Class 9 SELINA Solutions Maths Chapter 18  Statistics
Statistics Exercise Ex. 18
Solution 1(a)
Correct option: (ii) sizes of shoes
A variable that takes distinct, countable values is called a discrete variable.
Size of shoes can be 4, 5, ….. and so on but cannot take any value between 4 and 5, 5 and 6, and so on.
Solution 1(b)
Correct option: (ii) 54^{}
Maximum value = 90
Minimum value = 36
Therefore, range = maximum value  minimum value = 90  36 = 54
Solution 1(c)
Correct option: (iii) 40
Solution 1(d)
Correct option: (iv) 10.520.5^{}
Therefore, the class 1120 after adjustment is
Solution 1(e)
Correct option: (i) 12.517.5^{}
Class size = 15  10 = 5
Solution 2
(a)Discrete variable.
(b)Continuous variable.
(c)Discrete variable.
(d)Continuous variable.
(e)Discrete variable.
Solution 3
The frequency table for the given distribution is
Marks 
Tally Marks 
Frequency 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
_{} 
Solution 4
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.
Solution 5
(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 _{}
_{}
Solution 6
In case of frequency 10  19 the lower class limit is 10, upper class limit is 19 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
Solution 7
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
Solution 8
(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.
Solution 9
(i)The cumulative frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
(ii)The cumulative frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
Solution 10
(i)The frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} 
_{} _{} _{} _{} 
(ii)The frequency distribution table is
C.I 
c.f 
_{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} 
Solution 11
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.
Solution 12
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:
Solution 13
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:
Solution 14
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:
Statistics Exercise Test Yourself
Solution 1
The frequency table is
C.I 
c.f 
_{} _{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} _{} 
Solution 2
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 _{}
Solution 3
Class Interval 
Frequency 
Cumulative Frequency 
_{} _{} _{} _{} 6574 _{} 
_{} _{} _{} _{} _{} _{} 
_{} _{} _{} _{} _{} _{} 
Solution 4
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.
Solution 5(i)
(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 
Solution 5(ii)
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 
Solution 6
The frequency distribution table is as follows:
Solution 7
The cumulative frequency distribution table is as follows:
Class Interval 
Frequency 
Cumulative Frequency 
2029 
18 
18 
3039 
23 
18 + 23 = 41 
4049 
36 
41 + 36 = 77 
5059 
42 
77 + 42 = 119 
Solution 8
Steps:
1. Draw a histogram for the given data
2. Mark the midpoint at the top of each rectangle of the histogram drawn.
3. Also, mark the midpoint of the immediately lower classinterval (1020) and midpoint of the immediately higher classinterval (7080).
4. Join the consecutive midpoints marked by straight lines to obtain the required frequency polygon.