Find the standard deviation and coefficient of standard deviation:

X (less than) 10 20 30 40 50 60 70 80

Frequency 12 30 65 107 157 202 222 230

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Solution

Converting less than frequency distribution into simple frequency distribution:

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

0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80 12-0=12
30-12=18
65-30=35
107-65=42
157-107=50
202-157=45
222-202=20
230-222=8 5
15
25
35
45
55
65
75 60
270
875
1470
2250
2475
1300
600 25
225
625
1225
2025
3025
4225
5625 300
4050
21875
51450
101250
136125
84500
45000

Σf = 230 Σfm = 9300 Σfm² = 444550

$Mean (X) = \frac{Σ f m}{Σ f} = \frac{9300}{230} = 40.43 Standard Deviation (σ) = \sqrt{\frac{Σ f m^{2}}{Σ f} - {(X)}^{2}} = \sqrt{\frac{444550}{230} - {(40.43)}^{2}} = \sqrt{1932.83 - 1634.58} = \sqrt{298.25} = 17.26 Coefficient of standard deviation = \frac{σ}{X} = \frac{17.26}{40.47} = 0.43$

Suggest Corrections

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of persons	5	10	20	40	30	20	10	4

Class Interval	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	5	10	15	20	25	18	7

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	5	10	20	40	30	20	10	4

Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Frequency	3	61	132	153	140	51	2

Class	0−10	10−20	20−30	30−40	40−50	50−60	60−70
Frequency	2	4	6	8	6	4	2