Find the mean of each of the following frequency distributions :
Class Interval:
25−29
30−34
35−39
40−44
45−49
50−54
55−59
Frequency:
14
22
16
6
5
3
4
Answers
STEP DEVIATION METHOD:
Step deviation method is used in the cases where the deviation from the assumed mean 'A' are multiples of a common number. If the values of ‘di’ for each class is a multiple of ‘h’ the calculation become simpler by taking ui= di/h = (xi - A )/h
Here, h is the class size of each class interval.
★★ Find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi.
★★ Class marks (xi) = ( lower limit + upper limit) /2
★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn
MEAN = A + h ×(Σfiui /Σfi) , where ui = (xi - A )/h
[‘Σ’ Sigma means ‘summation’ ]
FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT
★★ Here,the frequency table is given in inclusive form. So we first change it into exclusive form by subtracting and adding h/2 to the lower and upper limits respectively of each class ,where ‘h’ denotes the difference of lower limit of a class and upper limit of the previous class.
Here, h/2 = ½ = 0.5
From the table : Σfiui = -79, Σfi = 70
Let the assumed mean, A = 42, h = 5
MEAN = A + h ×(Σfiui /Σfi)
MEAN = 42 + 5 (-79 /70)
= 42 - 79/14
= 42 - 5.6428
= 36.357
Mean = 36.357
Hence, the mean is 36.357
HOPE THIS ANSWER WILL HELP YOU….
Class interval mid-point of CI(x) frequency f fx
15-19. 17 3 51
20-24 22 12 264
25-29 27 21 567
30-34 32 15 480
35-39 37 5 185
40-45 42 4 168
45-49 47 2 94
62 1809
mean of the given data is
1809/62= 29.18
OK ☺☺☺