Question

To find out the concentration of $S{{O}_{2}}$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in certain city and is presented below:

Concentration of $S{{O}_{2}}$ (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Find out the mean concentration of $S{{O}_{2}}$ in the air.

Answer 1

Hint: First, we will start by defining what a mean is and then we will define the frequency and after that we will see from the table and take a value as assumed mean and then we will see what a class size is, then we will draw a table with class intervals, mid values, frequency, \[{{u}_{i}}=\dfrac{{{x}_{i}}-a}{h}\] where $a$ is assumed mean and $h$ is class size and ${{f}_{i}}{{u}_{i}}$ . After that we will put the value in the following formula to find the sample mean: $\overline{X}=a+h\left( \dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)$.

Complete step by step answer:
First, let’s understand what is meant by the mean of the given data. Now, sample mean is the mean of the collected data. To find the mean or the average of a collection of the sample, then we use the sample mean formula. The sample mean of the collected data is calculated by adding the numbers and then dividing the number of data collected.
Sample mean formula is given below:
$\overline{x}=\dfrac{\sum\nolimits_{i=1}^{n}{{{x}_{i}}}}{n}$
Where,
$\overline{x}=$ sample mean, \[\sum\nolimits_{i=1}^{n}{{{x}_{i}}}={{x}_{1}}+{{x}_{2}}+......+{{x}_{n}}\] , $n=$ Total number of terms.
Here, ${{x}_{1}},{{x}_{2}},.......,{{x}_{n}}$ are different values.
Now, let’s see what is meant by frequency and frequency table. Now, frequency refers to the number of times an event or a value occurs. A frequency table is a table that lists items and shows the number of times the items occur. We represent the frequency by the English alphabet ‘$f$ ’.
Now, let’s take our question we have been given the following table:

Concentration of $S{{O}_{2}}$ (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Let the assumed mean be $a=0.14$ and the class fit that is class size of this data $h=0.04$.
Now we will apply the following formula to find the class mark ${{x}_{i}}$ for each interval :
${{x}_{i}}=\dfrac{\text{Upper Class limit + Lower Class limit}}{2}$
Now, let’s make the following table:

Class Interval	Frequency (\[{{f}_{i}}\])	Class Mark $\left( {{x}_{i}} \right)$	\[{{u}_{i}}=\dfrac{{{x}_{i}}-a}{h}\]	\[{{f}_{i}}{{u}_{i}}\]
0.00-0.04	4	0.02	-3	-12
0.04-0.08	9	0.06	-2	-18
0.08-0.12	9	0.10	-1	-9
0.12-0.16	2	0.14	0	0
0.16-0..20	4	0.18	1	4
0.20-0.24	2	0.22	2	4
Total	\[\sum{{{f}_{i}}=30}\]			\[\sum{{{f}_{i}}{{u}_{i}}=-31}\]

Now, we know that the formula for mean: $\Rightarrow \overline{X}=a+h\left( \dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\Rightarrow \overline{X}=0.14+0.14\left( \dfrac{-31}{30} \right)\Rightarrow \overline{X}=0.09867\simeq 0.099$
Therefore, the mean concentration of $S{{O}_{2}}$ is $0.099$ppm.

Note:
In questions like these, always draw a table so that we will have a better understanding and the calculations will be easy. In the last column of the second table while calculating ${{f}_{i}}{{u}_{i}}$ , be careful as the values can be either negative or zero or positive. So, students can make a mistake while adding them.