Find mean, standard deviation and coefficient of variation.

Class-interval 0−4 4−8 8−12 12−16 16−20 20−24

Frequency 10 15 20 25 20 10

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Solution

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

0 − 4
4 − 8
8 − 12
12 − 16
16 − 18
20 − 24 10
15
20
25
20
10 2
6
10
14
18
22 20
90
200
350
360
220 4
36
100
196
324
484 40
540
2000
4900
6480
4840

Σf = 100 Σfm = 1240 Σfm² = 18800

$Mean (X) = \frac{Σ f m}{Σ f} = \frac{1240}{100} = 12.4 S . D . (σ) = \sqrt{\frac{Σ f m^{2}}{Σ f} - {(X)}^{2}} = \sqrt{\frac{18800}{100} - {(12.4)}^{2}} = \sqrt{188 - 153.76} = \sqrt{34.24} = 5.85 C . V . = \frac{σ}{X} \times 100 = \frac{5.85}{12.4} \times 100 = 47.17 %$

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Calculation (i) Arithmetic Mean and (ii) Standard Deviation for the following frequency distribution.

Class Interval

Frequency

20 – 25

25 – 30

30 – 35

35 – 40

40 – 45

45 – 50

N = 50

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Class-interval	0−4	4−8	8−12	12−16	16−20	20−24
Frequency	10	15	20	25	20	10