Question

Find the mean of the following frequency distribution by the step deviation method: -
Take assumed mean A = 18.

Class Interval	1 – 5	6 – 10	11 – 15	16 – 20	21 – 25	26 – 30	31 – 35
Frequency	4	12	7	10	5	6	2

Answer 1

Draw a frequency distribution table containing six columns. In the first column enter the class intervals, in the second column enter the mid values $\left( {{x}_{i}} \right)$ of each class interval by taking the average of the upper and lower limit, in the third column mention the frequency ${{f}_{i}}$. Now, in the fourth column we have to calculate the deviation ${{d}_{i}}={{x}_{i}}-A$, where A is the assumed mean, for each case, in the fifth column we have to calculate ${{\mu }_{i}}=\dfrac{{{d}_{i}}}{h}$, where h is the class width calculated by taking the difference of higher limit or lower limit of two consecutive class intervals, in the last column calculate the value of ${{f}_{i}}{{\mu }_{i}}$. Finally use the formula of mean by the step deviation method given as $\overline{x}=A+\dfrac{h\times \sum{{{f}_{i}}{{\mu }_{i}}}}{\sum{{{f}_{i}}}}$ to get the answer.

Complete step by step answer:
Here we have been provided with the frequency distribution table and we are asked to find the mean using the step deviation method. Let us see the steps involved to find the mean.

Class Interval	1 – 5	6 – 10	11 – 15	16 – 20	21 – 25	26 – 30	31 – 35
Frequency	4	12	7	10	5	6	2

Now, in the step deviation method we draw a frequency distribution table containing six columns. In the first column we enter the class intervals, in the second column the mid values $\left( {{x}_{i}} \right)$ of each class interval by taking the average of the upper and lower limit is calculated, in the third column we mention the frequency ${{f}_{i}}$. In the fourth column we have to calculate the deviation ${{d}_{i}}={{x}_{i}}-A$, where A is the assumed mean, for each case, in the fifth column we have to calculate ${{\mu }_{i}}=\dfrac{{{d}_{i}}}{h}$, where h is the class width calculated by taking the difference of higher limits or lower limits of two consecutive class intervals, in the last column calculate the value of ${{f}_{i}}{{\mu }_{i}}$.
Clearly we can see that the provided assumed mean is A = 18 and the difference between either the lower class limits or the upper class limits of any two consecutive class intervals is 5, so we have h = 5. Let us create a table containing the following values.

Class Interval	Mid values$\left( {{x}_{i}} \right)$	Frequency$\left( {{f}_{i}} \right)$	Deviation$\left( {{d}_{i}}={{x}_{i}}-A \right)$	${{\mu }_{i}}=\dfrac{{{d}_{i}}}{h}$	${{f}_{i}}{{\mu }_{i}}$
1 – 5	3	4	3 – 18 = -15	-3	-12
6 – 10	8	12	8 – 18 = -10	-2	-24
11 – 15	13	7	13 – 18 = -5	-1	-7
16 – 20	18	10	18 – 18 = 0	0	0
21 – 25	23	5	23 – 18 = 5	1	5
26 – 30	28	6	28 – 18 = 10	2	12
31 – 35	33	2	33 – 18 = 15	3	6

Now, the mean according to the step deviation method is given by the formula $\overline{x}=A+\dfrac{h\times \sum{{{f}_{i}}{{\mu }_{i}}}}{\sum{{{f}_{i}}}}$, so substituting the values using the table we get,
$\begin{aligned} & \Rightarrow \overline{x}=18+\dfrac{5\times \left( \left( -12 \right)+\left( -24 \right)+\left( -7 \right)+0+5+12+6 \right)}{\left( 4+12+7+10+5+6+2 \right)} \\\ & \Rightarrow \overline{x}=18-\dfrac{100}{46} \\\ & \Rightarrow \overline{x}=18-2.174 \\\ & \therefore \overline{x}=15.826 \\\ \end{aligned}$
Hence, the mean of the given data is 15.826.

Note: Note that we have different methods for calculating the mean of grouped data like: - the assumed mean method, direct method, step deviation method. We have to remember the steps of all these methods as any of the mentioned methods can be asked to use to get the mean. Remember all the basic terms used in the above solution.