Exercise 14.4 Solutions – Class X Mathematics
Exercise 14.4 Solutions
From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Question 1
The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rupees) |
250-300 |
300-350 |
350-400 |
400-450 |
450-500 |
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:
Step 1: Convert to less than type cumulative frequency distribution:
Daily income less than (in Rupees) |
300 |
350 |
400 |
450 |
500 |
Cumulative frequency |
12 |
12+14=26 |
26+8=34 |
34+6=40 |
40+10=50 |
Question 2
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg) |
Less than 38 |
Less than 40 |
Less than 42 |
Less than 44 |
Less than 46 |
Less than 48 |
Less than 50 |
Less than 52 |
Number of students |
0 |
3 |
5 |
9 |
14 |
28 |
32 |
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:
Step 1: The data is already in less than type cumulative frequency form.
Step 2: Drawing the ogive:
Ogive (Less than type) for weights of students
Step 3: Finding median from the graph:
Total number of students (n) = 35
Median position = n/2 = 17.5
From the graph, the x-coordinate corresponding to y=17.5 is approximately 46.5 kg.
Step 4: Verifying using the formula:
The median class is 46-48 (since 17.5 falls in the cumulative frequency of 28)
Using the formula:
\[
\text{Median} = L + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h
\]
Where:
L = 46 (lower limit of median class)
cf = 14 (cumulative frequency before median class)
f = 14 (frequency of median class)
h = 2 (class width)
\[
\text{Median} = 46 + \left(\frac{17.5 – 14}{14}\right) \times 2 = 46 + \left(\frac{3.5}{14}\right) \times 2 = 46 + 0.5 = 46.5 \text{ kg}
\]
This matches our graphical estimate.
Question 3
The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (Quintal/Hectare) |
50-55 |
55-60 |
60-65 |
65-70 |
70-75 |
75-80 |
Number of farmers |
2 |
8 |
12 |
24 |
38 |
16 |
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
Step 1: Convert to more than type cumulative frequency distribution:
Production yield more than (Quintal/Hectare) |
50 |
55 |
60 |
65 |
70 |
75 |
Cumulative frequency |
100 |
100-2=98 |
98-8=90 |
90-12=78 |
78-24=54 |
54-38=16 |
Step 2: Drawing the ogive (more than type):
Ogive (More than type) for production yield
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M.Sc,B.Ed:
SDVR ZPHS B.Gangaram
Sathupally(md)
Khammam(dt)
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