Class 9 NCERT Solutions 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.
Statistics Exercise Ex. 14.1
Solution 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.
Solution 2
Statistics Exercise Ex. 14.2
Solution 1
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.
Solution 3
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
Solution 2
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.
Solution 4
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.
Solution 5
Concentration of SO_{2} (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
Solution 6
Number of heads 
Number of times (frequency) 
0 
6 
1 
10 
2 
9 
3 
5 
Total 
30 
Solution 7
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.
Solution 8
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).
Solution 9
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 
Statistics Exercise Ex. 14.3
Solution 8
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 
Solution 7
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 
Solution 9
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.
Solution 1
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.
Solution 2
Here all the rectangle bars are of same width and have equal spacing in between them.
Solution 3
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.
Solution 4
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.
Solution 5
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)
Solution 6
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.
Statistics Exercise Ex. 14.4
Solution 1
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.
Solution 2
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 = 8^{th} observation while data is arranged in an ascending or descending order
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.
Solution 3
Solution 4
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.
Solution 5
Salary (in Rs) (x_{i}) 
Number of workers (f_{i}) 
f_{i}x_{i} 
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 


Solution 6
(ii) Mean is not suitable in cases where there are very high and low values for example salary in a company.
CBSE Class 9 Frequently Asked Questions
NCERT Solutions for Class 9 Maths Chapter 14 is on Statistics and includes Grouping and Tabulation of data, Graphical representation of data and Measures of central tendency. Each concept has NCERT solutions as well as concept and problemsolving videos. Students prefer to revise from these concepts and refer to the chapter notes and other features, including Sample papers, tests, and Ask a Doubt. Students use this UnDoubt service to see top user queries on specific concepts.
NCERT Solutions for Class 9 Maths Chapter 14 on statistics carries six marks and is crucial for exam perspective. This chapter deals with a graphical representation of data and teaches Measures of central tendency. Students use TopperLearning question bank to practice the Multiplechoice and Shortanswer type of questions. The bank also includes the CaseStudy based questions to revise and practice.
Accordingly to NCERT Solutions for Class 9 Maths Chapter 14, Statistics deals with collecting, presenting, analysing, and interpreting data. Data can be either ungrouped or grouped. Further, grouped data could be categorised into:
(a) Discrete frequency distribution and (b) Continuous frequency distribution.
Data can be represented in tables or the form of graphs. Common graphical forms are bar charts, pie diagrams, histograms, frequency polygons, ogives, etc.
The first order of comparison for a given data is the measures of central tendencies. Commonly used measures are (i) Arithmetic mean (ii) Median (iii) Mode.
Median is the measure that divides the data into two equal parts. The median is the middle term when the data is sorted. In the case of odd observations, the middle observation is the median. In the case of actual observations, the median is the average of the two middle observations. The median can be determined graphically. It does not take into account all the observations.
Use TopperLearning’s Ask a Doubt service to get answers to your academic queries. Get quick access to the content resources from TopperLearning Class 9 Plans.
The relationship between the mean, median, and mode of a data set is crucial to understanding statistics. The difference between mean and mode equals three times the difference between the mean and median.
3 Median = 2 Mean + Mode
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