NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14

Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less-than-type cumulative frequency distribution and draw its ogive.

Solution:

According to the question, we convert the given distribution to a less than type cumulative frequency distribution, 

Daily income	Frequency	Cumulative Frequency
Less than 120	12	12
Less than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50

Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve. 

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

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:

According to the given table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.

Now, 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 this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:

Class interval	Number of students(Frequency)	Cumulative Frequency
Less than 38	0	0
Less than 40	3 – 0 = 3	3
Less than 42	5 – 3 = 2	8
Less than 44	9 – 5 = 4	9
Less than 46	14 – 9 = 5	14
Less than 48	28 – 14 = 14	28
Less than 50	32 – 28 = 4	32
Less than 52	35 – 22 = 3	35

The class 46 – 48 has the maximum frequency, hence, this is the modal class

l = 46, h = 2, f₁ = 14, f₀ = 5 and f₂ = 4

Now we find the mode:

Mode = 

On substituting the values in the given formula, we get

= 46 + 0.95 = 46.95

Hence, the mode is verified.

Question 3. The following tables give the 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:

According to the question, we change the distribution to a more than type distribution.

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

Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.

NCERT Solutions Class 10 Maths Chapter 14 Statistics – This article is a useful resource containing free NCERT Solutions for Class 10 Maths Chapter 14 Statistics. These NCERT solutions have been developed by the subject matter experts at GFG to assist students in easily solving questions related to Statistics from the NCERT textbook.

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics cover all four exercises of the NCERT Class 10 Maths Chapter 14 according to the latest CBSE syllabus 2023-24 and guidelines, which are as follows:

NCERT Class 10 Maths Chapter 14 Statistics will help the students learn important statistical concepts like mean, mode, standard deviation, and the graphical depiction of cumulative frequency distribution.

The solutions to all the problems in this Chapter 14 Statistics exercises from the NCERT textbook have been properly covered in the NCERT Solutions for Class 10 Maths.