Statistics Chapter 14 Class 10 Maths NCERT Textbook With Solutions PDF

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NCERT Class 10 Maths Textbook Chapter 14 With Answer Book PDF Free Download


Chapter 14: Statistics

14.1 Introduction

In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions.

You have also learned to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths), and frequency polygons.

In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median, and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median, and mode from ungrouped data to that of grouped data.

We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution, and how to draw cumulative frequency curves, called ogives.

14.3 Mode of Grouped Data

Recall from Class IX, that a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.

Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data.

It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only.

14.5 Graphical Representation of Cumulative Frequency Distribution

As we all know, pictures speak better than words. A graphical representation helps us in understanding given data at a glance.

In Class IX, we have represented the data through bar graphs, histograms, and frequency polygons. Let us now represent a cumulative frequency distribution graphically.

Language English
No. of Pages35
PDF Size2.3 MB

NCERT Solutions Class 10 Maths Chapter 14 Statistics

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants0-22-44-66-88-1010-1212-14
Number of Houses1215623

Which method did you use for finding the mean, and why?


In order to find the mean value, we will use the direct method because the numerical value of fi and xi are small.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of plants(Class interval)No. of house frequency (fi)Mid-point (xi)fixi
Sum f= 20Sum fixi = 162

The formula to find the mean is:

Mean = x̄ = ∑fxi /∑f

= 162/20

= 8.1

Therefore, the mean number of plants per house is 8.1

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)100-120120-140140-160160-180180-200
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.


Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of the mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

So, u= (xi – A)/h = ui  = (xi – 150)/20

Substitute and find the values as follows:

Daily wages(Class interval)Number of workersfrequency (fi)Mid-point (xi)u= (xi – 150)/20fiui
TotalSum f= 50Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = A + h∑fiui /∑f=150 + (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wage of the workers = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowance(in c)11-1313-1515-1717-1919-2121-2323-35
Number of children76913f54


To find out the missing frequency, use the mean formula.

Here, the value of mid-point (xi)  mean x̄ = 18

Class intervalNumber of children (fi)Mid-point (xi)    fixi    
17-191318 = A234
Totalfi = 44+fSum fixi = 752+20f

The mean formula is

Mean = x̄ = ∑fixi /∑f= (752+20f)/(44+f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752+20f)/(44+f)

⇒ 18(44+f) = (752+20f)

⇒ 792+18f = 752+20f

⇒ 792+18f = 752+20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute65-6868-7171-7474-7777-8080-8383-86
Number of women2438742


From the given data, let us assume the mean as A = 75.5

x= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the uand fiui as follows:

Class IntervalNumber of women (fi)Mid-point (xi)ui = (xi – 75.5)/hfiui
Sum fi= 30Sum fiu= 4

Mean = x̄ = A + h∑fiui /∑f

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

Therefore, the mean heartbeats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes50-5253-5556-5859-6162-64
Number of boxes1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?


Since the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between the two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class IntervalNumber of boxes (fi)Mid-point (xi)di = xi – Afidi
55.5-58.513557 = A00
Sum fi = 400Sum fidi = 75

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi /∑f

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

NCERT Class 10 Maths Textbook Chapter 14 With Answer Book PDF Free Download

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