*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.
NCERT Solutions for Class 10 Maths Chapter 14 – Statistics Exercise 14.4 has been provided here, prepared as per the latest syllabus and guidelines prescribed by CBSE. These solutions of Class 10 NCERT are designed by our subject experts and hence give the best solutions for all the exercise problems.
The solutions of 10th Class Maths provided here are the updated version of the upcoming session of 2023-24. These are helpful for students for revision at the time of the board exam and also to score well. Click on the link given below to download the PDF of Exercise 14.4 NCERT Class 10 Maths solutions.
NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.4
Access Other Exercise Solutions of Class 10 Maths Chapter 14 – Statistics
Exercise 14.1 Solutions 9 Questions (Long)
Exercise 14.2 Solutions 6 Questions (Long)
Exercise 14.3 Solutions 7 Questions (Long)
Access Answers to NCERT Class 10 Maths Chapter 14 – Statistics Exercise 14.4
1. The following distribution gives the daily income of 50 workers in 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
Convert the given distribution table to a less than type cumulative frequency distribution, and we get
| Daily Income | Cumulative Frequency
(or) Number of workers |
| Less than 120 | 12 |
| Less than 140 | 26 |
| Less than 160 | 34 |
| Less than 180 | 40 |
| Less than 200 | 50 |
From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve
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 and verify the result by using the formula.
Solution:
From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them to get a smooth curve. The curve obtained is known as less than type ogive.
Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the median by making a table.
| Class interval | Number of students(Frequency) | Cumulative Frequency | |
| Less than 38 | 0 – 38 | 0 | 0 |
| Less than 40 | 38 – 40 | 3 – 0 = 3 | 3 |
| Less than 42 | 40 – 42 | 5 – 3 = 2 | 5 |
| Less than 44 | 42 – 44 | 9 – 5 = 4 | 9 |
| Less than 46 | 44 – 46 | 14 – 9 = 5 | 14 |
| Less than 48 | 46 – 48 | 28 – 14 = 14 | 28 |
| Less than 50 | 48 – 50 | 32 – 28 = 4 | 32 |
| Less than 52 | 50 – 52 | 35 – 22 = 3 | 35 |
Here, N = 35 and N/2 = 35/2 = 17.5
Median class = 46 – 48
Here, l = 46, h = 2, cf = 14, f = 14
The mode formula is given as:
= 46 + [(17.5 – 14)/ 14] × 2
= 46 + 0.5
= 46 + 0.5 = 46.5
Thus, median is verified.
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 its ogive.
Solution:
Converting the given distribution to a more than type distribution, we get
| Production Yield (kg/ha) | Number of farms |
| 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 |
From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper. The graph obtained is known as more than type ogive curve.
In the last exercise, students will learn to represent the cumulative frequency distribution in graphs as a cumulative frequency curve, for the given set of data. Also, learn to represent the median of grouped data, graphically as the x-coordinate of the point of intersection if the two gives for this data. The problems in Exercise 14.4 are solved as per the methods, explained in the example questions before the exercise. Get complete solutions for Chapter 14 of Class 10 Maths here and solve all statistics related problems in the best way.
Get other learning materials such as notes, books, previous year question papers, etc. to practise well. Solving sample papers are also of great help for students to understand the paper pattern and score well. The NCERT solutions are considered the best materials for students to practice with, for all the subjects class-wise and chapter-wise.
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