries I1 and I2 respectively, then Median – I1 ----------------------------- = ------------------------------

I2 - I1

N/2 - F1 F2 - F1

I1 and I2 are the lower and upper boundaries of the median class.

Illustration - 1 Find the median of the given data, Class Boundaries 15 ----------- 25 25 ----------- 35 35 ----------- 45 45 ----------- 55 55 ----------- 65 65 ----------- 75

Frequencies 4 11 19 14 0 2

Class Boundary

Cumulative frequency (less-than)

15 25 35 Median -----------45 55 65 75

or

Median – 35 25 - 15 ------------------ = ---------------------45 – 35 34 – 15

or

Median – 35 10 ---------------------- = -----------------10 19 Median = 40.26 (Ans)

0 4 15 -------- N/2 = 25 34 48 48 50 = N

Illustration - 2 Find the median of the following table:Wages

50 - 59

60 – 69

70 - 79

80 – 89

90 - 99

100 – 109 110 - 119

No of employee

8

10

16

14

10

5

2

Class Limits

Class Boundaries

Frequency

50 - 59

49.5 – 59.5

8

60 – 69

59.5 – 69.5

10

70 – 79

69.5 – 79.5

16

80 – 89

79.5 – 89.5

14

90 – 99

89.5 – 99.5

10

100 – 109

99.5 – 109.5

5

110 - 119

109.5 – 119.5

2

Class boundary

Cumulative frequency “lessthan” type

49.5 59.5 69.5 Median ------------79.5 89.5 99.5 109.5 119.5

or

0 8 18 --------- N/2 =32.5 34 48 58 63 65 = N

Median – 69.5 32.5 - 18 ------------------ = ---------------------79.5 – 69.5 34 – 18

Or Median = 69.5 + (14.5/16) x 10 = 78.56(Ans)

Most suitable average for frequency distribution with open end classes The most suitable average for frequency distribution with open end classes is Median. For example, consider the following distribution

Value

Frequency

Less than 100

40

100 - 200

89

200 - 300

148

300 - 400

64

400 and above

39

Total

380

Class boundary

100 200 Median-------300 400 -

Cumulative Frequency (less-than) 40 129 ----- N/2 = 190 277 341 380 = N

or

or

Median – 200 190 - 129 ------------------ = ---------------------300 – 200 277 – 129 Median = 200 + 100 x (61/148) = 241.2 (Ans)

Quartiles, Deciles and Percentiles Besides median, there are other measures which divide a series into a equal number of parts. Important amongst these are quartiles,deciles,and percentiles.

Quartile Quartiles are such values which divide the total observation into four equal parts. Obviously, there are 3-quartiles i) First quartile ( or lower quartile) : Q1 ii) Second quartile ( or Middle quartile) : Q2 iii)Third quartile (or upper quartile) : Q3 •

This means Q1, Q2 and Q3 are values of the variable corresponding to “less-than” cumulative frequencies N/4, 2N/4, 3N/4 respectively. Since , 2N/4 = N/2, it is evident that the second quartile Q2 is the same as Median.

Deciles Deciles are such values which divide the total number of observations in to 10 equal parts. There are 9 deciles D1,D2,……..,D9 called the first decile, second decile, etc. D1,D2,……..,D9 correspond to the cumulative frequencies N/10,2N/10,……,9N/10 respectively.

Percentiles Percentiles are such values which divide the total number of observations in to 100 equal parts. There are 99 percentiles P1,P2,……..,P99 called the first percentile, second percentile, etc. P1,P2, ……..,P99 correspond to the cumulative frequencies N/100,2N/100,……,99N/10 respectively.

Illustration The profit earned by 100 companies during 20032004 are given below:Calculate Q1, median , D4 and P80 interpret the values. Profits (Rs.lakhs)

Number of companies

Profits (Rs.lakhs)

Number of companies

20 – 30 30 – 40 40 – 50 50 - 60

4 8 18 30

60 – 70 70 – 80 80 – 90 90 - 100

15 10 8 7

Profits (Rs.lakhs) 20 30 40 Q1 ------------50 D4 ----------------------Median = Q2---------60 70 P80 -----------------80 90 100

Cumulative frequency (Less than type) 0 4 12 -------- N/4 = 25 30 ---------- 4N/10 = 40 --------- N/2 = 50 60 75 ----------- 80N/100 = 80 85 93 100 = N

or

Q1 – 40 25 - 12 ------------------ = ---------------------50 – 40 30 – 12

or

Q1 = 47.22 (Ans)

 25% of the companies earn annual profit of Rs. 47.22 lakhs and less. or

Median – 50 50 - 30 ------------------ = ---------------------60 – 50 60 – 30

or

Median = 56.67 (Ans)

 50% of the companies earn annual profit of Rs. 56.67 lakhs and less.

or

D4 – 50 40 - 30 ------------------ = ---------------------60 – 50 60 – 30

Or

D4

= 53.33 (Ans)

 40% of the companies earn annual profit of Rs. 53.33 lakhs and less.

or Or

P80 – 70 80 - 75 ------------------ = ---------------------80 – 70 85 – 75 P80 = 75 (Ans)

 80% of the companies earn annual profit of Rs. 75 lakhs and less.

Advantages 1) The median is superior to arithmetic mean in certain respects:b) It is especially useful in case of open-end distribution. c) It is not influenced by presence of extreme values. In fact when extreme values are present in the data, the median is a more satisfactory measure of central tendency than a mean.

c) Even when all the observations are not known, median can be calculated, provided the general location of all observations and the values near the middle are available. 2) Median is applicable to qualitative data in psychological and social studies, where numerical measurements may not be available but it is possible to rank the objects in some order.

Disadvantages 1) Unlike A.M it can not be treated algebraically. Given the median of several groups of observations, median of the composite group can not be determined. 2) Median is affected more by sampling fluctuations than the arithmetic mean and is therefore less reliable.