JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3

Jharkhand Board JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 14 Statistics Ex 14.3

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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.NoCausesFemale fatality rate (%)
1.Reproductive health conditions31.8
2.Neuropsychiatric conditions25.4
3.Injuries12.4
4.Cardiovascular conditions4.3
5.Respiratory conditions4.1
6.Other causes22.0

(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.
Answer:
(i) The data is represented below graphically.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 1
(ii) From the above graphical data, we observe that reproductive health conditions is the major cause of womens ill health and death worldwide.
(iii) Two factors responsible for cause in (ii)

  • Lack of proper care and under-standing.
  • Lack of medical facilities.

JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3

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.

SectionNumber of girls per thousand boys
Scheduled Caste (SC)940
Scheduled Tribe (ST)970
Non SC/ST920
Backward districts950
Non backward districts920
Rural930
Urban910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:
(i)
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 2
(ii) It can be observed from the above graph that the maximum number of girls per thousand boys is in ST. Also, the backward districts and rural areas have more number of girls per thousand boys than non¬backward districts and urban areas.

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Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political partyABCDEF
Seats won755537291037

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Answer:
(i)
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 3
(ii)The party named A has won the maximum number of seats.

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 – 1263
127 – 1355
136 – 1449
145 – 15312
154 – 1625
163 – 1714
172 – 1802

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(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) The data is represented in a discontinuous class interval. So, first we will make it continuous.
The difference is 1, so we subtract \(\frac{1}{2}\) = 0.5 from lower limit and add 0.5 to the upper limit.

Length (in mm)Number of leaves
117.5 – 126.53
126.5 – 135.55
135.5 – 144.59
144.5 – 153.512
153.5 – 162.55
162.5 – 171.54
171.5 – 180.52

JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 4
(ii) Yes, the data can also be represented by frequency polygon.
(iii) No, it is incorrect to conclude that the maximum number of leaves are 153 mm long because maximum number of leaves are lying between the length of 144.5-153.5.

JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 5.
The following table gives the life times of 400 neon lamps:

Number of lampsLifetime (in hours)
300 – 40014
400 – 50056
500 – 60060
600 – 70086
700 – 80074
800 – 90062
900 – 100048

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
Answer:
(i)
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 5
(ii) 74 + 62 + 48 = 184 lamp’s have a life time of more than 700 hours.

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Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Section ASection B
MarksFrequencyMarksFrequency
0-1030-105
10-20910-2019
20-301720-3015
30-401230-4010
40-50940-501

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.
Answer:
The class mark can be found by (Lower limit + Upper limit)/2.
For section A

Section A
MarksFrequency
0-103
10-209
20-3017
30-4012
40-509

For section B

Section B
MarksFrequency
0-105
10-2019
20-3015
30-4010
40-501

Now, we draw frequency polygons for the given data.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 6
In the two polygons, we observed that most of the students of section A got higher marks. So the result of section A is better.

JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3

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 ballsTeam ATeam B
1 – 625
7 – 1216
13 – 1882
19 – 24910
25 – 3045
31 – 3656
37 – 4263
43 – 48104
49 – 5468
55 – 60210

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Answer:
The data given in the form of discontinu¬ous class intervals. So, first we will make them continuous. The difference is 1, so
we subtract \(\frac{1}{2}\) = 0.5 from lower limit and add 0.5 to the upper limit.

Number of ballsClass MarksTeam ATeam B
0.5 – 6.53.525
6.5 – 12.59.516
12.5 – 18.515.582
18.5 – 24.521.5910
24.5 – 30.527.545
30.5 – 36.533.556
36.5 – 42.539.563
42.5 – 48.545.5104
48.5 – 54.551.568
54.5 – 60.557.5210

Now, we draw frequency polygons for the given data.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 7

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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 – 25
2 – 33
3 – 56
5 – 712
7 – 109
10 – 1510
15 – 174

 

Draw a histogram to represent the data above.
Answer:
The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 1 is selected and the length of the rectangle is proportionate to it.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 8
Taking the age of children on x-axis and no. of children on v-axis, the histogram can be drawn as follows.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 9

JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3

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 lettersNumber of surnames
1 – 46
4 – 630
6 – 844
8 – 1216
12 – 204

 

(i) Draw a histogram to depict the gi ven information.
(ii) Write the class interval in which the maximum number of surnames lie.
Answer:
(i) The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 2 is selected and the length of the rectangle is proportionate to it.
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 10
JAC Class 9 Maths Solutions Chapter 14 Statistics Ex 14.3 - 11
(ii) The class interval in which the maximum number of surnames lie is 6-8.

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