Calculate Standard deviation from the following series:

Class 0−10 10−20 20−30 30−40 40−50 50−60 60−70

Frequency 2 4 6 8 6 4 2

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Solution

Class Interval
(X)
Frequency
(f)
Mid-Values
(m) fm m² f×m2=fm2
0−10 2 5 10 25 50
10−20 4 15 60 225 900
20−30 6 25 150 625 3750
30−40 8 35 280 1225 9800
40−50 6 45 270 2025 12150
50−60 4 55 220 3025 12100
60−70 2 65 130 4225 8450
Σf=32 Σfm=1120 Σfm²=47200

$M e a n = \frac{Σ f m}{Σ f} = \frac{1120}{32} = 35$
$S t a n d a r d d e v i a t i o n, σ = \sqrt{(\frac{(\sum f m^{2})}{(\sum f)} - {(\frac{(\sum f m)}{\sum f})}^{2})}$
$S t a n d a r d d e v i a t i o n, σ = \sqrt{(\frac{47200}{32} - {(\frac{1120}{32})}^{2})} = \sqrt{1475 - {(35)}^{2}} = \sqrt{1475 - 1225} = \sqrt{250}$
$= 15.81$

Hence, standard deviation of the above series is 15.81

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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

Calculate the mean for the following frequency distribution:

Marks	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	3	7	10	6	8	2	4

Class	0−10	10−20	20−30	30−40	40−50	50−60	60−70
Frequency	6	8	10	15	5	4	2

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

Class Interval	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	4	8	5	4	9	10

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