Saturday 1 March 2014

STATISTICS C1


Distribution of length of nails in mm is as shown below. 

Length(mm)
15 - 21
22- 28
29 - 35
36 - 42
43 - 49
50 - 56
57 - 63
Frequency
8
12
20
28
16
10
6

Calculate the median.

Solution

First we prepare the frequency distribution table.

Class interval
Frequency(f)
Cumulative frequency
15-21
8
8
22-28
12
20
29-35
20
40
36-42
28
68
43-49
16
84
50-56
10
94
57-63
6
100

∑f = 100


N = 100, N/2=25, Median class must fall in the cumulative frequency of 50. This has to be 36 – 42.
nb = 40, nw = 28, Upper boundary(U)= 42.5, Lower boundary(L) =35.5
 i = Upper boundary – Lower boundary
i = 42.5 – 35.5
i = 7

Median = L + (N/2 – nb)i/nw
                              

Median = 35.5 + (100/2 – 40)7
                                    28

Median = 35.5 + (50 – 40)7
                                     28

Median = 35.5 + (10)7
                                14

  Median = 35.5 + 70
                                14

  Median = 35.5 +  5
                               
 Median = 40.5

Hence the median is 40.5 
            
TRY THIS…..

Distribution of length of nails in mm is as shown below. 

Length(mm)
15 - 21
22- 28
29 - 35
36 - 42
43 - 49
50 - 56
57 - 63
Frequency
6
10
18
24
16
10
6


Calculate the median.

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