NCERT Solutions for Class 9 Maths Chapter 14 - Statistics
How is data collected? What are the ways to present data? Discover the answers for these important questions in our NCERT Solutions for CBSE Class 9 Mathematics Chapter 14 Statistics. Practise the presentation of data in the form of histograms and bar graphs with our model answers for questions in the NCERT book.
At TopperLearning, you can access our online NCERT textbook solutions for CBSE Class 9 Maths to revise mean, mode and median calculations. The chapter solutions also show you the right way to construct a frequency distribution table according to the given information. You can learn statistics at our learning portal through video lessons, online practice tests and other Maths resources as well.
Chapter 14 - Statistics Exercise Ex. 14.1
1. Number of females per 1000 males in various states of our country.
2. Height and Weights of students of our class.
3. Temperature of past 10 days in our city.
4. Number of plants in our locality.
5. Rain fall in our city.
6. Marks obtained by students of the class in a test.
7. Date of birth of students.
8. Subjects taught in various schools in class X.
Chapter 14 - Statistics Exercise Ex. 14.2
So, the table representing the data is as follows:
Blood group |
Number of students |
A |
9 |
B |
6 |
AB |
3 |
O |
12 |
Total |
30 |
As 12 students have the blood group O and 3 have their blood group as AB, clearly that the most common blood group and the rarest blood group among these students is O and AB respectively.
Required grouped frequency distribution table as following -
Distance (in km) |
Tally marks |
Number of engineers |
0 - 5 |
|
5 |
5 - 10 |
|
11 |
10 -15 |
|
11 |
15 - 20 |
|
9 |
20 - 25 |
|
1 |
25 - 30 |
|
1 |
30 - 35 |
|
2 |
Total |
|
40 |
Most of the engineers are having their workplace up to 15 km distance, from their homes.
Class intervals will be as follows 84 - 86, 86 - 88, and 88 - 90... ...
Relative humidity (in %) |
Number of days (frequency ) |
84 - 86 |
1 |
86 - 88 |
1 |
88 - 90 |
2 |
90 - 92 |
2 |
92 - 94 |
7 |
94 - 96 |
6 |
96 - 98 |
7 |
98 - 100 |
4 |
Total |
30 |
(iii) Range of data = maximum value - minimum value
= 99.2 - 84.9
Heights (in cm) |
Number of students (frequency ) |
150 - 155 |
12 |
155 - 160 |
9 |
160 - 165 |
14 |
165 - 170 |
10 |
170 - 175 |
5 |
Total |
50 |
(ii) From the table we can see that 50% of students are shorter than 165 cm.
Concentration of SO2 (in ppm) |
Number of days (frequency ) |
0.00 - 0.04 |
4 |
0.04 - 0.08 |
9 |
0.08 - 0.12 |
9 |
0.12 - 0.16 |
2 |
0.16 - 0.20 |
4 |
0.20 - 0.24 |
2 |
Total |
30 |
Number of days for which concentration SO2 is more than 0.11 is number of days for which concentration is in between 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24.
So, required number of days = 2 + 4 + 2 = 8
Number of heads |
Number of times (frequency) |
0 |
6 |
1 |
10 |
2 |
9 |
3 |
5 |
Total |
30 |
Digit |
Frequency |
0 |
2 |
1 |
5 |
2 |
5 |
3 |
8 |
4 |
4 |
5 |
5 |
6 |
4 |
7 |
4 |
8 |
5 |
9 |
8 |
Total |
50 |
(ii) From the above table the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. So, the most frequently occurring digits are 3 and 9 and the least occurring digit is 0.
The grouped frequency distribution table is as follows:
Hours |
Number of children |
0 - 5 |
10 |
5 - 10 |
13 |
10 - 15 |
5 |
15 - 20 |
2 |
Total |
30 |
(ii) The number of children, who watched TV for 15 or more hours a week
is 2 (i.e. number of children in class interval 15 - 20).
Lives of batteries (in hours) |
Number of batteries |
2 - 2.5 |
2 |
2.5 - 3.0 |
6 |
3.0 - 3.5 |
14 |
3.5 - 4.0 |
11 |
4.0 - 4.5 |
4 |
4.5 - 5.0 |
3 |
Total |
40 |
Chapter 14 - Statistics Exercise Ex. 14.3

All the rectangle bars are of same width and having equal spacing between them.
(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.
Here all the rectangle bars are of same width and have equal spacing in between them.
Here rectangle bars are of same width and have equal spacing in between them.
(ii). We may find that political party 'A' won maximum number of seats.
Length (in mm) |
Number of leaves |
117.5 - 126.5 |
3 |
126.5 - 135.5 |
5 |
135.5 - 144.5 |
9 |
144.5 - 153.5 |
12 |
153.5 - 162.5 |
5 |
162.5 - 171.5 |
4 |
171.5 - 180.5 |
2 |
Now taking length of leaves on x axis and number of leaves on y axis we can draw the histogram of this information as below -
Here 1 unit on y axis represents 2 leaves.
(ii). Other suitable graphical representation of this data could be frequency polygon.
(iii). No as maximum number of leaves (i.e. 12) have their length in between of 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
Here 1 unit on y axis represents 10 lamps.
(ii). Number of neon lamps having their lifetime more than 700 are sum of number of neon lamps having their lifetime as 700 - 800, 800 - 900, and 900 - 1000.
So number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 + 48 = 184)
Class mark
Section A |
Section B |
||||
Marks |
Class marks |
Frequency |
Marks |
Class marks |
Frequency |
0 - 10 |
5 |
3 |
0 - 10 |
5 |
5 |
10 - 20 |
15 |
9 |
10 - 20 |
15 |
19 |
20 - 30 |
25 |
17 |
20 - 30 |
25 |
15 |
30 - 40 |
35 |
12 |
30 - 40 |
35 |
10 |
40 - 50 |
45 |
9 |
40 - 50 |
45 |
1 |
From the graph we can see performance of students of section 'A' is better than the students of section 'B' as for good marks.

Also class mark of each interval can be found by using formula -
Number of balls |
Class mark |
Team A |
Team B |
0.5 - 6.5 |
3.5 |
2 |
5 |
6.5 - 12.5 |
9.5 |
1 |
6 |
12.5 - 18.5 |
15.5 |
8 |
2 |
18.5 - 24.5 |
21.5 |
9 |
10 |
24.5 - 30.5 |
27.5 |
4 |
5 |
30.5 - 36.5 |
33.5 |
5 |
6 |
36.5 - 42.5 |
39.5 |
6 |
3 |
42.5 - 48.5 |
45.5 |
10 |
4 |
48.5 - 54.5 |
51.5 |
6 |
8 |
54.5 - 60.5 |
57.5 |
2 |
10 |
Now by taking class marks on x axis and runs scored on y axis we can construct frequency polygon as following -
Age (in years) |
Frequency (Number of children) |
Width of class |
Length of rectangle |
1 - 2
|
5
|
1 |
|
2 - 3 |
3 |
1 |
|
3 - 5 |
6 |
2 |
|
5 - 7 |
12 |
2 |
|
7 - 10 |
9 |
3 |
|
10 - 15 |
10 |
5 |
|
15 - 17 |
4 |
2 |
|
Now taking age of children on x axis and proportion of children per 1 year interval on y axis we may draw histogram as below -
Number of letters |
Frequency (Number of surnames) |
Width of class |
Length of rectangle |
1 - 4 |
6 |
3 |
|
4 - 6 |
30 |
2 |
|
6 - 8 |
44 |
2 |
|
8 -12 |
16 |
4 |
|
12 - 20 |
4 |
8 |
|
Now by taking number of letters on x axis and proportion of number of surnames per 2 letters interval on y axis and choosing an appropriate scale (1 unit = 4 students for y axis) we will construct the histogram as below
(ii). The class interval in which the maximum number of surname lie is
6 - 8 as there are 44 number of surnames in it i.e. maximum for this data.
Chapter 14 - Statistics Exercise Ex. 14.4
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
As the number of observations is 10. 10 is an even number. So, median score will be
Mode of data is the observation with the maximum frequency in data.
So, mode score of data is 3 as it is having maximum frequency as 4 in the data.
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
As the number of observations is 15 that is odd so, median of data will be

So, median score of data = 52
Mode of data is the observation with the maximum frequency in data. So mode of this data is 52 having the highest frequency in data as 3.


14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Here observation 14 is having the highest frequency i.e. 4 in given data. So, mode of given data is 14.
Salary (in Rs) (xi) |
Number of workers (fi) |
fixi |
3000 |
16 |
3000 * 16 = 48000 |
4000 |
12 |
4000 * 12 = 48000 |
5000 |
10 |
5000 * 10 = 50000 |
6000 |
8 |
6000 * 8 = 48000 |
7000 |
6 |
7000 * 6 = 42000 |
8000 |
4 |
8000 * 4 = 32000 |
9000 |
3 |
9000 * 3 = 27000 |
10000 |
1 |
10000 * 1 = 10000 |
Total |
|
|
(ii) Mean is not suitable in cases where there are very high and low values for example salary in a company.
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