Lesson: Statistics

Exercise 14.1 (2)

Question: 1

Give five examples of data that you can collect from

the day-to-day life.

Solution:

In our day to day life, we can collect the following

data:

1. Election results from newspaper and television.

2. Heights of students in our class.

3. Data about production of rice in the last 5 years in

the country.

4. Number of fans in our school.

5. Rainfall in our city in the last 5 years.

Question: 2

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

data.

Solution:

The information which is collected by the investigator

himself with a definite objective in his mind is called

as primary data whereas when the information is

gathered or produced by others, it is called as

secondary data. It can be observed that #2 and #4 are

primary data. #1, #3 and #5 are secondary data.

EXERCISE 14.2 (9)

Question: 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?

Solution:

The blood group of 30 students of the class can be

represented as follows:

Blood group

Number of students

A

9

B

6

AB

3

O

12

Total

30

The most common blood group is O as 12 students

(maximum number of students) have their blood group

as O and the rarest blood group among these students

is AB as 3 students (minimum number of students)

have their blood group as AB.

Question: 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:

A grouped frequency distribution table can be

constructed as follows.

Distance (in km)

Tally mark

Number of

engineers

0 – 5

5

5 – 10

11

10 −15

11

15 – 20

9

20 – 25

1

25 – 30

1

30 – 35

2

Total

40

It can be observed that most of the engineers live

between 5 km and 15 km from their work places.

There are very few engineers whose homes are as far

as 20 km from their work places.

Also, the frequencies of the following intervals are

same:

5-10 and 10-15 (both are 11)

20-25 and 25-30 (both are 1).

Question: 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)

By observing the data, the required table can be

constructed as follows:

Relative humidity (in %)

Number of days

(frequency)

84 − 86

1

86 − 88

1

88 − 90

2

90 − 92

2

92 − 94

7

94 − 96

6

96 − 98

7

98 − 100

4

Total

30

(ii) It can be observed that the relative humidity is

high. Therefore, the data could be from June, July or

August (a month belonging to the rainy season).

(iii) Range of data = Maximum value − Minimum

value

= 99.2 − 84.9 = 14.3

Question: 4

The heights of 50 students, measured to the nearest

centimeters, 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 need to construct a grouped frequency

distribution table taking class intervals as 160 −

165, 165 − 170, etc. By observing the data given

above, the required table can be constructed as

follows.

Height (in cm)

Number of students

(frequency)

150 − 155

12

155 − 160

9

160− 165

14

165 − 170

10

170 − 175

5

Total

50

(ii) It can be concluded that the heights of maximum

number of students are in the interval of 160–165

and the heights of minimum number of students

are in the interval 170-175.

Question: 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:

A grouped frequency table can be constructed as

follows.

Concentration of SO

2

(in ppm)

Number of days

(frequency )

0.00 − 0.04

4

0.04 − 0.08

9

0.08 − 0.12

9

0.12 − 0.16

2

0.16 − 0.20

4

0.20 − 0.24

2

Total

30

The number of days for which the concentration of

SO

2

is more than 0.11 is the number of days for which

the concentration is in between 0.12 − 0.16, 0.16 −

0.20, 0.20 − 0.24.

Required number of days = 2 + 4 + 2 = 8

Therefore, for 8 days, the concentration of SO

2

is

more than 0.11 ppm.

Question: 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:

A frequency distribution table can be constructed as

follows.

Number of heads

Number of times

(frequency)

0

6

1

10

2

9

3

5

Total

30

Question: 7

The value of π up to 50 decimal places is given below:

3.141592653589793238462643383279502884197169

39937510

(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) The required table can be constructed as follows:

Digit

Frequency

0

2

1

5

2

5

3

8

4

4

5

5

6

4

7

4

8

5

9

8

Total

50

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

The digit 0 occurs the least number of times.

Question: 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) The class intervals will be 0 − 5, 5 − 10, 10 −15.....

The grouped frequency distribution table can be

constructed as follows.

Hours

Number of children

0 – 5

10

5 – 10

13

10 – 15

5

15– 20

2

Total

30

(ii) 2 children watched TV for 15 or more hours a

week (i.e., the number of children in class interval 15

− 20).

Question: 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

the data, using class intervals of size 0.5 starting from

the intervals 2 – 2.5.

Solution:

We have to construct a grouped frequency table of

class size 0.5, starting from class interval 2 – 2.5.

Therefore, the class intervals will be 2 – 2.5, 2.5 – 3,

3 – 3.5...

The required grouped frequency distribution table can

be constructed as follows:

Lives of batteries

(in hours)

Number of batteries

2 − 2.5

2

2.5 − 3.0

6

3.0 − 3.5

14

3.5− 4.0

11

4.0 − 4.5

4

4.5 − 5.0

3

Total

40

Exercise 14.3 (9)

Question: 1

A survey conducted by an organization 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) Let us represent causes on x-axis and female

fatality rate on y-axis and choose an appropriate scale

of 1 unit = 5% in the y-axis. The graph of the

information given above can be constructed as

follows:

All the rectangle bars are of the same width and have

equal spacing between them.

(ii) The major cause of women’s ill health and death

worldwide is the reproductive health condition which

is 31.8%. (Maximum) of women are affected by it.

(iii) The factors are as follows.

1. Lack of medical facilities

2. Lack of proper diet

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 Caste (SC)

Scheduled Tribe (ST)

Non SC/ST

Backward districts

Non-backward districts

Rural

Urban

940

970

920

950

920

930

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) Let us represent section (variable) on x-axis and the

number of girls per thousand boys on y-axis.

The graph can be constructed by choosing an

appropriate scale (1 unit = 10 girls in the y-axis).

Here, all the rectangle bars are of the same length and

have equal spacing in between them.

(ii) It can be observed that the maximum number of

girls per thousand boys is for ST (i.e., 970) and the

minimum number of girls per thousand boys is for

urban (i.e., 910).

Also, the number of girls per thousand boys is more in

rural areas than in urban areas, backward districts than

in non-backward districts, SC and ST categories than

in non SC/ST.

Question: 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) Let us take political party on x-axis and seats won

on y-axis and choose an appropriate scale (1 unit = 10

seats in y-axis).

The required graph can be constructed as follows.

(ii) Political party ‘A’ won maximum number of seats

(i.e., 75).

Question: 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

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?

Solution:

(i)

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

Let us take the length of leaves on x-axis and the

number of leaves on y-axis. The histogram can be

drawn as below.

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 lengths in between 144.5 mm and 153.5 mm. It is

not necessary that all have their lengths as 153 mm.

Question: 5

The following table gives the life time of neon lamps:

Life time (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 life time of more than

700 hours?

Solution:

(i) Let us take life time (in hours) of neon lamps on x-

axis and the number of lamps

on y-axis. The histogram can be drawn as follows.

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

(ii) 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, (74 + 62 + 48 =) 184 lamps have life time

more than 700 hours.

Question: 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

20 − 30

30 − 40

40 − 50

9

17

12

9

10 − 20

20 − 30

30 − 40

40 – 50

19

15

10

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:

Upper classlimit Lower class limit

Classmark

2

Secti

on

A

Sectio

n B

Marks

Class

mark

s

Frequenc

y

Marks

Class

marks

Frequenc

y

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

Let us take class marks on x-axis and frequency on y-

axis and choose an appropriate scale (1 unit = 3 in y-

axis), the frequency polygon can be drawn as follows:

It can be observed that the students of section ‘A’ have

performed better than the students of section ‘B’ in

terms of good marks.

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

7 − 12

13 − 18

19 − 24

25 − 30

31 − 36

37 − 42

43 − 48

49 − 54

55 − 60

2

1

8

9

4

5

6

10

6

2

5

6

2

10

5

6

3

4

8

10

Represent the data of both the teams on the same

graph by frequency polygons.

[Hint: First make the class intervals continuous.]

Solution:

Class intervals of the given data are not continuous.

To make data continuous

1

0.5

2

has to be added to

the upper class limits and same has to be subtracted

from the lower class limits.

Class mark of each interval can be found by using the

following formula:

upperlimit+lowerlimit

Classmark

2

Thus, table can be rearranged as follows.

Number of

balls

Class mark

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

Now, taking class marks on x-axis and runs scored on

y-axis, we can construct frequency polygon as follows:

Question: 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

5

3

6

12

9

10

15 − 17

4

Draw a histogram to represent the data above.

Solution:

The given data has class intervals of varying width.

Proportion of children per 1 year interval can be

calculated as follows:

Age

(in year)

Frequency

(Number of

children)

Width of

class

Length of

rectangle

1 − 2

5

1

5×1

=5

1

2 − 3

3

1

3×1

=3

1

3 − 5

6

2

6×1

=3

2

5 − 7

12

2

12×1

=6

2

7 − 10

9

3

9×1

=3

3

10 − 15

10

5

5

10×1

=2

15 − 17

4

2

4×1

=2

2

Now, taking the age (in years) on x-axis and

proportion of children per 1 year interval on y-axis, we

can draw the histogram 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

4 − 6

6 − 8

8 − 12

6

30

44

16

12 – 20

4

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

(ii) Write the class interval in which the maximum

number of surnames lies.

Solution:

(i) Here, 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

letter

Frequency

(Number of

surname)

Width of

class

Length of

rectangle

1 − 4

6

3

6×2

=4

3

4 − 6

30

2

30×2

=30

2

6 − 8

44

2

44×2

=44

2

8 − 12

16

4

16 2

8

4

12 − 20

4

8

42

1

8

Now, 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 (1

unit = 4 students for y axis), the histogram can be

constructed as follows.

(ii) Maximum number of surnames lies in the interval

6 − 8 as it has 44 surnames in it.

Exercise 14.4 (6)

Question: 1

The following number of goals was 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:

The number of goals scored by the team is

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

2 3 4 5 0 1 3 3 4 3 28

10 10

28

Sum of all the observations

Mean =

Total number of observations

Mean Score

goals.

Now, arranging the number of goals in ascending

order, we get

0, 1, 2, 3, 3, 3, 3, 4, 4, 5

Total number of observations = 10, which is an even

number.

Therefore, median Score will be the mean of

10

5

2

th

and

th

10

16

2

observation of the arranged data.

= =

th th

5 observation+6 observation 3+3

Median score

22

6

3

2

Mode of data is the observation with the maximum

frequency.

So, the mode is 3 as it has the maximum frequency

equal to 4.

Question: 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:

The marks of 15 students in mathematics test are

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

Sum of all observations

Mean of data =

Total number of observations

41 39 48 52 46 62 54 40

96 52 98 40 42 52 60

15

822

54.8

15

Now, let us arrange these scores in an ascending order.

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

Now, number of observations is 15 which is an odd

number.

Therefore, the median

=

th

8

15+1

2

observation

Therefore, median

=

52

Mode of data is the observation with the maximum

frequency.

Therefore, mode = 52 (As 52 has the highest

frequency equal to 3).

Question: 3

The following observations have been arranged in

ascending order. If the median of the data is 63, find

the value of x.

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Solution:

Total number of observations in the given data =10

(which is an even number).

Therefore,

The median of the data

= mean of

10 10

( =5th) and ( +1=6th)

22

observation

Therefore,

.

Median

th th

5 observation+6 observation

=

2

+2 2 +2 2( +1)

63=

2 2 2

63= +1

=62

x+x x x

x

x

Question: 4

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

14, 18.

Solution:

Let us arrange the data in an ascending order:

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

Now, we can observe that 14 has the highest

frequency, i.e. 4.

Therefore, the mode of the data = 14.

Question: 5

Find the mean salary of 60 workers in a factory from

the following table:

Salary (in Rs)

Number of workers

3000

16

4000

12

5000

10

6000

8

7000

6

8000

4

9000

3

10000

1

Total

60

Solution:

To calculate the mean of the salaries, we need to

arrange the table as follows:

Salary

(in Rs)

i

x

Number of

worker

()

i

f

ii

fx

3000

16

3000 16 48000

4000

12

4000 12 48000

5000

10

5000 10 50000

6000

8

6000 8 48000

7000

6

7000 6 42000

8000

4

8000 4 32000

9000

3

9000 3 27000

10000

1

10000 1 10000

Total

f60

i

f x 305000

ii

fx

305000

= 5083.33

f 60

Mean=

ii

i

Therefore, the mean salary of 60 workers is Rs

5083.33.

Question: 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:

Extreme values or outliers in the data affect the mean.

So, mean is not a good representative of the data if

few of the points are very far from the most of the

other points. In such cases, median and mode give a

better estimate.

Let us consider the following examples:

(i) The following data represents the salaries of the

team members of IT department in a company.

Rs. 25200, Rs. 28300, Rs. 22550, Rs. 26500, Rs.

31450.

In this case, the observations are close to each other.

Therefore, mean is an appropriate measure of central

tendency.

(ii) The following data represents the marks obtained

by 12 students in science test.

49, 69, 66, 82, 74, 56, 87, 92, 79, 88, 62, 89

In this case, some observations are very far from other

observations. Therefore, in this case, median is the

appropriate measure of central tendency.