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 |
d -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
Σfd2 = 1052
(Σfd2/N) = (1052/84) = 263/21
Σfd = -158
(Σfd/N)2 = (-158/84)2 = (79/42)2
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 |
d -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
Σfd2 = 78
(Σfd2/N) = (78/50) = 39/25
Σfd = 8
(Σfd/N)2 = (8/50)2 = (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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