Question

Find the mean of the following data by using step-deviation method

Class	10-14	15-19	20-24	25-29	30-34
Frequency	15	110	135	115	25

Answer 1

Hint: First of all, the mean of any data using standard deviation method can be given by the formula, $Mean,\bar{X}=l+{\displaystyle \frac{\sum _{i=1}^n{f}_i{u}_i}{\sum _{i=1}^n{f}_i}}\times h$. Now, we will find the values of summation of frequencies, summation of product of $f_i$ and $u_i$ by making a table of distribution and then based on the values of that table and summation we will find our answer.

Complete step-by-step answer:
First of all, we will convert this table into vertical and based on the formula of mean i.e. $Mean,\bar{X}=l+{\displaystyle \frac{\sum _{i=1}^n{f}_i{u}_i}{\sum _{i=1}^n{f}_i}}\times h$, where, l is middle value of the given data that is assumed as mean for calculation, h is class size, so the table can be given as,

Class	Frequency ($f_i$)
10-14	15
15-19	110
20-24	135
25-29	115
30-34	25

Now, $x_i$ is the middle term in each class such as, to find the value of $x_i$ let’s consider the class 10-14, so the terms in 10-14 are 10, 11, 12, 13 and 14. So, $x_i$ will be 12. In the same way for class 15-19, $x_i$ will be 17, for 20-24 it will be 22, for 25-29 it will be 27 and for 30-34 it will be 32. So, we will substitute the values in the table.

Class	Frequency ($f_i$)	$x_i$
10-14	15	12
15-19	110	17
20-24	135	22
25-29	115	27
30-34	25	32

Now, $d_i$ is the difference between $x_i$ and the middle term of the terms of $x_i$ i.e. 22, which can be seen mathematically as,
$d_i=x_i-22$
Now, substituting the values of $x_i$ related to each class we will find the value of $d_i$, so, for class 10-14, value of $d_i$ will be,
$d_i=12-22=-10$
Same way, for class 15-19, value of $d_i$ will be, $17-22=-5$, for class 20-24 value of $d_i$ will be, $22-22=0$, for class 25-29, value of $d_i$ will be, $27-22=5$ and for class 30-34 value of $d_i$ will be, $32-22=10$. So, now, we will substitute these values in table so the table will be as,

Class	Frequency ($f_i$)	$x_i$	$d_i$
10-14	15	12	-10
15-19	110	17	-5
20-24	135	22	0
25-29	115	27	5
30-34	25	32	10

Now, value of $u_i$ can be given by the formula, $u_i={\displaystyle \frac{d_i}{5}}$, now, on substituting the values of $d_i$ we will get values of $u_i$ as,
For, $d_i=-10$, $u_i={\displaystyle \frac{-10}{5}}=-2$
For, $d_i=-5$, $u_i={\displaystyle \frac{-5}{5}}=-1$
For, $d_i=0$, $u_i={\displaystyle \frac{0}{5}}=0$
For, $d_i=5$, $u_i={\displaystyle \frac{5}{5}}=1$
For, $d_i=10$, $u_i={\displaystyle \frac{10}{5}}=2$
On substituting these values, we will get table as,

Class	Frequency ($f_i$)	$x_i$	$d_i$	$u_i={\displaystyle \frac{d_i}{5}}$
10-14	15	12	-10	-2
15-19	110	17	-5	-1
20-24	135	22	0	0
25-29	115	27	5	1
30-34	25	32	10	2

Now, summation of frequencies can be given as,
$\sum f_i=15+210+135+115+25=400$
Now, product of $f_i$ and $u_i$ can be given as,

Class	Frequency ($f_i$)	$x_i$	$d_i$	$u_i={\displaystyle \frac{d_i}{5}}$	$f_i{u}_i$
10-14	15	12	-10	-2	-30
15-19	110	17	-5	-1	-110
20-24	135	22	0	0	0
25-29	115	27	5	1	115
30-34	25	32	10	2	50

Now, summation of product of $f_i$ and $u_i$ can be given as
$\sum f_i{u}_i=-30-110+0+115+50=-140+165$
$\Rightarrow \sum f_i{u}_i=25$
Now, equation of mean can be given as,
$Mean,\bar{X}=l+{\displaystyle \frac{\sum _{i=1}^n{f}_i{u}_i}{\sum _{i=1}^n{f}_i}}\times h$
Where, $h=5$ as the difference between the size of class for all the classes is 5, which can be calculated by reducing 0.5 from the value of class then taking their subtraction which can be given as $13.5-9.5=5$, and $l=22$. On substituting all the values in equation, we will get,
$\bar{X}=22+{\displaystyle \frac{25}{400}}\times 5$
$\Rightarrow \bar{X}=22+{\displaystyle \frac{25\times 5}{400}}=22+0.3125$
$\Rightarrow \bar{X}=22+0.3125=22.3125$
Hence, the mean of the given data using step deviation method is 22.3125.

Note: Here, in question it was mentioned to find the mean by step deviation method so we solved the problem using the formula, $Mean,\bar{X}=l+{\displaystyle \frac{\sum _{i=1}^n{f}_i{u}_i}{\sum _{i=1}^n{f}_i}}\times h$. Now, students might use the formula, $Mean,\bar{X}={\displaystyle \frac{\sum _{i=1}^n{f}_i{x}_i}{\sum _{i=1}^n{f}_i}}$, where, product of $f_i$ and $x_i$ for each class will be,

Class	Frequency ($f_i$)	$x_i$	$f_i{x}_i$
10-14	15	12	225
15-19	110	17	1870
20-24	135	22	2970
25-29	115	27	3105
30-34	25	32	800

So, the summation of product of $f_i$ and $x_i$ will be, $\sum f_i{u}_i=225+1870+2970+2530+675=$, on substituting the values we will get, $$\bar{X}={\displaystyle \frac{8225}{400}}=22.425$$ . As we can see that the answer remains the same with minute variation in it, so, we can consider this method as an alternative method to the problem.