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Chapter 12: Statistics > Exercise 12.1

Exercise 12.1

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(%)
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?

can be considered as 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.

Lack of medical facilities
56=7×2×2×2
70=2×5×7
Lack of correct knowledge of treatment
True
False

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.

From the above graph, we can conclude that the number of girls per thousand boys is present in section ST. We can aslo observe that the backward districts and rural areas have girls per thosand boys than non-backward districts and urban areas.

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?

Party won the maximum number of seats.

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?

The data given in the quetsion can also be represented by a frequency polygon.

(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

We cannot conclude that the maximum number of leaves is 153 mm long bacause the number of leaves are lying in between the length of 144.5 - 153.5

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

Life time (in hours)Number of lamps
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?

700 lies in the class intervals 700 - 800 , 800 - 900 , 900 - 1000.

Hence, their corresponding frequencies when added up will be ( 74 + 62 + 48 ) = lamps.

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.

[-]Red Indicate Section A

[-]Green Indicate Section B

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

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]

[-]Blue Indicate Team A

[-]Yellow Indicate Team B

Solution:

Since the given intervals (e.g., 1–6, 7–12) are not continuous, we need to adjust them by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit:

Original IntervalAdjusted IntervalMidpointRuns (Team A)Runs (Team B)
1 - 60.5 - 6.525
7 - 126.5 - 12.516
13 - 1812.5 - 18.582
19 - 2418.5 - 24.5910
25 - 3024.5 - 30.545
31 - 3630.5 - 36.556
37 - 4236.5 - 42.563
43 - 4842.5 - 48.5104
49 - 5448.5 - 54.568
55 - 6054.5 - 60.5210