Question

Use step deviation method to find the mean of the following table.

Number of letters	120	240	360	480
Receivers	20	40	60	80

(a) 120
(b) 240
(c) 360
(d) 480

Answer 1

Hint: We start solving the problem by finding the sum of the Receivers and the length of each class interval. We then assume the mean and then find the deviation from the mean by using $\text{deviation from mean}\left( {{d}_{i}} \right)\text{=}\dfrac{\text{Number of letters - Assumed mean}}{\text{Length of class interval}}$. We then multiply the deviation with the receivers and find the sum of them. We then substitute all the required values in the formula $mean={{A}_{i}}+\dfrac{\sum\limits_{{}}^{{}}{{{R}_{i}}\times {{d}_{i}}}}{\sum\limits_{{}}^{{}}{{{R}_{i}}}}\times I$ and make necessary calculations to get the required value of mean.

Complete step by step answer:
According to the problem, we are given a distribution table as shown below and we need to find the mean of it.

Number of letters	120	240	360	480
Receivers	20	40	60	80

Let us find the length of each class interval and sum of all the receivers.

Number of letters (N)	Receivers $\left( {{R}_{i}} \right)$	I = length of class interval
120	20	120
240	40	120
360	60	120
480	80	120
	$\sum\limits_{{}}^{{}}{{{R}_{i}}}=200$

Let us assume the mean as 240. So, assumed mean ${{A}_{i}}=240$. Let us find the deviation of the number of letters from the mean, which is as defined below.
$\Rightarrow \text{deviation from mean}\left( {{d}_{i}} \right)\text{=}\dfrac{\text{Number of letters - Assumed mean}}{\text{Length of class interval}}$.

Number of letters (N)	Receivers $\left( {{R}_{i}} \right)$	I = length of class interval	Deviation $\left( {{d}_{i}} \right)$
120	20	120	$\dfrac{120-240}{120}=-1$
240	40	120	$\dfrac{240-240}{120}=0$
360	60	120	$\dfrac{360-240}{120}=1$
480	80	120	$\dfrac{480-240}{120}=2$
	$\sum\limits_{{}}^{{}}{{{R}_{i}}}=200$

Now, let us find the sum of the product of receivers and deviations i.e., ${{R}_{i}}\times {{d}_{i}}$.

Number of letters (N)	Receivers $\left( {{R}_{i}} \right)$	I = length of class interval	Deviation $\left( {{d}_{i}} \right)$	${{R}_{i}}\times {{d}_{i}}$
120	20	120	$\dfrac{120-240}{120}=-1$	-20
240	40	120	$\dfrac{240-240}{120}=0$	0
360	60	120	$\dfrac{360-240}{120}=1$	60
480	80	120	$\dfrac{480-240}{120}=2$	160
	$\sum\limits_{{}}^{{}}{{{R}_{i}}}=200$			$\sum\limits_{{}}^{{}}{{{R}_{i}}\times {{d}_{i}}}=200$

We know that the mean by step deviation method is defined as $mean={{A}_{i}}+\dfrac{\sum\limits_{{}}^{{}}{{{R}_{i}}\times {{d}_{i}}}}{\sum\limits_{{}}^{{}}{{{R}_{i}}}}\times I$.
Where ${{A}_{i}}$ = Assumed mean = 240.
I = Length of the class interval = 120.
Let us substitute all the remaining values to find the mean.
So, we get $mean=240+\dfrac{200}{200}\times 120$.
$\Rightarrow mean=240+120$.
$\therefore $mean = 360.
So, we have found the mean of the given table as 360.
The correct option for the given problem is (c).

Note:
We should confuse the formulas of deviation and mean while solving this problem. We get the same result for the mean by assuming any number as mean while solving this problem. Here we assumed that the values given in the Number of letters column is the upper limit of the boundary which will in turn give the length of the class interval. We can solve for mean without using the step deviation method which can be used for verification of result. Similarly, we can expect problems to find the median and mode of the given data.