NCERT solutions for class 10 maths chapter 14 Statistics  Today's world is highly dependent on data. Each and every sector has a group of data that represents various information. In the previous classes, you have studied the classification of the given data into grouped and ungrouped frequency distributions and have learnt to represent given data in the form of graphs such as histograms, bar graphs, etc and mean mode and median. Solutions of NCERT class 10 maths chapter 14 Statistics consists of the solutions for the questions based on the study of measures of central tendency, namely, mean, mode and median from the ungrouped data to that of grouped data. CBSE NCERT solutions for class 10 maths chapter 14 Statistics, consists detailed explanation of all 4 exercises comprising 25 questions. Statistics is the branch of applied mathematics that deals with the collection of data and represent it in a meaningful way. In NCERT class 10 maths chapter 14 Statistics you will also get the solutions of cumulative frequency, its distribution and how to draw cumulative frequency curves questions. Students face many reallife problems where the fundamentals of statistics are used to represent data in graphs or in tabular form. For classwise and subjectwise NCERT solutions, you can click on the given link.
Which method did you use for finding the mean, and why?
Number of plants

Number of houses

Classmark


02 
1 
1 
1 
24 
2 
3 
6 
46 
1 
5 
5 
68 
5 
7 
35 
810 
6 
9 
54 
1012 
2 
11 
22 
1214 
3 
13 
39 

=20 

=162 
Mean,
We used the direct method in this as the values of are small.
Let the assumed mean be a = 550
Daily Wages 
Number of workers 
Classmark



500520 
12 
510 
40 
480 
520540 
14 
530 
20 
280 
540560 
8 
550 
0 
0 
560580 
6 
570 
20 
120 
580600 
10 
590 
40 
400 

=50 


=240 
Mean,
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20
Daily pocket allowance 
Number of children 
Classmark


1113 
7 
12 
84 
1315 
6 
14 
84 
1517 
9 
16 
144 
1719 
13 
18 
234 
1921 
f 
20 
20f 
2123 
5 
22 
110 
2325 
4 
24 
96 

=44+f 

=752+20f 
Mean,
Therefore the missing f = 20
Let the assumed mean be a = 75.5
No. of heartbeats per minute 
Number of women 
Classmark



6568 
2 
66.5 
9 
18 
6871 
4 
69.5 
6 
24 
7174 
3 
72.5 
3 
9 
7477 
8 
75.5 
0 
0 
7780 
7 
78.5 
3 
21 
8083 
4 
81.5 
6 
24 
8386 
2 
84.5 
9 
18 

=30 


=12 
Mean,
Therefore, the mean heartbeats per minute of these women are 75.9
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Let the assumed mean be a = 57
Number of mangoes 
Number of boxes 
Classmark



5052 
15 
51 
6 
90 
5355 
110 
54 
3 
330 
5658 
135 
57 
0 
0 
5961 
115 
60 
3 
345 
6264 
25 
63 
6 
150 

=400 


=75 
Mean,
Therefore, the mean number of mangoes kept in a packing box is approx 57.19
Let the assumed mean be a = 225 and h = 50
Daily Expenditure 
Number of households 
Classmark




100150 
4 
125 
100 
2 
8 
150200 
5 
175 
50 
1 
5 
200250 
12 
225 
0 
0 
0 
250300 
2 
275 
50 
1 
2 
300350 
2 
325 
100 
2 
4 

=25 



= 7 
Mean,
Therefore, the mean daily expenditure on food is Rs. 211
Find the mean concentration of in the air.
Class Interval 
Frequency

Classmark


0.000.04 
4 
0.02 
0.08 
0.040.08 
9 
0.06 
0.54 
0.080.12 
9 
0.10 
0.90 
0.120.16 
2 
0.14 
0.28 
0.160.20 
4 
0.18 
0.72 
0.200.24 
2 
0.22 
0.44 

=30 

=2.96 
Mean,
Therefore, the mean concentration of in the air is 0.099 ppm
Number of days 
Number of Students 
Classmark


06 
11 
3 
33 
610 
10 
8 
80 
1014 
7 
12 
84 
1420 
4 
17 
68 
2028 
4 
24 
96 
2838 
3 
33 
99 
3840 
1 
39 
39 





=40 

=499 
Mean,
Therefore, the mean number of days a student was absent is 12.48 days.
Let the assumed mean be a = 75 and h = 10
Literacy rates 
Number of cities 
Classmark




4555 
3 
50 
20 
2 
6 
5565 
10 
60 
10 
1 
10 
6575 
11 
70 
0 
0 
0 
7585 
8 
80 
10 
1 
8 
8595 
3 
90 
20 
2 
6 

= 35 



= 2 
Mean,
Therefore, the mean mean literacy rate is 69.43%
Q1 The following table shows the ages of the patients admitted in a hospital during a year:
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
The class having maximum frequency is the modal class.
The maximum frequency is 23 and hence the modal class = 3545
Lower limit (l) of modal class = 35, class size (h) = 10
Frequency ( ) of the modal class = 23, frequency ( ) of class preceding the modal class = 21, frequency ( ) of class succeeding the modal class = 14.
Now,
Age

Number of patients 
Classmark


515 
6 
10 
60 
1525 
11 
20 
220 
2535 
21 
30 
630 
3545 
23 
40 
920 
4555 
14 
50 
700 
5565 
5 
60 
300 

=80 

=2830 
Mean,
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.
Determine the modal lifetimes of the components.
The class having maximum frequency is the modal class.
The maximum frequency is 61 and hence the modal class = 6080
Lower limit (l) of modal class = 60, class size (h) = 20
Frequency ( ) of the modal class = 61 frequency ( ) of class preceding the modal class = 52, frequency ( ) of class succeeding the modal class = 38.
Thus, the modal lifetime of 225 electrical components is 65.62 hours
The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 15002000
Lower limit (l) of modal class = 1500, class size (h) = 500
Frequency ( ) of the modal class = 40 frequency ( ) of class preceding the modal class = 24, frequency ( ) of class succeeding the modal class = 33.
Thus, the Mode of the data is Rs. 1847.82
Now,
Let the assumed mean be a = 2750 and h = 500
Expenditure 
Number of families 
Classmark




10001500 
24 
1250 
1500 
3 
72 
15002000 
40 
1750 
1000 
2 
80 
20002500 
33 
2250 
500 
1 
33 
25003000 
28 
2750 
0 
0 
0 
30003500 
30 
3250 
500 
1 
30 
35004000 
22 
3750 
1000 
2 
44 
40004500 
16 
4250 
1500 
3 
48 
45005000 
7 
4750 
2000 
4 
28 

=200 



= 35 
Mean,
Thus, the Mean monthly expenditure is Rs. 2662.50
The class having maximum frequency is the modal class.
The maximum frequency is 10 and hence the modal class = 3035
Lower limit (l) of modal class = 30, class size (h) = 5
Frequency ( ) of the modal class = 10 frequency ( ) of class preceding the modal class = 9, frequency ( ) of class succeeding the modal class = 3
Thus, Mode of the data is 30.625
Now,
Let the assumed mean be a = 32.5 and h = 5
Class 
Number of states 
Classmark




1520 
3 
17.5 
15 
3 
9 
2025 
8 
22.5 
10 
2 
16 
2530 
9 
27.5 
5 
1 
9 
3035 
10 
32.5 
0 
0 
0 
3540 
3 
37.5 
5 
1 
3 
4045 
0 
42.5 
10 
2 
0 
4550 
0 
47.5 
15 
3 
0 
5055 
2 
52.5 
20 
4 
8 

=35 



= 23 
Mean,
Thus, the Mean of the data is 29.22
The class having maximum frequency is the modal class.
The maximum frequency is 18 and hence the modal class = 40005000
Lower limit (l) of modal class = 4000, class size (h) = 1000
Frequency ( ) of the modal class = 18 frequency ( ) of class preceding the modal class = 4, frequency ( ) of class succeeding the modal class = 9
Thus, Mode of the data is 4608.70
The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 4050
Lower limit (l) of modal class = 40, class size (h) = 10
Frequency ( ) of the modal class = 20 frequency ( ) of class preceding the modal class = 12, frequency ( ) of class succeeding the modal class = 11
Thus, Mode of the data is 44.70
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
Edit Q
Let the assumed mean be a = 130 and h = 20
Class 
Number of consumers 
Cumulative Frequency 
Classmark




6585 
4 
4 
70 
60 
3 
12 
85105 
5 
9 
90 
40 
2 
10 
105125 
13 
22 
110 
20 
1 
13 
125145 
20 
42 
130 
0 
0 
0 
145165 
14 
56 
150 
20 
1 
14 
165185 
8 
64 
170 
40 
2 
16 
185205 
4 
68 
190 
60 
3 
12 


= 68 



= 7 
MEDIAN:
Median class = 125145; Cumulative Frequency = 42; Lower limit, l = 125;
c.f. = 22; f = 20; h = 20
Thus, the median of the data is 137
MODE:
The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency () of the modal class = 20; frequency () of class preceding the modal class = 13, frequency () of class succeeding the modal class = 14.
Thus, Mode of the data is 135.76
MEAN:
Mean,
Thus, the Mean of the data is 137.05
Q2 If the median of the distribution given below is 28.5, find the values of x and y.
Class 
Number of consumers 
Cumulative Frequency 
010 
5 
5 
1020 
x 
5+x 
2030 
20 
25+x 
3040 
15 
40+x 
4050 
y 
40+x+y 
5060 
5 
45+x+y 

= 60 

Now,
Given median = 28.5 which lies in the class 2030
Therefore, Median class = 2030
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10
Also,
Therefore, the required values are: x=8 and y=7
Class 
Frequency

Cumulative Frequency 
1520 
2 
2 
2025 
4 
6 
2530 
18 
24 
3035 
21 
45 
3540 
33 
78 
4045 
11 
89 
4550 
3 
92 
5055 
6 
98 
5560 
2 
100 
Therefore, Median class = 3545
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10
Thus, the median age is 35.75 years.
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5  126.5, 126.5  135.5, . . ., 171.5  180.5.)
The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.
Class 
Frequency

Cumulative Frequency 
117.5126.5 
3 
3 
126.5135.5 
5 
8 
135.5144.5 
9 
17 
144.5153.5 
12 
29 
153.5162.5 
5 
34 
162.5171.5 
4 
38 
171.5180.5 
2 
40 
Therefore, Median class = 144.5153.5
Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17
Thus, the median length of the leaves is 146.75 mm
Class 
Frequency

Cumulative Frequency 
15002000 
14 
14 
20002500 
56 
70 
25003000 
60 
130 
30003500 
86 
216 
35004000 
74 
290 
40004500 
62 
352 
45005000 
48 
400 
Therefore, Median class = 30003500
Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130
Thus, the median lifetime of a lamp is 3406.97 hours
Thus, the median length of the leaves is 146.75 mm
Class 
Number of surnames 
Cumulative Frequency 
Classmark


14 
6 
6 
2.5 
15 
47 
30 
36 
5.5 
165 
710 
40 
76 
8.5 
340 
1013 
16 
92 
11.5 
184 
1316 
4 
96 
14.5 
51 
1619 
4 
100 
17.5 
70 


= 100 

= 825 
MEDIAN:
Median class = 710; Lower limit, l = 7;
Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
Thus, the median of the data is 8.05
MODE:
The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 710
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency () of the modal class = 40; frequency () of class preceding the modal class = 30, frequency () of class succeeding the modal class = 16
Thus, Mode of the data is 7.88
MEAN:
Mean,
Thus, the Mean of the data is 8.25
Class 
Number of students 
Cumulative Frequency 
4045 
2 
2 
4550 
3 
5 
5055 
8 
13 
5560 
6 
19 
6065 
6 
25 
6570 
3 
28 
7075 
2 
30 
MEDIAN:
Median class = 5560; Lower limit, l = 55;
Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
Thus, the median weight of the student is 56.67 kg
Daily Income (UpperClass Limit) 
Cumulative Frequency 
Less than 120 
12 
Less than 140 
26 
Less than 160 
34 
Less than 180 
40 
Less than 200 
50 
Now,
Taking upperclass interval on the xaxis and their respective frequencies on the yaxis,
Taking upperclass interval on the xaxis and their respective frequencies on the yaxis,
Marking a point on the curve whose ordinate is 17.5 gives an xordinate= 46.5.
Hence, the Median of the data is 46.5
Now,
Weight (Class) 
Frequency 
Cumulative Frequency 
>38 
0 
0 
3840 
3 
3 
4042 
2 
5 
4244 
4 
9 
4446 
5 
14 
4648 
14 
28 
4850 
4 
32 
5052 
3 
35 
Median class = 4648; Lower limit, l = 46;
Cumulative frequency of preceding class, c.f. = 14; f = 14; h = 2
Thus, the median using formula is 46.5 which verifies the result.
Production yield (UpperClass Limit) 
Cumulative Frequency 
More than or equal to 50 
100 
More than or equal to 55 
98 
More than or equal to 60 
90 
More than or equal to 65 
78 
More than or equal to 70 
54 
More than or equal to 75 
16 
Now,
Taking lower class limit on the xaxis and their respective frequencies on the yaxis,
Chapter No. 
Chapter Name 
Chapter 1 
CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers 
Chapter 2 

Chapter 3 
Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4 
CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5 
NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6 

Chapter 7 
CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8 
NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9 
Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10 

Chapter 11 

Chapter 12 
Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13 
CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14 
NCERT solutions for class 10 maths chapter 14 Statistics 
Chapter 15 
Solutions of NCERT class 10  subject wise
First of all, go through NCERT Textbook to have an understanding of data.
Learn the techniques to calculate the mean, median, mode.
Once you learn the techniques and formulae, then move to the practice exercises available in the chapter.
While practicing, if you get stuck anywhere then you can take the help of NCERT solutions for class 10 maths chapter 14 Statistics.
Keep working hard & happy learning!