Find the mean of the following frequency distribution using step - deviation method:
Class | 84 - 90 | 90 - 96 | 96 - 102 | 102 - 108 | 108 - 114 | 114 - 120 |
Frequency | 15 | 22 | 20 | 18 | 20 | 25 |
We will find the mean using step - deviation method, where A = Assumed mean and h = length of class interval.
Here, let A = 99 and h = 6
Since, the class intervals are inclusive type, we’ll first convert it into exclusive type by extending the class interval from both the ends.
CLASS | MID - POINT(xi) | DEVIATION(di) di = xi – 99 | FREQUENCY(fi) | ui = di/h | fiui |
84 - 90 | 87 | - 12 | 15 | - 2 | - 30 |
90 - 96 | 93 | - 6 | 22 | - 1 | - 22 |
96 - 102 | 99 = A | 0 | 20 | 0 | 0 |
102 - 108 | 105 | 6 | 18 | 1 | 18 |
108 - 114 | 111 | 12 | 20 | 2 | 40 |
114 - 120 | 117 | 18 | 25 | 3 | 75 |
TOTAL | 120 | 81 |
We have got
A = 99, h = 6, Σfi = 120 & Σfiui = 81
∵ mean is given by
⇒ 
⇒ 
Thus, mean is 103.05.
18