EXERCISE 14.3
Q 1, Ex. 14.3, Page No 258 - Statistics - NCERT Solutions Class 9th Maths
Page No 258:
Question 1:
A survey conducted by an organisation 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 (%)1.2.3.4.5.6.Reproductive health conditionsNeuropsychiatric conditions
Injuries
Cardiovascular conditions
Respiratory conditions
Other causes31.825.412.44.34.122.0
(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.
Answer:
(i) 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.
(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.
(iii) The factors are as follows.
1. Lack of medical facilities
2. Lack of correct knowledge of treatment
Q 2, Ex. 14.3, Page No 258 - Statistics - Class 9th Mathematics
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
|
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:
(i) 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)
(ii) It can be observed that 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 that in urban areas, backward districts than that in non-backward districts, SC and ST than that in non-SC/ST.
Q 3, Ex. 14.3, Page No 258 - Statistics - Class 9th NCERT Mathematics
Page No 259:
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
|
(ii) Which political party won the maximum number of seats?
Answer:
(i) By taking polling results on x-axis and seats won as y-axis and choosing an appropriate scale (1 unit = 10 seats for y-axis), the required graph of the above information can be constructed as follows.
(ii) Political party ‘A’ won maximum number of seats.
Q 4, Ex. 14.3, Page No 258 - Statistics - Class 9th Maths
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
127 − 135
136 − 144
145 − 153
154 − 162
163 − 171
172 − 180
|
3
5
9
12
5
4
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?
Answer:
(i) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them. Therefore,
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.- Length (in mm)Number of leaves117.5 − 126.53126.5 − 135.55135.5 − 144.59144.5 − 153.512153.5 − 162.55162.5 − 171.54171.5 − 180.52
Taking the length of leaves on x-axis and the number of leaves on y-axis, the histogram of this information can be drawn as above.
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 length in between 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
Q 5, Ex. 14.3, Page No 259 - Statistics - Mathematics Class 9th
Question 5:
The following table gives the life times of neon lamps:
Length (in hours)
|
Number of lamps
|
300 − 400
400 − 500
500 − 600
600 − 700
700 − 800
800 − 900
900 − 1000
|
14
56
60
86
74
62
48
|
(ii) How many lamps have a lifetime of more than 700 hours?
Answer:
(i) 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.
(ii) 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.
Therefore, the number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 + 48 = 184)
Q 6, Ex 14.3, Page No 260 - Statistics - Class 9th Mathematics
Page No 260:
Question 6:
The following table gives the distribution of students of two sections according to the mark obtained by them:
Section A
|
Section B
| ||
Marks
|
Frequency
|
Marks
|
Frequency
|
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
|
3
9
17
12
9
|
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
|
5
19
15
10
1
|
Answer:
We can find the class marks of the given class intervals by using the following formula.Class mark $=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}$
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
|
Q 7, Ex 14.3, Page No 260 - Statistics - Maths Class 9th
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
|
[Hint: First make the class intervals continuous.]
Answer:
It can be observed that the class intervals of the given data are not continuous. There is a gap of 1 in between them. Therefore, $\frac{1}{2}=0.5$ 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 $=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}$
Continuous data with class mark of each class interval can be represented 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
|
Q 8, Ex. 14.3, Page No 261 - Statistics - Maths Class 9th
Page No 261:
Question 8:
A random survey of the number of children of various age groups playing in park was found as follows:
Age (in years)
|
Number of children
|
1 − 2
2 − 3
3 − 5
5 − 7
7 − 10
10 − 15
15 − 17
|
5
3
6
12
9
10
4
|
Answer:
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.
Age (in years)
|
Frequency (Number of children)
|
Width of class
|
Length of rectangle
|
1 − 2
|
5
|
1
|
$\frac{5 \times 1}{1}=5$
|
2 − 3
|
3
|
1
|
$\frac{3 \times 1}{1}=3$
|
3 − 5
|
6
|
2
|
$\frac{6 \times 1}{2}=3$
|
5 − 7
|
12
|
2
|
$\frac{12 \times 1}{2}=6$
|
7 − 10
|
9
|
3
|
$\frac{9 \times 1}{3}=3$
|
10 − 15
|
10
|
5
|
$\frac{10 \times 1}{5}=2$
|
15 − 17
|
4
|
2
|
$\frac{4 \times 1}{2}=2$
|
Q 9, Ex 14.3, Page No 261 - Statistics - Class 9th Maths
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
12 − 20
|
6
30
44
16
4
|
(ii) Write the class interval in which the maximum number of surname lie.
Answer:
(i) Here, it can be observed that 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 lettersFrequency (Number of surnames)Width of classLength of rectangle1 − 463$\frac{6 \times 2}{3}=4$4 − 6302$\frac{30 \times 2}{2}=30$6 − 8442$\frac{44 \times 2}{2}=44$8 − 12164$\frac{16 \times 2}{4}=8$12 − 2048$\frac{4 \times 2}{8}=1$
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