The annual profits earned by $30$ shops of a shopping complex in a locality give rise to the following distribution:
Profit (Rs in lakhs) Number of shops (frequency) More than or equal to $5$ $30$ More than or equal to $10$ $28$ More than or equal to $15$ $16$ More than or equal to $20$ $14$ More than or equal to $25$ $10$ More than or equal to $30$ $7$ More than or equal to $35$ $3$
Draw both ogives for the data above. Hence,obtain the median profit.
| Profit (Rs in lakhs) | Number of shops (frequency) |
| More than or equal to $5$ | $30$ |
| More than or equal to $10$ | $28$ |
| More than or equal to $15$ | $16$ |
| More than or equal to $20$ | $14$ |
| More than or equal to $25$ | $10$ |
| More than or equal to $30$ | $7$ |
| More than or equal to $35$ | $3$ |
Draw both ogives for the data above. Hence,obtain the median profit.
(17.5) To draw the 'more than' ogive,we plot the points $(5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7),$ and $(35, 3)$ on a graph where the horizontal axis represents the lower limits of profit and the vertical axis represents the cumulative frequency. We join these points with a smooth curve.
Next,we convert the given data into a frequency distribution table:
To draw the 'less than' ogive,we plot the points $(10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),$ and $(40, 30)$ on the same axes.
The abscissa of the point of intersection of the two ogives gives the median. From the graph,the point of intersection is at $x = 17.5$. Therefore,the median profit is $Rs. 17.5$ lakhs.
Next,we convert the given data into a frequency distribution table:
| Classes | $5-10$ | $10-15$ | $15-20$ | $20-25$ | $25-30$ | $30-35$ | $35-40$ |
| No. of shops | $2$ | $12$ | $2$ | $4$ | $3$ | $4$ | $3$ |
| Cumulative frequency | $2$ | $14$ | $16$ | $20$ | $23$ | $27$ | $30$ |
To draw the 'less than' ogive,we plot the points $(10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),$ and $(40, 30)$ on the same axes.
The abscissa of the point of intersection of the two ogives gives the median. From the graph,the point of intersection is at $x = 17.5$. Therefore,the median profit is $Rs. 17.5$ lakhs.
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Similar Questions
To find out the concentration of $SO_{2}$ in the air (in $parts\, per\, million$,i.e.,$ppm$),the data was collected for $30$ localities in a certain city and is presented below:
Concentration of $SO_{2}$ (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 the mean concentration of $SO_{2}$ in the air.
| Concentration of $SO_{2}$ (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 the mean concentration of $SO_{2}$ in the air.
Easy
View SolutionDuring the medical check-up of $35$ students of a class,their weights were recorded as follows:
Weight (in kg) Number of students Less than $38$ $0$ Less than $40$ $3$ Less than $42$ $5$ Less than $44$ $9$ Less than $46$ $14$ Less than $48$ $28$ Less than $50$ $32$ Less than $52$ $35$
Draw a less than type ogive for the given data. Hence,obtain the median weight from the graph and verify the result by using the formula.
| Weight (in kg) | Number of students |
|---|---|
| Less than $38$ | $0$ |
| Less than $40$ | $3$ |
| Less than $42$ | $5$ |
| Less than $44$ | $9$ |
| Less than $46$ | $14$ |
| Less than $48$ | $28$ |
| Less than $50$ | $32$ |
| Less than $52$ | $35$ |
Draw a less than type ogive for the given data. Hence,obtain the median weight from the graph and verify the result by using the formula.
Medium
View SolutionIf the median of the distribution given below is $28.5,$ find the values of $x$ and $y$.
Class interval Frequency $0-10$ $5$ $10-20$ $x$ $20-30$ $20$ $30-40$ $15$ $40-50$ $y$ $50-60$ $5$ Total $60$
| Class interval | Frequency |
| $0-10$ | $5$ |
| $10-20$ | $x$ |
| $20-30$ | $20$ |
| $30-40$ | $15$ |
| $40-50$ | $y$ |
| $50-60$ | $5$ |
| Total | $60$ |
Difficult
View Solution$A$ class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of $days$ a student was absent.
$\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Number of days} & 0-6 & 6-10 & 10-14 & 14-20 & 20-28 & 28-38 & 38-40 \\ \hline \text{Number of students} & 11 & 10 & 7 & 4 & 4 & 3 & 1 \\ \hline \end{array}$
$\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Number of days} & 0-6 & 6-10 & 10-14 & 14-20 & 20-28 & 28-38 & 38-40 \\ \hline \text{Number of students} & 11 & 10 & 7 & 4 & 4 & 3 & 1 \\ \hline \end{array}$
Medium
View Solution$A$ survey regarding the heights (in $cm$) of $51$ girls of Class $X$ of a school was conducted and the following data was obtained:
Height (in $cm$) Number of girls Less than $140$ $4$ Less than $145$ $11$ Less than $150$ $29$ Less than $155$ $40$ Less than $160$ $46$ Less than $165$ $51$
Find the median height.
| Height (in $cm$) | Number of girls |
| Less than $140$ | $4$ |
| Less than $145$ | $11$ |
| Less than $150$ | $29$ |
| Less than $155$ | $40$ |
| Less than $160$ | $46$ |
| Less than $165$ | $51$ |
Find the median height.
Easy
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