NCERT Maths Book Class 9 Solutions provides you precise and elaborate answers to the problems on statistics. Description of statistics in this chapter is explained simply as the collection of data on different aspects of the life of people, which is useful to the State and interpretation and drawing of inferences from the data. With a total of four chapters, Introduction to statistics includes the presentation of data collected in a raw form. Building blocks of this chapter are a presentation of data in tabular form by grouping them in a regular interval, histogram or polygon, bar graph drawing. Topics like how to find the measure of central tendency mean and mode and median of raw data.
NCERT Solutions for Class 9 Maths Chapter 14 Statistics by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 9 Maths Chapter 14 Statistics are designed as per the CBSE Class 9 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 9 Chapter 14 Statistics.
NCERT Solutions for Class 9 Maths Chapter 14 Statistics: EX 14.1
Give five examples of data that you can collect from day to day life.
1. Maximum or minimum temperature of particular month
2. Weights of students of our class
3. Production of paddy in the last 5 years in our country
4. Number of plants in our locality
5. Temperature of our city in the last 2 months.
Classify the data in Q1 above as primary or secondary data.
The information collected by investigator himself with a definite objective in is called as primary data whereas when information that is gathered from a source which already had the information stored, it is called as secondary data.
It can be observed that the data in 1, 3, and 5 is secondary data and the data in 2 and 4 is primary data.
NCERT Solutions for Class 9 Maths Chapter 14 Statistics: EX 14.2
The blood groups of 30 students of Class VIII are recoded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is
the most common, and which is the rarest, blood group among these
According to data , 9 students have their blood group as A, 6 have B, 3 have AB, and 12 have O. Hence, the blood group of 30 students of the class can be represented as follows.
It can be observed that O is the most common blood group and AB is the rarest among these students is respectively as 12 and 3.
The distance (in km) of 40 engineers from their residence to their place of
work were found as follows:
5 3 10 20 25 11 13 7 12 31 19 10 12
17 18 11 32 17 16 2 7 9 7 8 3 5 12
15 18 3 12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the
data given above taking the first interval as 0–5 (5 not included). What
main feature do you observe from this tabular representation?
Since, given data is very large we will construct a group frequency of class size 5. Hence the class interval will be 0-5, 5-10 and so on.
The classes are not overlapping.
Most of the engineers are living with in the 20km from the place of work .only few are living on longer distance.
The relative humidity (in %) of a certain city for a month of 30 days was as
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1 89.2
92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 94.2 95.1 89.2
(i) Construct a grouped frequency distribution table with classes 84−86,
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
According to data, minimum humidity is 84.9 and the maximum is 99.2.Therfore the class interval are 84- 86, 86 – 88 so on.
Since relative humidity is high so the data must be of a month of rainy season.
Range of data given above = maximum value − minimum value
The heights of 50 students, measured to the nearest centimeters, have been
found as follows:
161 150 154 165 168 161 154 162 150 151 162 164 171
165 158 154 156 172 160 170 153 159 161 170 162 165
166 168 165 164 154 152 153 156 158 162 160 161 173
166 161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution
table, taking the class intervals as 160−165,165−170, etc.
(ii) What can you conclude about their heights from the table?
Minimum height is 150 cm and the maximum height is 173 cm.
A grouped frequency distribution table can be constructed taking class intervals 150 – 155, 155-160, 160−165, 165−170, etc.
(ii) Clearly from table we see that more than 50% of students are shorter than 165 cm.
A study was conducted to find out the concentration of Sulphur dioxide in
the air in parts per million (ppm) of a certain city. The data obtained for 30
days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class
intervals as 0.00−0.04,0.04−0.08, and so on.
(ii) For how many days, was the concentration of Sulphur dioxide more
than 0.11 parts per million?
To construct grouped frequency table class intervals can be taken as 0.00−
Number of days in which concentration of SO2 was more than 0.11 is = 2 +4+2 =8
Three coins were tossed 30 times simultaneously. Each time the number of
heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0 1 3 1 1 2 2 0 1 2 1 3 0
0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
The data given above following frequency distribution table can be constructed as shown below:
The value of 𝜋𝜋 up to 50 decimal places is given below:
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal
(ii) What are the most and the least frequently occurring digits?
From given information following table can be constructed as,
In the table the least frequency is 2 and the maximum frequency is 8 of digit . Hence, the most frequently occurring digits are 3 and 9 and the least occurring digit is 0.
Thirty children were asked about the number of hours they watched TV
programmers in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8 10 3 4 12 2 8 15 1 17
6 3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class
width 5 and one of the class intervals as 5−10.
(ii) How many children watched television for 15 or more hours a week?
(i) The possible class intervals are 0−5,5−10,10−15… The grouped frequency distribution table is as follows:
(ii) The total number of children, who watched TV for 15 or more hours a week are 2 (i.e. the number of children in class interval 15−20).
A company manufactures car batteries of a particular type. The lives (in
years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5 3.5 2.3 3.2 3.4 3.8 3.2
4.6 3.7 2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8 3.5 3.2 3.9 3.2
3.2 3.1 3.7 3.4 4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class
intervals of size 0.5 starting from the intervals 2−2.5.
To construct a grouped frequency table of class size 0.5 and starting from class interval 2−2.5. So, our class intervals will be as 2−2.5,2.5−3,3− 3.5 ……. Required grouped frequency distribution table is as below
NCERT Solutions for Class 9 Maths Chapter 14 Statistics: EX 14.3
A survey conducted by an organization for the cause of illness and death
among the women between the ages 15−44 (in years) worldwide, found the
following figures (in %):
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death
(iii) Try to find out, with the help of your teacher, any two factors which
play a major role in the cause in (ii) above being the major cause.
By representing causes on x-axis and family fatality rate on y-axis and choosing an appropriate scale (1 unit = 5% for y axis), the graph of the information given above can be constructed as follows.
All the rectangle bars are of same width and having equal spacing between them.
Reproductive health condition is the major cause of women’s ill health and death worldwide as 31.8% of women are affected by it.
The factors are as follows:
a. Lack of medical facilities
b. Lack of correct knowledge of treatment
The following data on the number of girls (to the nearest ten) per thousand
boys in different sections of Indian society is given below.
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the
By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be constructed by choosing an appropriate scale
(1 unit = 100 girls for y-axis)
It can be observed that the maximum number of girls per thousand boys (i.e., 970) is for ST and minimum number of girls per thousand boys (i.e., 910) is for urban. Also, the number of girls per thousand boys is greater in rural areas than in urban areas, backward districts than that in non-backward districts, SC and ST than in non SC/ST.
Given below are the seats won by different political parties in the polling
outcome of a state assembly elections:
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
We take polling results on x axis and seats won as y axis and choose an appropriate scale (1 unit =10 seats for y axis) so that we are able to draw the required graph of representing information as below.
We may find that political party ‘A’ won maximum number of seats.
The length of 40 leaves of a plant are measured correct to one millimetre,
and the obtained data is represented in the following table:
(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153
mm long? Why?
It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them.
Hence, 0.5 has to be added to each upper class limit and also have to subtract 0.5 from the lower class limits so as to make the class intervals continuous.
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:
Other suitable graphical representation of this data could be frequency polygon.
No, as maximum number of leaves (i.e. 12) has their length in between of 144.5 mm and 153.5 mm.
It is not necessary that all have their lengths as 153 mm.
The following table gives the life times of 400 neon lamps:
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
By taking life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn as follows.
It can be concluded that the number of neon lamps having their lifetime more than 700 is the sum of the number of neon lamps having their lifetime as 700−800, 800−900, and 900−1000.
Hence, the number of neon lamps having their lifetime more than 700 hours is 184 (74 + 62 + 48 = 184).
The following table gives the distribution of students of two sections
according to the mark obtained by them:
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the
performance of the two sections.
We can find the class marks of the given class intervals by using the following formula Class mark = Upper class limit+Lower class limit/2
It can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.
The runs scored by two teams A and B on the first 60 balls in a cricket
match are given below:
Represent the data of both the teams on the same graph by frequency
polygons. [Hint: First make the class intervals continuous.]
It can be observed that the class intervals of the given data are not continuous. There is a gap f 1 in between them.
Hence, 0.5 has to be added to the upper class limits and 0.5 has to be subtracted from the lower class limits. Also, class mark of each interval can be found by using the following formula
Class mark = Upper class limit+Lower class limit/2
Continuous data with class mark of each class interval can be represented as follows:
The frequency polygon for the above data can be constructed by
(i) Number of balls on x-axis
(ii) Runs scored on y-axis with an approximate scale of “1unit=1run” as the lowest run was at 1 and the highest was at 10.
A random survey of the number of children of various age groups playing
in a park was found as follows:
Draw a histogram to represent the data above.
Here, it can be observed that the data has class intervals of varying width. The proportion of children per 1 year interval can be calculated as follows.
Will take the age of children on x-axis and proportion of children per 11 year interval per year on y-axis, the histogram can be:
100 surnames were randomly picked up from a local telephone directory
and a frequency distribution of the number of letters in the English
alphabet in the surnames was found as follows:
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
The proportion of number of surnames per 2 letter interval (class interval of min class size for reference) can be made.
(ii) The class interval in which the maximum number of surnames lies is 6−8 as it has 44 surnames in it i.e., the maximum for this data.
NCERT Solutions for Class 9 Maths Chapter 14 Statistics: EX 14.4