Question

For the following distribution, find the mean using step deviation method (Round off the answer to the nearest whole number):

Marks	Frequency
0 – 5	3
5 – 10	5
10 – 15	7
15 – 20	8
20 – 25	10
25 – 30	11
30 – 35	14
35 – 40	19
40 – 45	15
45 – 50	13

(a) 29
(b) 31
(c) 35
(d) 37

Answer 1

Hint:
Here, we will use the step deviation method to find the mean. First, we will calculate the class marks and take one of the middle values as the assumed mean. Then, we will calculate the deviations of the class marks from the assumed mean. Next, we will calculate the step deviations using the common factor of the deviations. Then, we will multiply the frequencies with the respective step deviations. Finally, we will substitute all the required values into the formula for the mean of a continuous series using step deviation and simplify the expression to obtain the required value of the mean.

Formula Used:
The mean of a continuous series is given by the formula \[A + \dfrac{{\sum {fd'} }}{{\sum f }} \times h\], where \[A\] is the assumed mean, \[\sum {fd'} \] is the sum of the product of the frequencies and the step deviations, \[\sum f \] is the sum of all frequencies, and \[h\] is the class size of the mean class.

Complete step by step solution:
First, we will find the class marks of the given class intervals.
The class marks of the class intervals is the average of the lower limit and the upper limit.
Therefore, we get the class marks of the first two class intervals as
Class marks of the class interval 0 – 5 \[ = \dfrac{{0 + 5}}{2} = \dfrac{5}{2} = 2.5\]
Class marks of the class interval 5 – 10 \[ = \dfrac{{5 + 10}}{2} = \dfrac{{15}}{2} = 7.5\]
Similarly, we can find the class marks of the other class intervals.
We will add the mid values to a new column in the table.
Thus, we get the table

Marks	Frequency	Class marks
0 – 5	3	\[2.5\]
5 – 10	5	\[7.5\]
10 – 15	7	\[12.5\]
15 – 20	8	\[17.5\]
20 – 25	10	\[22.5\]
25 – 30	11	\[27.5\]
30 – 35	14	\[32.5\]
35 – 40	19	\[37.5\]
40 – 45	15	\[42.5\]
45 – 50	13	\[47.5\]

We will assume one of the middle class marks as assumed mean.
Let us take \[27.5\] as the assumed mean \[A\].
Now, we will find the deviations \[d\] of the class marks from the assumed mean \[A\].
The deviations show how far the class marks are from the assumed mean.
The deviations are calculated using the formula \[d = {x_i} - A\], where \[A\] is the assumed mean and \[{x_i}\] are the respective class marks.
Therefore, we get the deviations of the first two class intervals as
Deviation for the class interval 0 – 5 \[ = 2.5 - 27.5 = - 25\]
Deviation for the class interval 5 – 10 \[ = 7.5 - 27.5 = - 20\]
Similarly, we can find the deviations of the other class intervals.
We will add the deviations to a new column in the table.
Thus, we get the table

Marks	Frequency \[f\]	Class marks \[{x_i}\]	Deviations \[d\]
0 – 5	3	\[2.5\]	\[ - 25\]
5 – 10	5	\[7.5\]	\[ - 20\]
10 – 15	7	\[12.5\]	\[ - 15\]
15 – 20	8	\[17.5\]	\[ - 10\]
20 – 25	10	\[22.5\]	\[ - 5\]
25 – 30	11	\[27.5\]	0
30 – 35	14	\[32.5\]	5
35 – 40	19	\[37.5\]	10
40 – 45	15	\[42.5\]	15
45 – 50	13	\[47.5\]	20

Now, we will calculate the step deviations using the deviations.
We can observe that all the deviations are divisible by 5.
The step deviations are calculated using the formula \[d' = \dfrac{d}{c}\], where \[d\] are the respective deviations and \[c\] is the common factor of the deviations.
Therefore, we get the step deviations of the first two class intervals as
Step deviation for the class interval 0 – 5 \[ = \dfrac{{ - 25}}{5} = - 5\]
Step deviation for the class interval 5 – 10 \[ = \dfrac{{ - 20}}{5} = - 4\]
Similarly, we can find the step deviations of the other class intervals.
We will add the step deviations to a new column in the table.
Thus, we get the table

Marks	Frequency \[f\]	Class marks \[{x_i}\]	Deviations \[d\]	Step deviations \[d'\]
0 – 5	3	\[2.5\]	\[ - 25\]	\[ - 5\]
5 – 10	5	\[7.5\]	\[ - 20\]	\[ - 4\]
10 – 15	7	\[12.5\]	\[ - 15\]	\[ - 3\]
15 – 20	8	\[17.5\]	\[ - 10\]	\[ - 2\]
20 – 25	10	\[22.5\]	\[ - 5\]	\[ - 1\]
25 – 30	11	\[27.5\]	0	0
30 – 35	14	\[32.5\]	5	1
35 – 40	19	\[37.5\]	10	2
40 – 45	15	\[42.5\]	15	3
45 – 50	13	\[47.5\]	20	4

Next, we will multiply the frequency of the class intervals with the respective step deviations of the same class interval to find \[fd'\].
Therefore, we get
\[fd'\] for the class interval 0 – 5 \[ = 3 \times \left( { - 5} \right) = - 15\]
\[fd'\] for the class interval 5 – 10 \[ = 5 \times \left( { - 4} \right) = - 20\]
Similarly, we can find the \[fd'\] of the other class intervals.
We will add the \[fd'\] to a new column in the table.
Thus, we get the table

Marks	Frequency \[f\]	Class marks \[{x_i}\]	Deviations \[d\]	Step deviations \[d'\]	\[fd'\]
0 – 5	3	\[2.5\]	\[ - 25\]	\[ - 5\]	\[ - 15\]
5 – 10	5	\[7.5\]	\[ - 20\]	\[ - 4\]	\[ - 20\]
10 – 15	7	\[12.5\]	\[ - 15\]	\[ - 3\]	\[ - 21\]
15 – 20	8	\[17.5\]	\[ - 10\]	\[ - 2\]	\[ - 16\]
20 – 25	10	\[22.5\]	\[ - 5\]	\[ - 1\]	\[ - 10\]
25 – 30	11	\[27.5\]	0	0	0
30 – 35	14	\[32.5\]	5	1	14
35 – 40	19	\[37.5\]	10	2	38
40 – 45	15	\[42.5\]	15	3	45
45 – 50	13	\[47.5\]	20	4	52

Now, we will find the mean of the given data.
We will use the formula for mean of a continuous series to solve the given problem.
The mean of a continuous series is given by the step deviation method formula \[A + \dfrac{{\sum {fd'} }}{{\sum f }} \times h\], where \[A\] is the assumed mean, \[\sum {fd'} \] is the sum of the product of the frequencies and the step deviations, \[\sum f \] is the sum of all frequencies, and \[h\] is the class size of the mean class.
The assumed mean is \[A = 27.5\].
The sum of \[fd'\] is given by
\[\sum {fd'} = \left( { - 15} \right) + \left( { - 20} \right) + \left( { - 21} \right) + \left( { - 16} \right) + \left( { - 10} \right) + 0 + 14 + 38 + 45 + 52\]
Adding the terms, we get
\[\sum {fd'} = 67\]
The sum of the frequencies is given by
\[\sum f = 3 + 5 + 7 + 8 + 10 + 11 + 14 + 19 + 15 + 13\]
Adding the terms, we get
\[\sum f = 105\]
The class size of the assumed mean class 25 – 30 is 5.
Thus, \[h = 5\].
Substituting \[A = 27.5\], \[\sum {fd'} = 67\], \[\sum f = 105\], and \[h = 5\] in the step deviation method formula for mean of a continuous series \[A + \dfrac{{\sum {fd'} }}{{\sum f }} \times h\], we get
\[ \Rightarrow {\rm{Mean}} = 27.5 + \dfrac{{67}}{{105}} \times 5\]
Multiplying the terms, we get
\[\begin{array}{l} \Rightarrow {\rm{Mean}} = 27.5 + \dfrac{{335}}{{105}}\\ \Rightarrow {\rm{Mean}} = 27.5 + \dfrac{{67}}{{21}}\end{array}\]
Simplifying the expression, we get
\[ \Rightarrow {\rm{Mean}} = 27.5 + 3.19\]
Adding the terms, we get
\[ \Rightarrow {\rm{Mean}} = 30.69\]
Rounding the mean to the nearest whole number, we get
\[ \Rightarrow {\rm{Mean}} = 31\]
Therefore, the mean of the given series is 31.

Thus, the correct option is option (b).

Note:
We used the term ‘class size’ in the solution to find the value of \[h\].
Class size is the difference in the upper limit and the lower limit of a class interval. It is also known as class width. The formula to calculate class size of a class interval is the upper limit of the class interval \[ - \] lower limit of the class interval.
Thus, the class size of the assumed mean class is \[h = 30 - 25 = 5\].