Maths NCERT solutions class 9 chapter 14 STATISTICS

14. STATISTICS

Maths NCERT solutions class 9 chapter 14 STATISTICS

14.1 Introduction
14.2 Collection of Data
14.3 Presentation of Data
14.4 Graphical Representation14.5 Measures of Central Tendency
14.6 Summary

Important formula

Exercise 14.1

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 Q.1. Give five examples of data that you can collect from your day-to-day life.

Five examples from day-to-day life:
(i) Daily expenditures of household.
(ii) Amount of rainfall.
(iii) Bill of electricity.
(iv) Poll or survey results.
(v) Marks obtained by students.

Q.2. Classify the data in Q.1 above as primary or secondary data.

Primary Data: (i) (iii) and (v)
Secondary Data: (ii) and (iv)

Exercise 14.2

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Q.1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

The frequency means the number of students having same blood group. We will represent the data in table:

Blood GroupNumber of Students
(Frequency)
A9
B6
O12
AB3
Total30

Most common Blood Group (Highest frequency): O
Rarest Blood Group (Lowest frequency): AB

Q.2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?

The given data is very large. So, we construct a group frequency of class size 5. Therefore, class interval will be 0-5, 5-10, 10-15, 15-20 and so on. The data is represented in the table as:

The classes in the table are not overlapping. Also, 36 out of 40 engineers have their house below 20 km of distance.

Q.3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?

(i) The given data is very large. So, we construct a group frequency of class size 2. Therefore, class interval will be 84-86, 86-88, 88-90, 90-92 and so on. The data is represented in the table as:

Relative humidity (in %)Frequency
84-861
86-881
88-902
90-922
92-947
94-966
96-987
98-1004
Total30

(ii) The humidity is very high in the data which is observed during rainy season. So, it must be rainy season.
(iii) Range of data = Maximum value of data – Minimum = 99.2 − 84.9 = 14.3

Q.4. The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table?

(i) The data with class interval 160-165, 165-170 and so on is represented in the table as:

Height (in cm)No. of Students
(Frequency)
150-15512
155-1609
160-16514
165-17010
170-1755
Total50

(ii) From the given data, it can be concluded that 35 students i.e. more than 50% are shorter than 165 cm.

Q.5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

(i) The data with class interval 0.00 – 0.04, 0.04 – 0.08 and so on is represented in the table as:

Concentration of sulphur dioxide in air
(in ppm)
Frequency
0.00 − 0.044
0.04 − 0.089
0.08 − 0.129
0.12 − 0.162
0.16 − 0.204
0.20 − 0.242
Total30

(ii) 2 + 4 + 2 = 8 days have the concentration of sulphur dioxide more than 0.11 parts per million.

Q.6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.

The frequency distribution table for the data given above can be prepared as follow:

Number of HeadsFrequency
06
110
29
35
Total30

Q.7. The value of π upto 50 decimal places is given below:
3.1415926535897
932384626433832
795028841971693
9937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?

(i)The frequency is given as follow:

DigitsFrequency
02
15
25
38
44
55
64
74
85
98
Total30

(ii) The digit having the least frequency occurs the least and the digit with highest frequency occurs the most. 0 has frequency 2 and thus occurs least frequently while 3 and 9 have frequency 8 and thus occur most frequently.

Q.8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?

(i) The distribution table for the given data, taking class width 5 and one of the class intervals as 5-10 is as follows:

Number of HoursFrequency
0-510
5-1013
10-155
15-202
Total30

(ii) We observed from the given table that 2 children television for 15 or more hours a week.

Q.9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.

A grouped frequency distribution table using class intervals of size 0.5 starting from the interval 2 – 2.5 is constructed.

Lives of batteries (in years)No. of batteries
(Frequency)
2-2.52
2.5-36
3-3.514
3.5-411
4-4.54
4.5-53
Total40

Exercise 14.3

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Q.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.CausesFemale 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.

(i) The data is represented below graphically.

(ii) From the above graphical data, we observe that reproductive health conditions is the major cause of women’s ill health and death worldwide.
(iii) Two factors responsible for cause in (ii)
• Lack of proper care and understanding.
• Lack of medical facilities.

Q.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.

S.No.SectionNumber of girls per thousand boys
1.Scheduled Caste (SC)940
2.Scheduled Tribe (ST)970
3.Non SC/ST920
4.Backward districts950
5.Non-backward districts920
6.Rural930
7.Urban910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.

(i)

(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.

Q.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?

(i)

(ii) The party named A has won the maximum number of seat.

Q.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:

S.No.Length (in mm)Number of leaves
1.118 – 1263
2.127 – 1355
3.136 – 1449
4.145 – 15312
5.154 – 1625
6.163 – 1714
7.172 – 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?

(i) The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

S.No.Length (in mm)Number of leaves
1.117.5 – 126.53
2.126.5 – 135.55
3.135.5 – 144.59
4.144.5 – 153.512
5.153.5 – 162.55
6.162.5 – 171.54
7.171.5 – 180.52

(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

Q.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?

(i)

(ii) 74 + 62 + 48 = 184 lamps have a life time of more than 700 hours.

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

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.

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

MarksClass MarkFrequency
0-1053
10-20159
20-302517
30-403512
40-50459

For section B,

MarksClass MarkFrequency
0-1055
10-201519
20-302515
30-403510
40-50451

Now, we draw frequency polygon for the given data.

Q.7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]

The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

Number of ballsTeam ATeam B
0.5-6.525
6.5-12.516
12.5-18.582
18.5-24.5910
24.5-30.545
30.5-36.556
36.5-42.563
42.5-48.5104
48.5-54.568
54.5-60.5210

Now, we draw frequency polygon for the given data.

Q.8. A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

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.

Age (in years)Number of children (frequency)Width of classLength of rectangle
1-251(5/1)×1 = 5
2-331(3/1)×1 = 3
3-562(6/2)×1 = 3
5-7122(12/2)×1 = 6
7-1093(9/3)×1 = 3
10-15105(10/5)×1 = 2
15-1742(4/2)×1 = 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

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:

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.

(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.
The proportion of the surnames per 2 letters interval can be calculated as:

Number of lettersNumber of surnamesWidth of classLength of rectangle
1-463(6/3)×2 = 4
4-6302(30/2)×2 = 30
6-8442(44/2)×2 = 44
8-12164(16/4)×2 = 8
12-2048(4/8)×2 = 1

(ii) The class interval in which the maximum number of surnames lie is 6-8.

Maths NCERT solutions class 9 chapter 14 STATISTICS