STANDARD DEVIATION FOR CONTINUOUS DATA

Mean Method :

where xi = Middle value of the ith class.

fi = Frequency of the ith class

Step Deviation Method : 

To make the calculation simple, we provide the following formula. Let A be the assumed mean, xi be the middle value of the ith class and c is the width of the class interval.

Example 1 :

The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.

Solution :

20.5 - 24.5

15

24.5 - 28.5

18

28.5 - 32.5

20

32.5 - 36.5

16

36.5 - 40.5

8

40.5 - 44.5

7

x is the mid value of the given set.

d = (x - 34.5)/2

x


22.5

26.5

30.5

34.5

38.5

42.5


-6

-4

-2

0

2

4

f


15

18

20

16

8

7

d2


36

16

4

0

4

16

fd2


540

288

80

0

32

112

fd


-90

-72

-40

0

16

28

Σf  =  N  =  84

Σfd =  1052

(Σfd2/N)  =  (1052/84)  =  263/21

Σfd  =  -158

(Σfd/N)2  =  (-158/84)=  (79/42)

σ  =  √(22092-6241)/1764

σ  =  (√15851/1764) ⋅ 2

σ  =  √8.98 ⋅ 2

σ  =  2.99 ⋅ 2

σ  =  5.99

Hence the required standard deviation is 6.

Example 2 :

The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation

Solution :

x is the mid value of the given set.

d = (x - 22)/2

x


18

20

22

24

26


-2

-1

0

1

2

f


6

8

17

10

9

d2


4

1

0

1

4

fd2


24

8

0

10

36

fd


-12

-8

0

10

18

Σf  =  N  =  50

Σfd =  78

(Σfd2/N)  =  (78/50)  =  39/25

Σfd  =  8

(Σfd/N)2  =  (8/50)=  (4/25)2   =  16/625

σ  =  √(975 - 16)/625

σ  =  √959/625

σ  =  (√1.533) ⋅ 2

σ  =  1.23 ⋅ 2

σ  =  2.47

Hence the required standard deviation is 2.47.

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