Question

Find the mean salary of 80 workers of a factory from the following tables:\[\]

Salary(in Rs)	Number of workers
5000	22
6000	18
7000	15
8000	10
9000	8
10000	7

Answer 1

Hint: We use the formula for mean $\mu $ with frequency which is given $\mu =\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}$ where ${{x}_{i}}$any data value as salaries in rupees as given in the question and ${{f}_{i}}$ is the frequency that corresponds to ${{x}_{i}}$ as number of workers in the factory. We first find $\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}$ and then divide it by $\sum\limits_{i=1}^{n}{{{f}_{i}}}$.\[\]

Complete step-by-step solution:
We know that the mean is a one of the measures of central tendency. The mean with frequenting data is the sum of product of data values and frequencies divided by sum of frequencies. If there are $n$ data values say ${{x}_{1}},{{x}_{2}},{{x}_{3}}...,{{x}_{n}}$ and with corresponding frequencies ${{f}_{1}},{{f}_{2}},{{f}_{3}}...,{{f}_{n}}$ then the mean is given by
\[\mu =\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
We see that there are a total 80 workers in the factory. The salary in the first column is data values ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}},{{x}_{6}}$ and the number of workers who get a particular salary is the corresponding frequency of the data value. Let us consider the first row in the table, the data value ${{x}_{1}}$ is 5000 and frequency ${{f}_{1}}$is 22. So we calculate the product of frequency and data value as,
\[{{x}_{1}}{{f}_{1}}=5000\times 22=110000\]
Let us consider the second row, where the data value ${{x}_{2}}$ is 6000 and frequency ${{f}_{2}}$is 18. So we calculate the product of frequency and data value as,
\[{{x}_{2}}{{f}_{2}}=6000\times 18=108000\]
Let us consider the third row ,where the data value ${{x}_{3}}$ is 7000 and frequency ${{f}_{3}}$is 55. So we calculate the product of frequency and data value as,
\[{{x}_{3}}{{f}_{3}}=7000\times 15=105000\]
Let us consider the fourth row here the data value ${{x}_{4}}$ is 8000 and frequency ${{f}_{4}}$is 10. So we calculate the product of frequency and data value as,
\[{{x}_{4}}{{f}_{4}}=8000\times 10=80000\]
Let us consider the fifth row ,where the data value ${{x}_{5}}$ is 9000 and frequency ${{f}_{5}}$is 8. So we calculate the product of frequency and data value as,
\[{{x}_{5}}{{f}_{5}}=9000\times 8=72000\]
Let us consider the sixth row, where the data value ${{x}_{6}}$ is 10000 and frequency ${{f}_{6}}$is 7. So we calculate the product of frequency and data value as,
\[{{x}_{6}}{{f}_{6}}=10000\times 7=70000\]
Let us find the sum of product of frequencies and data values.
\[\begin{align}
&\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}={{x}_{1}}{{f}_{1}}}+{{x}_{2}}{{f}_{2}}+{{x}_{3}}{{f}_{3}}+{{x}_{4}}{{f}_{4}}+{{x}_{5}}{{f}_{5}}+{{x}_{6}}{{f}_{6}} \\
&\Rightarrow\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}=110000+108000+105000+80000+72000+70000}=545000 \\
\end{align}\]
The sum of frequency is
\[\begin{align}
  & \sum\limits_{i=1}^{n}{{{f}_{i}}=}{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}+{{f}_{6}} \\
 & \Rightarrow \sum\limits_{i=1}^{n}{{{f}_{i}}=}22+18+15+10+8+7=80 \\
\end{align}\]
So the mean in rupees is
\[\mu =\dfrac{\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{6}{{{f}_{i}}}}=\dfrac{585000}{80}=6812.5\]

Note: The mean is also known as average or expectation. When we are given grouped data with classes say ${{I}_{1}},{{I}_{2}},...,{{I}_{n}}$ , then we first find the midpoints of classes ${{p}_{1}},{{p}_{2}},...,{{p}_{n}}$ and then find the mean as $\mu =\dfrac{\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}$. If we are given large frequencies then we can find the mean with step deviation method.