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Question 13 Marks
In a particular section of Class $IX, 40$ students were asked about the months of their birth and the following graph was prepared for the data so obtained:

Observe the bar graph given above and answer the following questions:
$i.$ How many students were born in the month of November?
$ii.$ In which month were the maximum number of students born?
Answer
from the graph it is clear that.
$i. 4$ students were born in the month of November.
$ii.$ The Maximum number of students were born in the month of August.
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Question 23 Marks
Consider a small unit of a factory where there are $5$ employees : a supervisor and four labourers. The labourers draw a salary of $₹ 5,000$ per month each while the supervisor gets $₹15,000 $ per month. Calculate the mean, median and mode of the salaries of this unit of the factory.
Answer
From the given data we have,Mean = $\begin{equation} \frac{5000+5000+5000+5000+15000}{5}=\frac{35000}{5} \end{equation} = 7000$
So, the mean salary is ₹ $7000$ per month.
To obtain the median, we arrange the salaries in ascending order:
$5000, 5000, 5000, 5000, 15000$
Since the number of employees in the factory is $5$, the median is given by the $\begin{equation} \left(\frac{5+1}{2}\right)+h=\frac{6}{2} \text { th } \end{equation} = 3rd$ observation.
Therefore, the median is ₹ $5000 $per month.
To find the mode of the salaries, i.e., the modal salary, we see that $5000 $occurs the maximum number of times in the data $5000, 5000, 5000, 5000, 15000$. So, the modal salary is ₹ $5000 $per month.
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Question 33 Marks
Form the frequency distribution table and Find the mean of the marks obtained by $30$ students of Class $IX$ of a school, as given below:
$10, 20, 36, 92, 95, 40, 50, 56, 60, 70, 92, 88, 80, 70, 72, 70, 36, 40, 36, 40, 92, 40, 50, 50, 56, 60, 70, 60, 60, 88$
Answer
The frequency distribution table is given below.
Marks ($x_i$) Number of students ($f_i$) $f_ix_i$
$10$ $1$ $10$
$20$ $1$ $20$
$36$ $3$ $108$
$40$ $4$ $160$
$50$ $3$ $150$
$56$ $2$ $112$
$60$ $4$ $240$
$70$ $4$ $280$
$72$ $1$ $72$
$80$ $1$ $80$
$88$ $2$ $176$
$92$ $3$ $276$
$95$ $1$ $95$
  $\begin{equation} \sum_\limits{i=1}^{13} f_{i}=30 \end{equation}$ $\begin{equation} \sum_\limits{i=1}^{13} f_{i} x_{i}=1779 \end{equation}$
So, the mean $\begin{equation} \bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}=\left(\frac{\sum_\limits{i=1}^{13} f_{i} x_{i}}{\sum_\limits{i=1}^{13} f_{i}}\right) \end{equation}$
$\begin{equation} =\frac{1779}{30}=59.3 \end{equation}$
 
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