STATISTICS B12

Distribution of French test marks was given as hereunder.

Length(mm)	15 - 19	20- 24	25 - 29	30 - 34	35 - 39	40 - 44	45 - 49
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
15-19	5	5
20- 24	6	11
25 - 29	9	20
30 - 34	12	32
35 - 39	7	39
40 - 44	4	43
45 - 49	2	45
	∑f = 45

N = 45, N/2=22.5, Median class must fall in the cumulative frequency of 20. This has to be 30 – 34.

n_b = 20, n_w = 12, Upper boundary(U)= 34.5, Lower boundary(L) =29.5

 i = Upper boundary – Lower boundary

i = 34.5 – 29.5

i = 5

Median = L + (N/2 – n_b)i/n_w

                              

Median = 29.5 + (45/2 – 20)5

                                     12

Median = 29.5 + (22.5 – 20)5

                                    12

Median = 29.5 + (2.5)5

                               12

  Median = 29.5 + 12.5

                               12

  Median = 29.5 +  1.0417

                               

 Median = 30.54167

Hence the median is 30.54