STATISTICS B13

Distribution of French test marks was given as hereunder.

Length(mm)	33 - 37	38- 42	43 - 47	48-52	53 - 57	58 - 62	63 - 67
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
33 - 37	5	5
38- 42	7	12
43 - 47	10	22
48 - 52	12	34
53 - 57	8	42
58 - 62	5	47
63 - 67	3	50
	∑f = 50

N = 45, N/2=22.5, Median class must fall in the cumulative frequency of 25. This has to be 48 – 52.

n_b = 22, n_w = 12, Upper boundary(U)= 52.5, Lower boundary(L) =47.5

i = Upper boundary – Lower boundary

i = 52.5 – 47.5

i = 5

Median = L + (N/2 – n_b)i/n_w

                              

Median = 47.5 + (50/2 – 22)5

                                  12

Median = 47.5 + (25 – 22)5

                                 12

Median = 47.5 + (3)5

                             12

Median = 47.5 + 15

                             12

Median = 47.5 +  1.25

                               

Median = 48.75

Hence the median is 48.75