## Exercise 14.3

### Question 1:

A survey conducted by an organisation far 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 Cardivasculor 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 ill the cause in (ii) above being the major causes.

### Answer:

(i) By representing causes on x-axis and family fatality rate on y-axis and choosing an appropriate scale (1 unit 5% far y axis), the graph of the information given above can be constructed as follows:

In the graph, the rectangle bars are of the same width and have equal spacing among them.

(ii) As shown, Reproductive health condition is the major cause of women's ill health and death worldwide.

(iii) The factors are as follows.
1. Lack of medical facilities
2. Lack of awareness of  right medical treatment

### 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 Cast(SC) 940
Scheduled Tribe 970
Non SC/ST 920
Backward districts 950
Non-Backward districts 920
Rural 930
Urban 910

(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) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above is constructed by choosing an appropriate scale ( i.e. 1 unit=100 girls for y-axis).

Here, all the rectangle bars between are of same length and have equal spacing in between them.

(ii) It is seen 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 from urban.

Also, the number of girls per thousand boys is greater in rural areas as compared to that in urban areas, backward districts than that in non-backward districts, SC and ST than that in non-SC/ ST.

### Question 3:

Given below are the sea s won by different political parties in the polling outcome of a state assembly elections

 Political Party Seats Won A B C D E F 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?

### Answer:

(i) By taking polling results on x-axis and seats won as y-axis. The  appropriate scale chosen is (1 unit 10 seats for y-axis), the required graph of the above Information can be constructed as follows.
In the above graph, the rectangle bars are of same length and have equal spacing in between them.

(ii) Political party 'A' won maximum number of seats.

### Question 4:

The length of 40 leaves of a plant are measured correct to one millimetre, 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?

### 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,1/2 = 0.5 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 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

Taking the length of leaves on x-axis and the number of leaves on y-axis, histogram can be drawn as shown above. In the graph, 1 unit on y-axis represents 2 leaves.

(i) Other suitable graphical representation of this data is Frequency polygon.

(ii) No, as maximum number of leaves ( i.e., 12) has their length in between 144.5mm and 153.5 mm.
It is not necessary that all have their lengths as 153 mm

### Question 5:

The following table gives the life times of neon lamps:

Length(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 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 an y-axis,
the histogram of the given information can be drawn as follows.

Here, 1 unit on y-axis represents 10 lamps.

(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.
i.e. (74 + 62 + 48 = 184)

### 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 3 0 - 10 5
10 - 20 9 0 - 10 19
20 - 30 17 0 - 10 15
30 - 40 12 0 - 10 10
40 - 50 9 0 - 10 1

Represent the marks of the students of bath the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections

### Answer:

We can find the class marks of the given class intervals by using the following formula.
class marks = (upper limit + lower limit) /  2
Section A
Section B
Marks Class Marks Frequency Marks 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
Taking class scale (1 Limit marks on x-axis and frequency on y axis 3 for y-axis), the frequency polygon can and choosing an appropriate be drawn as follows.

It is seen that the performance of students of section 'A' is better than the students of section 'B',

### 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 2 5
7 - 12 1 6
13 - 18 8 2
19 - 24 9 10
25 - 30 4 5
31 - 36 5 6
37 - 42 6 3
43 - 48 10 4
49 - 54 6 8
55 - 60 2 10
Represent the data of both teams on the same graph by frequency polygons.
[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,
1/ = 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 marks = (upper class limit + lower class limit)/ 2

Continuous data with class mark of each class internal can be represented as follows.

Number of balls Class marks 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

By taking class marks on x-axis and runs scored on y-axis, a frequency polygon is  constructed as shown as follows:

### 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 5
2 - 3 3
3 - 5 6
5 - 7 12
7 - 10 9
10 - 15 10
15 - 17 4

### Answer:

Here, it can be observed that the data has class intervals of varying width. proportion of chidren per 1 year interva can be calculated as follows.

Age(in years) Frequency(number of children) Width of class Length of rectangle
1 - 2 5 1
5x1/1
= 5
2 - 3 3 1
3x1/1
= 3
3 - 5 6 2
6x1/2
= 3
5 - 7 12 2
12x1/2
= 6
7 - 10 9 3
9x1/ 3
= 3
10 - 15 10 5
10x1/5
= 2
15 - 17 4 2
4x1/1
= 2

Taking the age of children proportion of children per 1 year interval on y-axis,the histogram can be drawn 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 6
4 - 6 30
6 - 8 44
8 - 12 16
12 - 20 4

(i) Draw a histogram to depict the g•ven information.
(ii) Write the class int—wal 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 letters Frequency(number of surnames) Width of class Length of rectangle
1 - 2 6 3
6x2/3
= 4
4 - 6 30 2
30x2/2
= 30
6 - 8 44 2
44x2/2
= 44
8 - 12 16 4
16x2/4
= 8
12 - 20 4 8
4x2/8
= 1

By taking the number of letters on x-axis and the proportion af 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 is drawn  as follows.

(ii) The class interval in which the maximum number of surnames lies is 6 - 8 as it has 44 surnames in it.

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