Question

Find the arithmetic mean using the assumed mean method:

Class Interval	Frequency
100-120	10
120-140	20
140-160	30
160-180	15
180-200	5

Answer 1

Hint: To solve this problem, we should know the assumed mean method. In the assumed mean method, we will assume a certain number within the data given as the mean and is denoted by a. We will calculate the deviation of different classes from the assumed mean and we will calculate the weighted average of the deviations with the weights being the frequencies and the average is added to the assumed mean. If a is the assumed mean ${{f}_{i}}$ denotes the frequency of the ${{i}^{th}}$ class which is having a deviation of ${{d}_{i}}$ from the assumed mean, the formula for the mean is $\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}$. Using this method, we can assume a mean of 150 as it is in the middle and has the highest frequency and applying the formula gives the answer.

Complete step by step answer:
We can write the formula for assumed mean method as
If a is the assumed mean ${{f}_{i}}$ denotes the frequency of the ${{i}^{th}}$ class which is having a deviation of ${{d}_{i}}$ from the assumed mean, the formula for the mean is $\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}$.
Whenever we are given the classes as an interval, we should take the middle value as the representative of the class. That is, for a class of 100-120, we take the class representative value as 110. Likewise, restructuring the data, we get

Class Interval	Class representative ${{x}_{i}}$	Frequency
100-120	110	10
120-140	130	20
140-160	150	30
160-180	170	15
180-200	190	5

Let us assume the assumed mean as 150. We can write the deviations of different classes as
Class -110
Deviation ${{d}_{1}}=110-150=-40$
Frequency ${{f}_{1}}=10$
Class -130
Deviation ${{d}_{2}}=130-150=-20$
Frequency ${{f}_{2}}=20$
Class -150
Deviation ${{d}_{3}}=150-150=0$
Frequency ${{f}_{3}}=30$
Class -170
Deviation ${{d}_{4}}=170-150=20$
Frequency ${{f}_{4}}=15$
Class -190
Deviation ${{d}_{5}}=190-150=40$
Frequency ${{f}_{5}}=5$
It can be written in the tabular format as

Class Interval	Class representative ${{x}_{i}}$	Frequency${{f}_{i}}$	Deviation ${{d}_{i}}={{x}_{i}}-150$	${{f}_{i}}\times {{d}_{i}}$
100-120	110	10	$110-150=-40$	$10\times \left( -40 \right)=-400$
120-140	130	20	$130-150=-20$	$20\times \left( -20 \right)=-400$
140-160	150	30	$150-150=0$	$30\times 0=0$
160-180	170	15	$170-150=20$	$15\times 20=300$
180-200	190	5	$190-150=40$	$5\times 40=200$
		$\sum{{{f}_{i}}=80}$		$\sum{{{f}_{i}}{{d}_{i}}=-300}$

Using the assumed mean formula, we get
$\overline{x}=150+\dfrac{{{f}_{1}}{{d}_{1}}+{{f}_{2}}{{d}_{2}}+{{f}_{3}}{{d}_{3}}+{{f}_{4}}{{d}_{4}}+{{f}_{5}}{{d}_{5}}}{{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}}$
Substituting the values, we get
\[\begin{align}
  & \overline{x}=150+\dfrac{10\times \left( -40 \right)+20\times \left( -20 \right)+30\times 0+15\times 20+40\times 5}{10+20+30+15+5} \\
 & \overline{x}=150+\dfrac{-400+-400+30\times 0+300+200}{80}=150+\dfrac{-300}{80}=150-3.75=146.25 \\
\end{align}\]
$\therefore $ The mean of the given data is 146.25

Note: The main purpose of assumed mean method is to reduce the calculation part. The main trick to use in assuming the mean is to assume the mean as the class having the highest frequency and assume a mean which is in the middle of the classes. In this way of assuming, the calculations will be easier than assuming a mean at the end points of the given range.