Question

Find the mean of each of the following frequency distributions :
Class interval:
10−30
30−50
50−70
70−90
90−110
110−130
Frequency:
5
8
12
20
3
2

Answer 1

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  

From the table : Σfiui = 14, Σfi = 50

Let the assumed mean, A = 60,  h = 20

MEAN = A + h ×(Σfiui /Σfi)

MEAN = 60  + 20(14/50)

= 60 + 28/5  

= 60 + 5.6

= 65.6

Mean = 65.6

Hence, the mean is 65.6

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

HIII BUDDY!!!


HERE'S YOUR ANSWER:-


$solution$



MEAN=A+h×sigma f1×u1/N



Mean=60+20×14/50



=60+280



=60+5.6



=65.6 ans



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