### NCERT Class 9 Maths Chapter 14 – Statistics

## Exercise 14.1
Page: 239

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

Solution:

Five examples
from day-to-day life are

- The number of students in our class
- The number of fans in our school
- Electricity bills of our house for the last two years
- Election results obtained from television or
newspapers
- Literacy rate figures obtained from Educational
Survey

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

Solution:

Primary data:
When the information was collected by the investigator themselves with a
definite objective in their mind, the data obtained is called primary data.

Primary data;
(i), (ii) and (iii)

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

Secondary
data; (iv) and (v)

## Exercise 14.2
Page: 245

**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:

Frequency is
the number of students having the same blood group. The frequency is
represented in the table or the frequency distribution table.

The most
common Blood Group is the blood group with the highest frequency: O

The rarest
Blood Group is the blood group with the lowest frequency: AB

**2. The distance (in km) of 40 engineers from their residence to
their place of work was 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:

Since the
given data is very large, we construct a grouped frequency distribution table
of class size 5. **∴**,
class intervals will be 0-5, 5-10, 10-15, 15-20 and so on. The data is
represented in the grouped frequency distribution table as

In the given
table, the classes do not overlap. Also, we find that the houses of 36 out of
40 engineers are below 20 km of distance.

**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) Since the
given data is very large, we construct a grouped frequency distribution table
of class size 2.

**∴**, class intervals will be 84-86,
86-88, 88-90, 90-92 and so on. The data is represented in the grouped frequency
distribution table as

(ii) The
humidity is very high in the given data. Since the humidity is observed to be
high during the rainy season, the data here must be about the rainy season.

(iii) The
range of a data = The maximum value of the data – minimum value of the data

= 99.2−84.9

= 14.3

**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) The data
given in the question can be represented by a grouped frequency distribution
table, taking the class intervals as 160 – 165, 165 – 170, etc., as

(ii) It can
be concluded from the given data and the table that 35 students, i.e., more
than 50% of the total 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) The
grouped frequency distribution table for the data given in the question with
class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on is given below.

(ii) The
number of days in which the concentration of sulphur dioxide was more than 0.11
parts per million = 2+4+ 2 = 8

**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:

The frequency
distribution table for the data given in the question is given below.

**7. The value of π up to 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) The
frequency distribution of the digits from 0 to 9 after the decimal point is
given in the table below.

(ii) The
digit having the least frequency occurs the least. Since 0 occurs only twice,
it has a frequency of 2. **∴**,
the least frequently occurring digit is 0.

The digit
with the highest frequency occurs the most. Since 3 and 9 occur eight times, it
has a frequency of 8. **∴**,
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 5-10.**

**(ii)** **How many children have watched television for 15 or
more hours a week?**

Solution:

(i) The
grouped frequency distribution table for the data given in the question, taking
class width 5 and one of the class intervals as 5-10, is given below.

(ii) From the
given table, we can conclude that 2 children watched television for 15 or more
hours a week.

**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 interval 2 – 2.5.**

Solution:

The grouped
frequency distribution table for the data given in the table, using class
intervals of size 0.5 starting from interval 2 – 2.5, is given below.

## Exercise 14.3
Page: 258

**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 %):**

**(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) The
information given in the question is represented below graphically.

(ii) We can
observe from the graph that reproductive health conditions are the major cause
of women’s ill health and death worldwide.

(iii) Two
factors responsible for the cause in (ii) are

- Lack of proper care and understanding.
- Lack of medical facilities.

**2. The following data on the number
of girls (to the nearest ten) per thousand boys in different sections of Indian
society are given below.**

**(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) The
information given in the question is represented below graphically.

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

**3. Given below are the seats won by
different political parties in the polling outcome of state assembly elections:**

**(i) Draw a bar graph to represent the
polling results.**

**(ii) Which political party won the
maximum number of seats?**

Solution:

(i) The bar
graph representing the polling results is given below.

(ii) From the
bar graph, it is clear that Party A won the maximum number of seats.

**4. The length of 40 leaves of a plant
is measured correctly to one millimetre, and the obtained data is represented
in the following table:**

**(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 is 153 mm long? Why?**

Solution:

(i) The data
given in the question is represented in the discontinuous class interval. So,
we have to make it in the continuous class interval. The difference is 1, so
taking half of 1, we subtract ½ = 0.5 from the lower limit and add 0.5 to the
upper limit. Then, the table becomes

(ii) Yes, the
data given in the question can also be represented by a frequency polygon.

(iii) No, we
cannot conclude that the maximum number of leaves is 153 mm long because the
maximum number of leaves are lying in-between the length of 144.5 – 153.5

**5. The following table gives the
lifetimes of 400 neon lamps.**

**(i) Represent the given information
with the help of a histogram.**

**(ii)** **How many lamps have a lifetime of more than 700
hours?**

Solution:

(i) The
histogram representation of the given data is given below.

(ii) The
number of lamps having a lifetime of more than 700 hours = 74+62+48 = 184

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

**Represent the marks of the students
of both sections on the same graph by two frequency polygons. From the two
polygons, compare the performance of the two sections.**

Solution:

The
class-marks = (lower limit + upper limit)/2

For section
A,

For section
B,

Representing
these data on a graph using two frequency polygon, we get

From the
graph, we can conclude that the students of Section A performed better than
Section B.

**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 teams on
the same graph by frequency polygons.**

**[Hint: First, make the class
intervals continuous.]**

Solution:

The data
given in the question is represented in the discontinuous class interval. So,
we have to make it in the continuous class interval. The difference is 1, so
taking half of 1, we subtract ½ = 0.5 = 0.5 from the lower limit and add 0.5 to
the upper limit. Then, the table becomes

The data of
both teams are represented on the graph below by frequency polygons.

**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.**

Solution:

The width of
the class intervals in the given data varies.

We know that,

The area of
the rectangle is proportional to the frequencies in the histogram.

Thus, the
proportion of children per year can be calculated as given in the table below.

Let x-axis =
the age of children

y-axis =
proportion of children per 1-year interval

**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.**

Solution:

(i) The width
of the class intervals in the given data is varying.

We know that,

The area of
the rectangle is proportional to the frequencies in the histogram.

Thus, the
proportion of the number of surnames per 2 letters interval can be calculated
as given in the table below.

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

## Exercise 14.4
Page: 269

**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:

Mean =
Average = Sum of all the observations/Total number of observations

=
(2+3+4+5+0+1+3+3+4+3)/10

= 28/10

= 2.8

Median

To find the
median, we first arrange the data in ascending order.

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

Here,

Number of
observations (n) = 10

Since the
number of observations is even, the median can be calculated as

= 3

Mode

To find the
mode, we first arrange the data in ascending order.

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

Here,

We find that
3 occurs most frequently (4 times).

∴ Mode = 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:

Mean=Average
= Sum of all the observations/Total number of observations

=
(41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/15

= 822/15

= 54.8

Median

To find the
median, we first arrange the data in ascending order.

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

Here,

Number of
observations (n) = 15

Since the
number of observations is odd, the median can be calculated as

Median =
[(n+1)/2]^{th} observation

= [(15+1)/2]^{th} observation

= (16/2)^{th} observation

= 8^{th} observation

= 52

Mode

To find the
mode, we first arrange the data in ascending order.

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

Here,

We find that
52 occurs most frequently (3 times).

∴ Mode = 52

**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:

Number of
observations (n) = 10

Given that
Median = 63

Since the
number of observations is even, the median can be calculated as

63 = [5^{th} observation+(5+1)^{th} observation]/2

63 = [5^{th} observation+6^{th} observation]/2

63 =
(x+x+2)/2

63 = (2x+2)/2

x = 63-1

x = 62

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

Solution:

Mode

To find the
mode, we first arrange the given data in ascending order.

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

Here,

We find that
14 occurs most frequently (4 times).

∴ Mode = 14

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

Solution:

The mean
salary is ₹5083.33

**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) Mean
marks obtained in the examination

(ii) Runs scored by Mahendra Singh Dhoni in 7 matches are

39, 51, 56,
102, 83, 48, 91

Here,

Mean =
(39+51+56+102+83+48+91)/7

= 470/7

= 67.1.

Median,

Arranging in
ascending order, we get 39, 48, 51, 56, 83, 91, 102

n = 7

Median =
[(n+1)/2]^{th} observation

= ( (7+1)/2)^{th }observation

= (8/2)^{th }observation

= 4^{th }observation

= 56

## No comments:

## Post a Comment