Question

Find the mean of each of the following frequency distributions:
Class interval:
0−6
6−12
12−18
18−24
24−30
Frequency:
7
5
10
12
6

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 = 5   , Σfi = 40

Let the assumed mean, A = 15,  h = 6

MEAN = A + h ×(Σfiui /Σfi)

MEAN = 15  + 6(5/40)

= 15 + 6/8

= 15 + ¾  

= 15 + 0.75

= 15.75

Mean = 15.75

Hence, the mean is 15.75  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution:
➖➖➖
Here I had solved mean with Direct Method:

1) Calculation for Class Mark:

$x_{i}=\frac{upper\:limit+lower\:limit}{2}$

by this way one can calculate class Mark for each class.

$\begin{table}[] \centering \begin{tabular}{|l|l|l|l|} \cline{1-4} C.I. & Freq. & Class Mark & Xifi \\ \cline{1-4} 0-6 & 7 & 3 & 21 \\ \cline{1-4} 6-12 & 5 & 9 & 45 \\ \cline{1-4} 12-18 & 10 & 15 & 150 \\ \cline{1-4} 18-24 & 12 & 21 & 252 \\ \cline{1-4} 24-30 & 6 & 27 & 162 \\ \cline{1-4} & \Sigma=40 & & \Sigma=630 \\ \cline{1-4} \end{tabular} \end{table}$

✔️Direct Mean:

$\bar x=\frac{\Sigma x_{i}f_{i}}{\Sigma f_{i}}\\\\\bar x=\frac{630}{40}\\\\\bar x=15.75$

Hope it helps you.