Using short cut method, find the mean, variation and standard deviation for the data :
Class | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Frequency | 3 | 7 | 12 | 15 | 8 | 3 | 2 |
Here, we apply the step deviation method with A = 65 and h = 10
To find: MEAN
Class | (fi) | Class Mark (xi) | di = xi - A |
| fiyi |
30 – 40 | 3 | 35 | 35 – 65 = -30 | -3 | -9 |
40 – 50 | 7 | 45 | 45 – 65 = -20 | -2 | -14 |
50 – 60 | 12 | 55 | 55 – 65 = -10 | -1 | -12 |
60 – 70 | 15 | 65 = A | 0 | 0 | 0 |
70 – 80 | 8 | 75 | 75 – 65 = 10 | 1 | 8 |
80 – 90 | 3 | 85 | 85 – 65 = 20 | 2 | 6 |
90-100 | 2 | 95 | 95 – 65 = 30 | 3 | 6 |
Total | ∑fi = 50 | ∑fiyi =-15 |
Now, 
To find: VARIANCE
(fi) | (xi) | yi | yi2 | fiyi | fiyi2 |
3 | 35 | -3 | (-3)2 = 9 | -9 | 27 |
7 | 45 | -2 | (-2)2 = 4 | -14 | 28 |
12 | 55 | -1 | (-1)2 = 1 | -12 | 12 |
15 | 65 = A | 0 | 0 | 0 | 0 |
8 | 75 | 1 | (1)2 = 1 | 8 | 8 |
3 | 85 | 2 | (2)2 = 4 | 6 | 12 |
2 | 95 | 3 | (3)2 = 9 | 6 | 18 |
N= 50 | ∑fiyi =-15 | ∑fiyi2 = 105 |
= 201
To find: STANDARD DEVIATION
= 14.17
1