Exercise - 14.4
The following distribution gives the daily income of 50 workers of a factory.
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
| Daily income (in Rs.)(upper class limits) | Number of workers | Cumulative Frequency (cf) |
| Less than 120 | 12 | 12 |
| Less than 140 | 14 | 12 + 14 = 26 |
| Less than 160 | 8 | 26 + 8 = 34 |
| Less than 180 | 6 | 34 + 6 = 40 |
| Less than 200 | 10 | 40 + 10 = 50 |
| Total | n = 50 | |
Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
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)(upper class limits) | Number of students (cf) |
| 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 |
Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.
%20U-14.4.png)
Here, n = 35. So, n/2 = 35/2 = 17.5
Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5.
Therefore, median of this data is 46.5.
%20U-14.4.png)
It can be observed that the difference between two consecutive upper class limits is 2.
| Weight (in kg)(upper class limits) | Number of students | Cumulative Frequency (cf) |
| Less than 38 | 0 | 0 |
| 38 - 40 | 3 - 0 = 3 | 3 |
| 40 - 42 | 5 - 3 = 2 | 5 |
| 42 - 44 | 9 - 5 = 4 | 9 |
| 44 - 46 | 14 - 9 = 5 | 14 |
| 46 - 48 | 28 - 14 = 14 | 28 |
| 48 - 50 | 32 - 28 = 4 | 32 |
| 50 - 52 | 35 - 32 = 3 | 35 |
| Total | n = 35 | |
Now, n = 35
n/2 = 35/2 = 17.5 is 28. So, this observation lies in the class 46 - 48.
Then, l (the lower limit) = 46,
cf (the cumulative frequency of the class preceding 46 - 48) = 14,
f (the frequency of the median class 46 - 48) = 14,
h (the class size) = 2.
Therefore, Median = l + [(n/2 - cf)/f] x h
= 46 + [(35/2 - 14)/14] x 2
= 46 + [(17.5 - 14)/14] x 2
= 46 + (3.5/14 x 2)
= 46 + 3.5/7
= 46 + 0.5
= 46.5
Therefore, the median of data is 46.5.
Hence, the value of median is verified.
The following table gives production yield per hectare of wheat of 100 farms of a village.
Change the distribution to a more than type distribution, and draw its ogive.
| Production yeild (lower class limits) | Number of farms | Cumulative Frequency (cf) |
| Less than 120 | 2 | 100 |
| Less than 140 | 8 | 100 - 2 = 98 |
| Less than 160 | 12 | 98 - 8 = 90 |
| Less than 180 | 24 | 90 - 12 = 78 |
| Less than 200 | 38 | 78 - 24 = 54 |
| Less than 200 | 16 | 54 - 38 = 16 |
| Total | n = 100 | |
Taking lower class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.
