Chapter 18 : Statistics  Selina Solutions for Class 9 Maths ICSE
Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.
Chapter 18  Statistics Excercise Ex. 18(A)
State, which of the following variables are continuous and which are discrete:
(a) Number of children in your class.
(b) Distance travelled by a car.
(c) Sizes of shoes.
(d) Time.
(e) Number of patients in a hospital.
(a)Discrete variable.
(b)Continuous variable.
(c)Discrete variable.
(d)Continuous variable.
(e)Discrete variable.
Given below are the marks obtained by 30 students in an examination:
08 
17 
33 
41 
47 
23 
20 
34 
09 
18 
42 
14 
30 
19 
29 
11 
36 
48 
40 
24 
22 
02 
16 
21 
15 
32 
47 
44 
33 
01 


Taking class intervals 1  10, 11  20, ....., 41  50; make a frequency table for the above distribution.
The frequency table for the given distribution is
Marks 
Tally Marks 
Frequency 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
The marks of 24 candidates in the subject mathematics are given below:
45 
48 
15 
23 
30 
35 
40 
11 
29 
0 
3 
12 
48 
50 
18 
30 
15 
30 
11 
42 
23 
2 
3 
44 
The maximum marks are 50. Make a frequency distribution taking class intervals 0  10, 1020, ...... .
The frequency table for the given distribution is
Marks 
Tally Marks 
Frequency 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
In this frequency distribution, the marks 30 are in the class of interval 30  40 and not in 20  30. Similarly, marks 40 are in the class of interval 40  50 and not in 30  40.
Fill in the blanks:
(a) A quantity which can very from one individual to another is called a .............
(b) Sizes of shoes are ........... variables.
(c) Daily temperatures is ........... variable.
(d) The range of the data 7, 13, 6, 25, 18, 20, 16 is ............
(e) In the class interval 35  46; the lower limit is .......... and upper limit is .........
(f) The class mark of class interval 22  29 is .......... .
(a)Variable.
(b)Discrete variables.
(c)Continuous variable.
(d)The range is _{}
(e)Lower limit is _{} and upper limit is _{}
(f)The class mark is _{}
_{}
Find the actual lower class limits, upper class limits and the midvalues of the classes: 10  19, 20  29, 30  39 and 40  49.
In case of frequency 10  19 the lower class limit is 10, upper class limit is 20 and midvalue is
In case of frequency 20  29 the lower class limit is 20, upper class limit is 29 and midvalue is
In case of frequency 30  39 the lower class limit is 30, upper class limit is 39 and midvalue is
In case of frequency 40  49 the lower class limit is 40, upper class limit is 49 and midvalue is
Find the actual lower and upper class limits and also the class marks of the classes:
1.1  2.0, 2.1 3.0 and 3.1  4.0.
In case of frequency 1.1  2.0 the lower class limit is 1.1, upper class limit is 2.0 and class mark
is
In case of frequency 2.1  3.0 the lower class limit is 2.1, upper class limit is 3.0 and class mark
is
In case of frequency 3.1  4.0 the lower class limit is 3.1, upper class limit is 4.0 and class mark
is
Use the table given below to find:
(a) The actual class limits of the fourth class.
(b) The class boundaries of the sixth class.
(c) The class mark of the third class.
(d) The upper and lower limits of the fifth class.
(e) The size of the third class.
Class Interval 
Frequency 
30  34 
7 
35  39 
10 
40  44 
12 
45  49 
13 
50  54 
8 
55  59 
4 
(a)
The actual class limit of the fourth class will be:
44.549.5.
(b)
The class boundaries of the sixth class will be:
54.559.5
(c)
The class mark of the third class will be the average of the lower bound and the upper bound of the interval. Therefore class mark will be:
(d)
The upper and lower limit of the fifth class is 54 and 50 respectively.
(e)
The size of the third class will be: 44  40 + 1 =5.
Construct a cumulative frequency distribution table from the frequency table given below:
(i)
Class Interval 
Frequency 
0  8 
9 
8  16 
13 
16  24 
12 
24  32 
7 
32  40 
15 
(ii)
Class Interval 
Frequency 
1  10 
12 
11  20 
18 
21  30 
23 
31  40 
15 
41  50 
10 
(i)The cumulative frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
(ii)The cumulative frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
Construct a frequency distribution table from the following cumulative frequency distribution:
(i)
Class Interval 
Cumulative Frequency 
10  19 
8 
20  29 
19 
30  39 
23 
40  49 
30 
(ii)
C.I. 
C.F. 
5  10 
18 
10  15 
30 
15  20 
46 
20  25 
73 
25  30 
90 
(i)The frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} 
_{} _{} _{} _{} 
(ii)The frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
Construct a frequency table from the following data:
Marks 
No. of students 
less than 10 
6 
less than 20 
15 
less than 30 
30 
less than 40 
39 
less than 50 
53 
less than 60 
70 
The frequency table is
C.I 
c.f 
_{} _{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} _{} 
Construct the frequency distribution table from the following cumulative frequency table:
Ages 
No. of students 
Below 4 
0 
Below 7 
85 
Below 10 
140 
Below 13 
243 
Below 16 
300 
(i) State the number of students in the age group 10  13.
(ii) State the agegroup which has the least number of students.
The frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} 
_{} _{} _{} _{} 
(i)The number of students in the age group _{} is _{}
(ii)The age group which has the least number of students is _{}
Fill in the blanks in the following table:
Class Interval 
Frequency 
Cumulative Frequency 
25  34 
...... 
15 
35  44 
...... 
28 
45  54 
21 
...... 
55  64 
16 
...... 
65  74 
...... 
73 
75  84 
12 
...... 
Class Interval 
Frequency 
Cumulative Frequency 
_{} _{} _{} _{} 6574 _{} 
_{} _{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} _{} 
The value of _{} upto 50 decimal place is
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution table of digits from 0 to 9 after the decimal place.
(ii) Which are the most and least occurring digits?
X 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
F 
2 
5 
5 
8 
4 
5 
4 
4 
5 
8 
Most occurring digits are 3 and 9. Least occurring digits are 0.
Chapter 18  Statistics Excercise Ex. 18(B)
Construct a frequency polygon for the following distribution:
Classintervals 
0  4 
4  8 
8  12 
12  16 
16  20 
20  24 
Frequency 
4 
7 
10 
15 
11 
6 
The frequency polygon is shown in the following figure
Steps:
(i)Drawing a histogram for the given data.
(ii)Marking the midpoint at the top of each rectangle of the histogram drawn.
(iii)Also, marking midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
(iv)Joining the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
Construct a combined histogram and frequency polygon for the following frequency distribution:
ClassIntervals 
10  20 
20  30 
30  40 
40  50 
50  60 
Frequency 
3 
5 
6 
4 
2 
Steps:
 Draw a histogram for the given data.
 Mark the midpoint at the top of each rectangle of the histogram drawn.
 Also, mark the midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
 Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
The required combined histogram and frequency polygon is shown in the following figure:
Construct a frequency polygon for the following data:
ClassIntervals 
10  14 
15  19 
20  24 
25  29 
30  34 
Frequency 
5 
8 
12 
9 
4 
The class intervals are inclusive. We will first convert them into the exclusive form.
ClassInterval 
Frequency 
9.5  14.5 
5 
14.5  19.5 
8 
19.5  24.5 
12 
24.5  29.5 
9 
29.5  34.5 
4 
Steps:
 Draw a histogram for the given data.
 Mark the midpoint at the top of each rectangle of the histogram drawn.
 Also, mark the midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
 Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
The required frequency polygon is as follows:
The daily wages in a factory are distributed as follows:
Daily wages (in Rs.) 
125  175 
175  225 
225  275 
275  325 
325  375 
Number of workers 
4 
20 
22 
10 
6 
Draw a frequency polygon for this distribution.
Steps:
 Draw a histogram for the given data.
 Mark the midpoint at the top of each rectangle of the histogram drawn.
 Also, mark the midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
 Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
The required frequency polygon is as follows:
Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram
C.I 
10  30 
30  50 
50  70 
70  90 
90  110 
110  130 
130  150 
ƒ 
4 
7 
5 
9 
5 
6 
4 
(a) Using Histogram:
C.I. 
f 
10  30 
4 
30  50 
7 
50  70 
5 
70  90 
9 
90  110 
5 
110  130 
6 
130  150 
4 
Steps:
 Draw a histogram for the given data.
 Mark the midpoint at the top of each rectangle of the histogram drawn.
 Also, mark the midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
 Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
(b) Without using Histogram:
Steps:

Find the classmark (midvalue) of each given classinterval.
 On a graph paper, mark classmarks along Xaxis and frequencies along Yaxis.

On this graph paper, mark points taking values of classmarks along Xaxis and the values of their corresponding frequencies along Yaxis.
 Draw line segments joining the consecutive points marked in step (3) above.
C.I. 
Classmark 
f 
10  10 
0 
0 
10  30 
20 
4 
30  50 
40 
7 
50  70 
60 
5 
70  90 
80 
9 
90  110 
100 
5 
110  130 
120 
6 
130  150 
140 
4 
150  170 
160 
0 
Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram
C.I 
5  15 
15  25 
25  35 
35  45 
45  55 
55  65 
ƒ 
8 
16 
18 
14 
8 
2 
Using Histogram:
C.I. 
f 
5  15 
8 
15  25 
16 
25  35 
18 
35  45 
14 
45  55 
8 
55  65 
2 
Steps:
 Draw a histogram for the given data.
 Mark the midpoint at the top of each rectangle of the histogram drawn.
 Also, mark the midpoint of the immediately lower classinterval and midpoint of the immediately higher classinterval.
 Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.
Without using Histogram:
Steps:

Find the classmark (midvalue) of each given classinterval.
 On a graph paper, mark classmarks along Xaxis and frequencies along Yaxis.
 On this graph paper, mark points taking values of classmarks along Xaxis and the values of their corresponding frequencies along Yaxis.
 Draw line segments joining the consecutive points marked in step (3) above.
C.I. 
Classmark 
f 
5  5 
0 
0 
5  15 
10 
8 
15  25 
20 
16 
25  35 
30 
18 
35  45 
40 
14 
45  55 
50 
8 
55  65 
60 
2 
65  75 
70 
0 
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