EXERCISE 15.4
Page No 293:
Question 1:
The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rs) | 100 − 120 | 120 − 140 | 140 − 160 | 160 − 180 | 180 − 200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Answer:
The frequency distribution table of less than type is as follows.
Daily income (in Rs) (upper class limits) | Cumulative frequency |
Less than 120 | 12 |
Less than 140 | 12 + 14 = 26 |
Less than 160 | 26 + 8 = 34 |
Less than 180 | 34 + 6 = 40 |
Less than 200 | 40 + 10 = 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.
Question 2:
During 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 verify the result by using the formula.
Answer:
The given cumulative frequency distributions of less than type are
Weight (in kg) upper class limits | Number of students (cumulative frequency) |
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.
Here, n = 35
So, = 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.
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
Weight (in kg) | Frequency (f) | Cumulative frequency |
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 |
The cumulative frequency just greater thanis 28, belonging to class interval 46 − 48.
Median class = 46 − 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Therefore, median of this data is 46.5.
Hence, the value of median is verified.
Question 3:
The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg/ha) | 50 − 55 | 55 − 60 | 60 − 65 | 65 − 70 | 70 − 75 | 75 − 80 |
Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |
Change the distribution to a more than type distribution and draw ogive.
Answer:
The cumulative frequency distribution of more than type can be obtained as follows.
Production yield (lower class limits) | Cumulative frequency |
more than or equal to 50 | 100 |
more than or equal to 55 | 100 − 2 = 98 |
more than or equal to 60 | 98 − 8 = 90 |
more than or equal to 65 | 90 − 12 = 78 |
more than or equal to 70 | 78 − 24 = 54 |
more than or equal to 75 | 54 − 38 = 16 |
Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows.