Statistics – Exercise 14.3 – Class IX

For previous exercise Statistics – Exercise 14.2  – Class IX


  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 %):
Sl. No.CausesFemale Fertility Rate(%)
1Reproductive health conditions31.8
2Neuropsychiatric conditions25.4
3Injuries12.4
4Cardiovascular conditions4.3
5Respiratory conditions4.1
6Other 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.

Solution:

(i) Here we are representing causes on x – axis and female fertility rate on y –axis, the graph of the information given above can be constructed as follows:

Statistics - Exercise 14.2 - Class IX

All the rectangular bars are of the same width and have equal spacing between them.

(ii) Reproductive health condition is the major cause of female infertility and death worldwide as 31.8% are affected by it.

(iii)The two main factors which play a major role in the women’s illness which is mentioned in (ii) above being the major cause are

  • Lack of medical facilities
  • Lack of proper knowledge about health issues

  1. 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.

Solution:

(i) Represent section 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)

Statistics - Exercise 14.2 - Class IX

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

(ii) It can be observed that maximum number of girls per thousand boys is ST and minimum number  of girls per thousand boys is Urban.

Also 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.


  1. 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?

Solution:

(i)Take polling results on x-axis and  seats won on y-axis and choose scale such as 1unit = 10seats for  y-axis, the required graph o the above information can be constructed as follows:

 

Statistics - Exercise 14.2 - Class IX

Here, all the rectangular bars are of the same length an have equal spacing in between them

(ii) Political party  named A won with maximum seats 75.


  1. 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.

(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?

Solution:

(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 0.5 has to be added to each upper class limit and also have to subtract 0.5 from the  lower claass limits so as to make the  class intervals continuous.

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

Statistics - Exercise 14.2 - Class IX

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 153.5mm. It is not necessary that all have their lengths as 153 mm.


  1. The following table gives the life times of 400 neon lamps:
Life time (in hours)Number of lumps
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?

Solution:

(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.

 

Statistics - Exercise 14.2 - Class IX

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  sum of the number of neon lamps having lifetime as 800 – 800, 800 – 900 and 900 – 1000

Therefore,  the number of neon lamps having their lifetime more than 700  hours is 74  + 62 +  48 =184.


  1. 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 – 40130 – 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.

Solution:

We can find the class marks of the given class intervals by using the following formula.

Class mark = Upper class limit + Lower class  limit/2

 Section A Section B
MarksClass marksFrequencyMarksClass marksFrequency
0 – 10530 – 1055
10 – 2015910 – 201519
20 – 30251720 – 302515
30 – 4035130 – 403510
40 – 5045940 – 50451

Taking class marks on x-axis and frequency on y-axis and choosing an appropriate scale  (1 unit = 3 for y-axis), the frequency polygon can be drawn as follows.

Statistics - Exercise 14.2 - Class IX

It can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.


  1. 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.]

Solution:

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, 1/2 = 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 = Upper class limit+Lower class limit/2

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

Number of ballsClass markTeam ATeam B
1 – 63.525
7 – 129.516
13 – 1815.582
19 – 2421.5910
25 – 3027.545
31 – 3633.556
37 – 4239.563
43 – 4845.5104
49 – 5451.568
55 – 6057.5210

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

Statistics - Exercise 14.2 - Class IX


  1. 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.

Solution:

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 classesLength of rectangle
1 – 2515×1/1 = 5
2 – 3313×1/1 = 3
3 – 5626×1/2 = 3
5 – 712212×1/2 = 6
7 – 10939×1/2 = 3
10 – 1510510×1/5 = 2
15 – 17424×1/2 = 2

Taking the age of children on x-axis and proportion of children per 1 year interval on y – axis, the histogram can be drawn as follows.

Statistics - Exercise 14.2 - Class IX


  1. 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 given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Solution:

(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 rectangle
1 – 4636×2/3= 4
4 – 630230×2/2  = 30
6 – 844244×2/2 = 44
8 – 1216416×2/4 = 8
12 – 20484×2/8 = 1

By taking the number of letters on x-axis and the proportion of the number of surnames per 2 letters interval on y-axis and choosing an appropriate scale such as 1 unit = 4 students  for y axis, the histogram can be constructed as follows.

Statistics - Exercise 14.2 - Class IX

(ii) The class interval in which the maximum number of surnames lies is 6 − 8 as it has 44 surnames in it i.e., the maximum for this data.


For next exercise – Statistics – Exercise 14.4 – Class IX


 

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