Statistics
NCERT Exercise 14.4
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.
Solution:
Daily income | Cumulative frequency |
---|---|
Less than 120 | 12 |
Less than 140 | 26 |
Less than 160 | 34 |
Less than 180 | 40 |
Less than 200 | 50 |
Question 2. During the medical checkup 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.
Solution:
Weight (in kg) | Frequency | Cumulative Frequency |
---|---|---|
36-38 | 0 | 0 |
38-40 | 3 | 3 |
40-42 | 2 | 5 |
42-44 | 4 | 9 |
44-46 | 5 | 14 |
46-48 | 14 | 28 |
48-50 | 4 | 32 |
50-52 | 3 | 35 |
Since `N = 35` and `n/2 = 17.5` hence median class = Less than 46-48
Here; `l = 46`, `cf = 14`, `f = 14` and `h = 2`
Median can be calculated as follows:
`text(Median)=l+(n/2-cf)/(f)xxh`
`=46+(17.5-14)/(14)xx2`
`=46+1/2=46.5`
This value of median verifies the median shown in ogive.
Question 3. The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
---|---|---|---|---|---|---|
Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |
Change this distribution to a more than type distribution, and draw its ogive.
Solution:
Production yield | Cumulative frequency |
---|---|
More than 50 | 100 |
More than 55 | 98 |
More than 60 | 90 |
More than 65 | 78 |
More than 70 | 54 |
More than 75 | 16 |