14. Statistics

Class 9 Mathematics Chapter 14. Statistics

Chapter 14. Statistics

Important Note :

1. When the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data.

2. When the information was gathered from a source which already had the information stored, the data obtained is called secondary data.

3. The difference of the highest and the lowest values in the data is called the range of the data.

The range highest values – lowest values .

4. Class size (class width) Upper class limit – Lower class limit .

5. Class mark

6. Mean

7. The median is that value of the given number of observations, which divides it into exactly two parts.

(i) When the number of observations () is odd, then the median observation.
(ii) When the number of observations () is even, then the median observations.

9. Mode : The mode is the most frequently occurring observation.

Class 9 Chapter 14. Statistics Exercise 14.1

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

Solution: There are five examples of data that we can collect from our day-to-day life :

(i) Heights of 20 students of the class IX .

(ii) Number of absentees in each day in our class for a month .

(iii) Number of members in the families of our classmates .

(iv) Heights of 15 plants in or around our school .

(v) Electricity bills of our house for last two years .

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

Solution: Primary data : (i) , (ii) , (iii) and (iv)

Secondary data : (v)

[ Note : Primary data : When the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data .

Secondary data : When the information was gathered from a source which already had the information stored, the data obtained is called secondary data .]

Class 9 Chapter 14. Statistics Exercise 14.2

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?

Solutino: We construct the frequency distribution table :

Blood group	Frequency
A	9
B	6
O	12
AB	3
Total	30

The most common blood group is O and the rarest blood group is AB .

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?

Solution: We construct the frequency distribution table :

Class interval

Frequency

0 – 5

5 – 10

10 – 15

15 – 20

20 – 25

25 – 30

30 – 35

Total

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?

Solution: (i) We construct the frequency distribution table :

Relative humidity (in %)

Frequency

84 – 86

86 – 88

88 – 90

90 – 92

92 – 94

94 – 96

96 – 98

98 – 100

Total

(ii) The data appears to be taken in the rainy season( as the relative humidity is high) .

(iii) The range

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?

Solution: (i) We construct the frequency distribution table :

Heights (in cm)

Frequency

150 – 155

155 – 160

160 – 165

165 – 170

170 – 175

Total

(ii) One conclusion that we can draw from the above table is that more than 50% of students are shorter than 165 cm .
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?

Solution: (i) We construct the frequency distribution table :

Concentration of sulphur dioxide

(in ppm)

Frequency

0.00 – 0.04

0.04 – 0.08

0.08 – 0.12

0.12 – 0.16

0.16 – 0.20

0.20 – 0.24

Total

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days .

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.

Solution: We construct the frequency distribution table :

Number of heads

Frequency

Total

7. The value of π upto 50 decimal places is given below : 3.14159265358979323846264338327950288419716939937510
(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?

Solution: (i) We construct the frequency distribution table :

Digits

Frequency

Total

(ii) The most frequently occurring digits are 3 and 9 .

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?

Solution: (i) We construct the frequency distribution table :

Number of hours

Frequency

0 – 5

5 – 10

10 – 15

15 – 20

Total

(ii) 2 children

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.

Solution: We construct the frequency distribution table :

Life of batteries (in years)

Frequency

2.0 – 2.5

2.5 – 3.0

3.0 – 3.5

3.5 – 4.0

4.0 – 4.5

4.5 – 5.0

Total

Class 9 Chapter 14. Statistics Exercise 14.3

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.	Causes	Female fatality rate (%)
1.	Reproductive health conditions	31.8
2.	Neuropsychiatric conditions	25.4
3.	Injuries	12.4
4.	Cardiovascular 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 in the cause in (ii) above being the major cause.

Solution: (i) We obtain the bar graph as given in figure :

(ii) Reproductive health conditions .

(iii) The major cause are :

(a) Bad condition of healthcare system ,

(b) Lack of Access to Family Planning Services .

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 Caste (SC)	940
Scheduled Tribe (ST)	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.

Solution: (i) We obtain the bar graph as given in figure :

(ii) The bar graph highlights the gender imbalance in various sections of Indian society. In all categories, the number of girls per thousand boys is consistently lower than 1000, indicating a higher male population.
3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political Party	A	B	C	D	E	F
Seats Won	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?

Solution : (i) We obtain the bar graph as given in figure :

(ii) The maximum number of seats is won by political party A (75) .
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 – 126

127 – 135

136 – 144

145 – 153

154 – 162

163 – 171

172 – 180

(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?

Solution: (i) We have,

The lower limit of 127 - 135 = 127

The upper limit of 118 - 126 = 126

The difference = 127 - 126 = 1

So, half difference

The lower limit of the class interval = 118 - 0.5 = 117.5

The upper limit of the class interval = 126 + 0.5 = 126.5

We construct the table :

Length (in mm)

Number of leaves

117.5 – 126.5

126.5 – 135.5

135.5 – 144.5

144.5 – 153.5

153.5 – 162.5

162.5 – 171.5

171.5 – 180.5

We draw the graph in given figure :

(ii) Yes, there are other suitable graphical representations for the same data. One such representation is a frequency polygon.A frequency polygon is a graph that uses line segments to connect the midpoints of each class interval.

(iii) No, it is not correct to conclude that the maximum number of leaves are 153 mm long. The histogram only represents the frequency of leaves falling within each class interval. The class interval "144.5 – 153.5" has the highest frequency (12 leaves), but it does not necessarily mean that the maximum length of leaves is 153 mm.

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

Life time (in hours)

Number of lamps

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

800 – 900

900 – 1000

(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) We draw the graph by given information :

(ii) The number of lamps is 184 (= 74 + 62 + 48)

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

Section A		Section B
Marks	Frequency	Marks	Frequency
0 – 10	3	0 – 10	5
10 – 20	9	10 – 20	19
20 – 30	17	20 – 30	15
30 – 40	12	30 – 40	10
40 – 50	9	40 – 50	1

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 construct the data table :

Section A			Section B
Marks	Class marks	Frequency	Marks	Class 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

We draw the graph by given information :

We draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis.

We plotting of section A and joining the points O(0,0) , A(5, 3), B(15, 9), C(25, 17), D(35, 12), E(45, 9) and F(55, 0) by line segments. So, the resultant frequency polygon will be OABCDEF .

We plotting of section B and joining the points O(0,0) , P(5, 5), Q(15, 19), R(25, 15), S(35, 10), T(45, 1) and U(55, 0) by line segments. So, the resultant frequency polygon will be OPQRSTU .

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

7 – 12

13 – 18

19 – 24

25 – 30

31 – 36

37 – 42

43 – 48

49 – 54

55 – 60

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

Solution: We have ,

The lower limit of the class interval 7 – 12 = 7

The upper limit of the class interval 1 – 6 = 6

The difference = 7 – 6 = 1

So , half difference

Now, The lower limit = 1 – 0.5 = 0.5

And the upper limit = 6 + 0.5 = 6.5

Class marks

We construct the data table :

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
43.5 – 48.5	45.5	10	4
48.5 – 54.5	51.5	6	8
54.5 – 60.5	57.5	2	10

We draw the graph by given information :

We draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis.

For Team A : We plotting and joining the points (0, 0), (3.5, 2), (9.5, 1), (15.5, 8), (21.5, 9) , (27.5, 4) , (33.5 , 5) , (39.5 , 6) , (45.5 , 10) , (51.5 , 6) , (57.5 , 2) and (63.5, 0) by line segments.

This is the resultant of frequency polygon graph for Team A .

For Team B : We plotting and joining the points (0, 0), (3.5, 5), (9.5, 6), (15.5, 2), (21.5, 10) , (27.5, 5) , (33.5 , 6) , (39.5 , 3) , (45.5 , 4) , (51.5 , 8) , (57.5 , 10) and (63.5, 0) by line segments.

This is the resultant of frequency polygon graph for Team B .

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 – 2

2 – 3

3 – 5

5 – 7

7 – 10

10 – 15

15 – 17

Draw a histogram to represent the data above.

Solution : We draw the graph in the given figure :

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

4 – 6

6 – 8

8 – 12

12 – 20

(i) Draw a histogram to depict the given information.

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

Solution: (i) We obtain the graph as shown in figure :

(ii) (ii) The class interval 6 - 8 ( the maximum number of surnames is 44 ) .

Class 9 Chapter 14. Statistics Exercise 14.4

1. The following number of goals were scored by a team in a series of 10 matches: 2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.

Solution: We arrange the data in ascending order : 0 , 1 , 2 , 3 , 3 , 3 , 3 , 4 , 4 , 5

Mean

For median : Here,

Median

For mode : Here 3 occurs most frequently (i.e., four times) . So, the mode is 3 .

2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded: 41 , 39 , 48 , 52 , 46 , 62 , 54 , 40 , 96 , 52 , 98 , 40 , 42 , 52 , 60
Find the mean, median and mode of this data.

Solution: We arrange the data in the following form :

39 , 40 , 40 , 41 , 42 , 46 , 48 , 52 , 52 , 52 , 54 , 60 , 62 , 96 , 98

For mean :

Mean

For median : Here,

Median

For mode :

39 , 40 , 40 , 41 , 42 , 46 , 48 , 52 , 52 , 52 , 54 , 60 , 62 , 96 , 98

Here, 52 occurs most frequently ,i.e., three times. So, the mode is 52 .

3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of .
29 , 32 , 48 , 50 , , , 72 , 78 , 84 , 95

Solution: Given , the data :

29 , 32 , 48 , 50 , , , 72 , 78 , 84 , 95

Here,

Median

Therefore, the value of is 62 .

4. Find the mode of 14 , 25 , 14 , 28 , 18 , 17 , 18 , 14 , 23 , 22 , 14 , 18.

Solution: We construct the table :

Data

Frequency

Here, 14 occurs most frequently, i.e., four times.

Therefore, the mode of the data is 14 .

5. Find the mean salary of 60 workers of a factory from the following table:

Salary (in Rs)

Number of workers

3000

4000

5000

6000

7000

8000

9000

10000

Total 60

Solution: We construct the table :

Salary (in Rs)

No. of workers

3000

4000

5000

6000

7000

8000

9000

10000

48000

50000

48000

42000

32000

27000

10000

Total

305000

We have ,

Mean

6. Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Solution: (i) The mean is an appropriate measure of central tendency when calculating the average score of a class in an exam. Add up all the individual scores of the students and then divide the sum by the total number of students to find the mean score.

(ii) The mean is not appropriate when measuring the central tendency of income in a city with a few extremely high earners, as it can be skewed by those outliers. In such cases, the median (middle value when data is arranged in ascending or descending order) is a better representation of the typical income.