Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.
JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.4
Question 1.
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.
Solution:
Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
| Weight (in kg) | No. 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 = cf |
| 46 – 48 | 14 = f | 28 |
| 48 – 50 | 4 | 32 |
| 50 – 52 | 3 | 35 |
| n = 35 |
\(\frac{\mathrm{n}}{2}=\frac{35}{2}=17.5\)
Plot the points (38, 0) (40, 3) (42, 5) (44, 9) (46, 14) (48, 28) (50, 32) (52, 35)
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 46 + \(\left[\frac{17.5-14}{14}\right]\) × 2
= 46 + \(\frac{3.5 \times 2}{14}\)
= 46 + 0.5= 46.5.
Question 3.
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.
Solution:
| Production yield (in kg/hec) | Number of farms | c.f. |
| More than 50 | 2 | 100 |
| More than 55 | 8 | 98 |
| More than 60 | 12 | 90 |
| More than 65 | 24 | 78 |
| More than 70 | 38 | 54 |
| More than 75 | 16 | 16 |
∴ Co-ordinate points are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).
