 Lesson: Statistics
Exercise 14.1 (2)
Question: 1
Give five examples of data that you can collect from
the day-to-day life.
Solution:
In our day to day life, we can collect the following
data:
1. Election results from newspaper and television.
2. Heights of students in our class.
3. Data about production of rice in the last 5 years in
the country.
4. Number of fans in our school.
5. Rainfall in our city in the last 5 years. Question: 2
Classify the data in Q.1 above as primary or secondary
data.
Solution:
The information which is collected by the investigator
himself with a definite objective in his mind is called
as primary data whereas when the information is
gathered or produced by others, it is called as
secondary data. It can be observed that #2 and #4 are
primary data. #1, #3 and #5 are secondary data.
EXERCISE 14.2 (9) Question: 1
The blood groups of 30 students of Class VIII are
recorded 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 students?
Solution:
The blood group of 30 students of the class can be
represented as follows:
Blood group
Number of students
A
9
B
6
AB
3
O
12
Total
30
The most common blood group is O as 12 students
(maximum number of students) have their blood group
as O and the rarest blood group among these students
is AB as 3 students (minimum number of students)
have their blood group as AB. Question: 2
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 features do
you observe from this tabular representation?
Solution:
A grouped frequency distribution table can be
constructed as follows.
Distance (in km)
Tally mark
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
It can be observed that most of the engineers live
between 5 km and 15 km from their work places.
There are very few engineers whose homes are as far
as 20 km from their work places.
Also, the frequencies of the following intervals are
same:
5-10 and 10-15 (both are 11)
20-25 and 25-30 (both are 1).
Question: 3
The relative humidity (in %) of a certain city for a
month of 30 days was as follows: 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 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table
with classes 84 86, 86 88, etc.
(ii) Which month or season do you think this data is
(iii) What is the range of this data?
Solution:
(i)
By observing the data, the required table can be
constructed as follows:
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
(ii) It can be observed that the relative humidity is
high. Therefore, the data could be from June, July or
August (a month belonging to the rainy season).
(iii) Range of data = Maximum value − Minimum
value
= 99.2 − 84.9 = 14.3
Question: 4
The heights of 50 students, measured to the nearest
centimeters, have been found to be 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? Solution:
(i) We need to construct a grouped frequency
distribution table taking class intervals as 160 −
165, 165 − 170, etc. By observing the data given
above, the required table can be constructed as
follows.
Height (in cm)
Number of students
(frequency)
150 − 155
12
155 − 160
9
160− 165
14
165 − 170
10
170 − 175
5
Total
50
(ii) It can be concluded that the heights of maximum
number of students are in the interval of 160165
and the heights of minimum number of students
are in the interval 170-175. Question: 5
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?
Solution:
A grouped frequency table can be constructed as
follows.
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
The number of days for which the concentration of
SO
2
is more than 0.11 is the number of days for which
the concentration is in between 0.12 − 0.16, 0.16 −
0.20, 0.20 − 0.24.
Required number of days = 2 + 4 + 2 = 8
Therefore, for 8 days, the concentration of SO
2
is
more than 0.11 ppm.
Question: 6
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. Solution:
A frequency distribution table can be constructed as
follows.
Number of times
(frequency)
0
6
1
10
2
9
3
5
Total
30
Question: 7
The value of π up to 50 decimal places is given below:
3.141592653589793238462643383279502884197169
39937510
(i) Make a frequency distribution of the digits from 0
to 9 after the decimal point. (ii) What are the most and the least frequently
occurring digits?
Solution:
(i) The required table can be constructed as follows:
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) The most frequently occurring digits are 3 and 9.
The digit 0 occurs the least number of times.
Question: 8 they watched TV programmes 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?
Solution:
(i) The class intervals will be 0 − 5, 5 − 10, 10 −15.....
The grouped frequency distribution table can be
constructed as follows.
Hours
Number of children
0 5
10
5 10
13
10 15
5
15 20
2
Total
30 (ii) 2 children watched TV for 15 or more hours a
week (i.e., the number of children in class interval 15
− 20).
Question: 9
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
the data, using class intervals of size 0.5 starting from
the intervals 2 2.5.
Solution:
We have to construct a grouped frequency table of
class size 0.5, starting from class interval 2 2.5.
Therefore, the class intervals will be 2 2.5, 2.5 3,
3 3.5...
The required grouped frequency distribution table can
be constructed as follows: 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
Exercise 14.3 (9) Question: 1
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 %):
S. No.
Causes
Female fatality
rate (%)
1.
Reproductive
health conditions
31.8
2.
Neuropsychiatric
conditions
25.4
3.
Injuries
12.4
4.
Cardiovascular
conditions
4.3
5.
Respiratory
conditions
4.1
6.
Other causes
22.0
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill
health and death worldwide?
(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. Solution:
(i) Let us represent causes on x-axis and female
fatality rate on y-axis and choose an appropriate scale
of 1 unit = 5% in the y-axis. The graph of the
information given above can be constructed as
follows:
All the rectangle bars are of the same width and have
equal spacing between them. (ii) The major cause of women’s ill health and death
worldwide is the reproductive health condition which
is 31.8%. (Maximum) of women are affected by it.
(iii) The factors are as follows.
1. Lack of medical facilities
2. Lack of proper diet
Question: 2
The following data on the number of girls (to the
nearest ten) per thousand boys in different sections of
Indian society is given below:
Section
Number of girls per
thousand boys
Scheduled Caste (SC)
Scheduled Tribe (ST)
Non SC/ST
Backward districts
Non-backward districts
Rural
Urban
940
970
920
950
920
930
910
(i) Represent the information above by a bar graph. (ii) In the classroom discuss what conclusions can be
arrived at from the graph.
Solution:
(i) Let us represent section (variable) on x-axis and the
number of girls per thousand boys on y-axis.
The graph can be constructed by choosing an
appropriate scale (1 unit = 10 girls in the y-axis).
Here, all the rectangle bars are of the same length and
have equal spacing in between them.
(ii) It can be observed that the maximum number of
girls per thousand boys is for ST (i.e., 970) and the minimum number of girls per thousand boys is for
urban (i.e., 910).
Also, the number of girls per thousand boys is more in
rural areas than in urban areas, backward districts than
in non-backward districts, SC and ST categories than
in non SC/ST.
Question: 3
Given below are the seats won by different political
parties in the polling outcome of a state assembly
elections:
Political
Party
A
B
C
D
E
F
Seats Won
75
55
37
29
10
37
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number
of seats?
Solution: (i) Let us take political party on x-axis and seats won
on y-axis and choose an appropriate scale (1 unit = 10
seats in y-axis).
The required graph can be constructed as follows.
(ii) Political party ‘A’ won maximum number of seats
(i.e., 75).
Question: 4 The length of 40 leaves of a plant are measured correct
to one millimetre, and the obtained data is represented
in the following table:
Length (in mm)
Number of leaves
118 − 126
3
127 − 135
5
136 − 144
9
145 − 153
12
154 − 162
5
163 − 171
4
172 − 180
2
(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?
Solution:
(i)
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
Let us take the length of leaves on x-axis and the
number of leaves on y-axis. The histogram can be
drawn as below.
Here, 1 unit on y-axis represents 2 leaves.
(ii) Other suitable graphical representation of this data
is frequency polygon. (iii) No, as maximum number of leaves (i.e., 12) has
their lengths in between 144.5 mm and 153.5 mm. It is
not necessary that all have their lengths as 153 mm.
Question: 5
The following table gives the life time of neon lamps:
Life time (in hours)
Number of lamps
300 − 400
14
400 − 500
56
500 − 600
60
600 − 700
86
700 − 800
74
800 − 900
62
900 − 1000
48
(i) Represent the given information with the help of a
histogram.
(ii) How many lamps have a life time of more than
700 hours?
Solution:
(i) Let us take life time (in hours) of neon lamps on x-
axis and the number of lamps on y-axis. The histogram can be drawn as follows.
Here, 1 unit on y-axis represents 10 lamps.
(ii) 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.
Therefore, (74 + 62 + 48 =) 184 lamps have life time
more than 700 hours.
Question: 6
The following table gives the distribution of students
of two sections according to the marks obtained by
them:
Section A
Section B
Marks
Frequency
Marks
Frequency
0 − 10
3
0 − 10
5 10 − 20
20 − 30
30 − 40
40 − 50
9
17
12
9
10 − 20
20 − 30
30 − 40
40 50
19
15
10
1
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.
Solution:
Upper classlimit Lower class limit
Classmark
2
Secti
on
A
Sectio
n B
Marks
Class
mark
s
Frequenc
y
Marks
Class
marks
Frequenc
y
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 Let us take class marks on x-axis and frequency on y-
axis and choose an appropriate scale (1 unit = 3 in y-
axis), the frequency polygon can be drawn as follows:
It can be observed that the students of section ‘A have
performed better than the students of section ‘B’ in
terms of good marks.
Question: 7
The runs scored by two teams A and B on the first 60
balls in a cricket match are given below: Number of balls
Team A
Team B
1 − 6
7 − 12
13 − 18
19 − 24
25 − 30
31 − 36
37 − 42
43 − 48
49 − 54
55 − 60
2
1
8
9
4
5
6
10
6
2
5
6
2
10
5
6
3
4
8
10
Represent the data of both the teams on the same
graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Solution:
Class intervals of the given data are not continuous.
To make data continuous
1
0.5
2
the upper class limits and same has to be subtracted
from the lower class limits.
Class mark of each interval can be found by using the
following formula: upperlimit+lowerlimit
Classmark
2
Thus, table can be rearranged as follows.
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, taking class marks on x-axis and runs scored on
y-axis, we can construct frequency polygon as follows: Question: 8
A random survey of the number of children of various
age groups playing in a park was found as follows:
Age (in years)
Number of children
1 − 2
2 − 3
3 − 5
5 − 7
7 − 10
10 − 15
5
3
6
12
9
10 15 − 17
4
Draw a histogram to represent the data above.
Solution:
The given data has class intervals of varying width.
Proportion of children per 1 year interval can be
calculated as follows:
Age
(in year)
Frequency
(Number of
children)
Width of
class
Length of
rectangle
1 − 2
5
1
5×1
=5
1
2 − 3
3
1
3×1
=3
1
3 − 5
6
2
1
=3
2
5 − 7
12
2
12×1
=6
2
7 − 10
9
3
1
=3
3
10 − 15
10
5
5
10×1
=2
15 − 17
4
2
4×1
=2
2 Now, taking the age (in years) on x-axis and
proportion of children per 1 year interval on y-axis, we
can draw the histogram as follows.
Question: 9
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:
Number of letters
Number of surnames
1 4
4 − 6
6 − 8
8 − 12
6
30
44
16 12 20
4
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum
number of surnames lies.
Solution:
(i) Here, the data has class intervals of varying
width. The proportion of the number of
surnames per 2 letters interval can be calculated
as follows.
Number of
letter
Frequency
(Number of
surname)
Width of
class
Length of
rectangle
1 − 4
6
3
2
=4
3
4 − 6
30
2
32
=30
2
6 − 8
44
2
42
=44
2
8 − 12
16
4
16 2
8
4
12 − 20
4
8
42
1
8 Now, taking the number of letters on x-axis and the
proportion of the number of surnames per 2 letters
interval on y-axis and choosing an appropriate scale (1
unit = 4 students for y axis), the histogram can be
constructed as follows.
(ii) Maximum number of surnames lies in the interval
6 − 8 as it has 44 surnames in it.
Exercise 14.4 (6) Question: 1
The following number of goals was scored by a team
in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Solution:
The number of goals scored by the team is
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
2 3 4 5 0 1 3 3 4 3 28
10 10
28
Sum of all the observations
Mean =
Total number of observations
Mean Score
goals.

Now, arranging the number of goals in ascending
order, we get
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Total number of observations = 10, which is an even
number.
Therefore, median Score will be the mean of
10
5
2
th
and
th
10
16
2

observation of the arranged data. = =

th th
5 observation+6 observation 3+3
Median score
22
6
3
2
Mode of data is the observation with the maximum
frequency.
So, the mode is 3 as it has the maximum frequency
equal to 4.
Question: 2
In a mathematics test given to 15 students, the
following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
The marks of 15 students in mathematics test are
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Sum of all observations
Mean of data =
Total number of observations
41 39 48 52 46 62 54 40
96 52 98 40 42 52 60
15
822
54.8
15




Now, let us arrange these scores in an ascending order. 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Now, number of observations is 15 which is an odd
number.
Therefore, the median
=
th
8
15+1
2
observation
Therefore, median
=
52
Mode of data is the observation with the maximum
frequency.
Therefore, mode = 52 (As 52 has the highest
frequency equal to 3).
Question: 3
The following observations have been arranged in
ascending order. If the median of the data is 63, find
the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Total number of observations in the given data =10
(which is an even number).
Therefore,
The median of the data
= mean of
10 10
( =5th) and ( +1=6th)
22
observation Therefore,
.
Median
th th
5 observation+6 observation
=
2
+2 2 +2 2( +1)
63=
2 2 2
63= +1
=62
x+x x x
x
x
Question: 4
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22,
14, 18.
Solution:
Let us arrange the data in an ascending order:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Now, we can observe that 14 has the highest
frequency, i.e. 4.
Therefore, the mode of the data = 14.
Question: 5 Find the mean salary of 60 workers in a factory from
the following table:
Salary (in Rs)
Number of workers
3000
16
4000
12
5000
10
6000
8
7000
6
8000
4
9000
3
10000
1
Total
60
Solution:
To calculate the mean of the salaries, we need to
arrange the table as follows:
Salary
(in Rs)
i
x
Number of
worker
()
i
f
ii
fx
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
f60
i
f x 305000
ii
fx
305000
= 5083.33
f 60
Mean=
ii
i
Therefore, the mean salary of 60 workers is Rs
5083.33.
Question: 6
Give one example of a situation in which
(i) The mean is an appropriate measure of central
tendency.
(ii) The mean is not an appropriate measure of central
tendency but the median is an appropriate measure of
central tendency.
Solution:
Extreme values or outliers in the data affect the mean.
So, mean is not a good representative of the data if
few of the points are very far from the most of the
other points. In such cases, median and mode give a
better estimate.
Let us consider the following examples:
(i) The following data represents the salaries of the
team members of IT department in a company.
Rs. 25200, Rs. 28300, Rs. 22550, Rs. 26500, Rs.
31450.
In this case, the observations are close to each other.
Therefore, mean is an appropriate measure of central
tendency.
(ii) The following data represents the marks obtained
by 12 students in science test.
49, 69, 66, 82, 74, 56, 87, 92, 79, 88, 62, 89
In this case, some observations are very far from other
observations. Therefore, in this case, median is the
appropriate measure of central tendency.