Calculate standard deviation and coefficient of dispersion from the data below:

Mid-Points 5 15 25 35 45 55 65 75

Frequency 5 8 7 12 28 20 10 10

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Solution

Mid-Values
(m) Frequency
(f) fm m² fm²

5
15
25
35
45
55
65
75 5
8
7
12
28
20
10
10 25
120
175
420
1260
1100
650
750 25
225
625
1225
2025
3025
4225
5625 125
1800
4375
14700
56700
60500
42250
56250

Σf = 100 Σfm= 4500 Σfm² = 236700

$Mean (X) = \frac{Σ f m}{Σ f} = \frac{4500}{100} = 45 Standard Deviation (σ) = \sqrt{\frac{Σ f m^{2}}{Σ f} - {(X)}^{2}} = \sqrt{\frac{236700}{100} - {(45)}^{2}} = \sqrt{2367 - 2025} = \sqrt{342} = 18.49 C . V = \frac{σ}{X} \times 100 = \frac{18.49}{45} \times 100 = 41.09 %$

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Frequency	6	10	16	15	24	8	7

Prepare a frequency distribution from the following data:

Mid-Points	25	35	45	55	65	75
Frequency	6	10	9	12	8	5

Size	5−10	10−15	15−20	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	6	10	18	30	15	12	10	6	4

Size	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	2	5	7	13	21	16	8	3

Marks	5−15	15−25	25−35	35−45	45−55	55−65
No. of Students	8	12	6	14	7	3

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