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.
AnswerFrom 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.
View full question & answer→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:
$10, 20, 36, 92, 95, 40, 50, 56, 60, 70, 92, 88, 80, 70, 72, 70, 36, 40, 36, 40, 92, 40, 50, 50, 56, 60, 70, 60, 60, 88$
AnswerThe 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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