STATISTICS B11

Distribution of French test marks was given as hereunder.

Length(mm)	20 - 27	28- 35	36 - 43	44-51	52 - 59	60 - 67	68 - 75
Frequency	3	5	9	12	6	3	2

Calculate the median.

Solution

First we prepare the frequency distribution table.

Class interval	Frequency(f)	Cumulative frequency
20 - 27	5	5
28- 35	6	11
36 - 43	9	20
44 - 51	12	32
52 - 59	7	39
60 - 67	4	43
68 - 75	2	45
	∑f = 45

N = 45, N/2=22.5, Median class must fall in the cumulative frequency of 22.5. This has to be 44 – 51.

n_b = 20, n_w = 12, Upper boundary(U)= 51.5, Lower boundary(L) =43.5

i = Upper boundary – Lower boundary

i = 51.5 – 43.5

i = 8

Median = L + (N/2 – n_b)i/n_w

                              

Median = 43.5 + (45/2 – 20)8

                                     12

Median = 43.5 + (22.5 – 20)8

                                   12

Median = 43.5 + (2.5)8

                               12

  Median = 43.5 + 20

                             12

  Median = 43.5 +  1.67

                               

 Median = 45.17

Hence the median is 45.17