NCERT Solutions for Class 9 Mathematics Chapter 14 – Statistics

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

Ans.

The five examples of data that we can collect from our day-to-day life:
(1) To know monthly expense of 20 houses in our society.
(2) To know number of members in the families of our classmates.
(3) To know the pet animal population in our society.
(4) To know the shoe number of 25 classmates in our class.
(5) To know the number of children below 5 years in Godwin Society.

Q.2 Classify the data in the previous question (Q.1) above as primary or secondary data.

Ans.

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.

We can see that data given in example 1, 4 and 5 are primary data. Data given in example 2 and 3 are secondary data.

Q.3 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?

Ans.

The frequency distribution table is as given below:

The most common blood group is O because there are 12 students of this blood group.

The rarest blood group is AB because there are only three students of this blood group.

Q.4 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?

Ans.

The frequency table having intervals 0 – 5, 5 – 10, 10 – 15, … is given below:

From the table, we observe that the residence of mostly engineers is 5 to 20 km away from their office. Few engineers have their homes more than 20 km away from their offices.

Q.5 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?

Ans.

(i) The grouped frequency distribution table is given below:

(ii) These data are about rainy season.

(iii) Range of this data
= Maximum data – Minimum data
= 99.2 – 84.9 = 14.3

Q.6 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?

Ans.

(i) The grouped frequency distribution table related to height of students is given below:

(ii) From the table it is clear that height of more than 50% students is less than 165 cm.

Q.7 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?

Ans.

(i) The grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08 and so on is given below:

(ii) The concentration of sulphur dioxide is more than 0.11 parts per million for 8 (2+4+2) days.

Q.8 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.

Ans.

The frequency distribution table for the data obtained by tossing the coin is given below:

Q.9 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?

Ans.

(i) Frequency distribution of the digits from 0 to 9 is given below in the form of table:

(ii) The most frequently occurring digits are 3 and 9. The least frequently occurring digit is 0.

Q.10 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?

Ans.

(i) The grouped frequency distribution table for given hours, taking class width 5 is given below:

(ii) 2 children watched television for 15 or more hours a week.

Q.11 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.

Ans.

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

Q.12 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 %):

(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.

Ans.

(i) The graphical representation of the given information is given below:

(ii) Reproductive health condition is the major cause of women’s ill health and death worldwide.
(iii) The two factors which play a major role in the cause of reproductive health conditions are:

(i) Improper diet (ii) Lack of medical facility.

Q.13 The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is 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.

Ans.

(i) Bar graph showing given information is given below:

(ii) From the graph we see that number of girls per thousand boys is less in urban areas.

Q.14 Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

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

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

Ans.

(i) A bar graph representing the polling results is given below:

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

Q.15 The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:

(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 numbers of leaves are 153 mm long? Why?

Ans.

(i) Difference of upper limit of 118 – 126 and lower limit of

127 – 135 = 127 – 16
= 1
Half of the difference = (1/2)
= 0.5

New class interval formed from 118 – 126 is
(118 – 0.5) – (126 + 0.5) i.e., 117.5 – 126.5.
New class interval formed from 127 – 135 is
(127 – 0.5) – (135 + 0.5) i.e., 126.5 – 135.5.
In the same manner, new table with new class interval is given below:

(ii) Frequency polygon is other suitable graphical representation for the same data.
(iii) No, it is not correct to conclude that the maximum numbers of leaves are 153 mm long because upper limit of a class interval is not included in the interval.

Q.16 The following table gives the life times of 400 neon lamps:

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

(ii) How many lamps have a life time of more than 700 hours?

Ans.

(i) The given information is shown by histogram in the figure given below:

(ii) Number of lamps has a life time of more than 700 hours is 184(74+62+48).

Q.17 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.

Ans.

$\begin{array}{l}\mathrm{Class}\text{mark}=\frac{\mathrm{Lower}\text{class limit}+\mathrm{Upper}\text{class limit}}{2}\\ \mathrm{For}\text{class interval}(0-10):\\ \mathrm{Class}\text{mark}=\frac{0+10}{2}=5\\ \mathrm{For}\text{class interval}(10-20):\\ \mathrm{Class}\text{mark}=\frac{10+20}{2}=15\\ \end{array}$

In this way, frequency table will be as given below:

The frequency polygons are drawn as given below:

Q.18 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.

Ans.

Difference of upper limit of 1 – 6 and lower limit of 7 – 12 = 7 – 6 = 1
Half of the difference = (1/2) = 0.5

New class interval formed from 1 – 6 is
(1 – 0.5) – (6 + 0.5) i.e., 0.5 – 6.5.
New class interval formed from 7 – 12 is
(7 – 0.5) – (12 + 0.5) i.e., 6.5 – 12.5.
In the same manner, new table with new class interval is given below:

Polygons for both teams are shown in the following graphs:

Q.19 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.

Ans.

The modified frequency table is given below:

The histogram based on the above data is given below:

Q.20 100 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.

Ans.

(i) The modified frequency table is given below:

The histogram of given data is given below:

(ii) Maximum number of surnames is in interval 6 – 8.

Q.21 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.

Ans.

$\begin{array}{l}\text{Mean=}\frac{\mathrm{Sum}\mathrm{of}\mathrm{data}}{\mathrm{Number}\mathrm{of}\mathrm{data}}\\ =\frac{2+3+4+5+0+1+3+3+4+3}{10}\\ =\frac{28}{10}\\ =2.8\\ \text{Thus},\text{the mean of the score is 2}\text{.8}\text{.}\\ \text{Ascending order of goals:}\\ 0,\text{1, 2, 3, 3, 3, 3, 4, 4, 5}\\ \text{Number of observations}\left(n\right)=10\left(even\right)\\ Median=\frac{{\left(\frac{n}{2}\right)}^{th}term+{(\frac{n}{2}+1)}^{th}term}{2}\\ =\frac{{\left(\frac{10}{2}\right)}^{th}term+{(\frac{10}{2}+1)}^{th}term}{2}\\ =\frac{{\left(5\right)}^{th}term+{\left(6\right)}^{th}term}{2}\\ =\frac{3+3}{2}\\ =3\\ \text{Thus, median of the goals is 3}\text{.}\\ \text{Mode:}\\ \text{Since},\text{the frequency of 3 is more}\text{.}\\ \text{So, mode of goals scored by the team}=\text{3}\text{.}\end{array}$

Q.22 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.

Ans.

$\begin{array}{l}\text{Mean of the marks obtained in test}\\ =\frac{\begin{array}{l}\text{41}+\text{39}+\text{48}+\text{52}+\text{46}+\text{62}+\text{54}+\text{4}0+\text{96}+\text{52}+\text{98}\\ +\text{4}0+\text{42}+\text{52}+\text{6}0\end{array}}{15}\\ =\frac{822}{15}\\ =54.8\\ \text{Ascending order of marks:}\\ \text{39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98}\\ \mathrm{n}=15\left(\mathrm{odd}\right)\\ \text{∴Median}={\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{th}}\mathrm{term}\\ ={\left(\frac{15+1}{2}\right)}^{\mathrm{th}}\mathrm{term}\\ =8^{\mathrm{th}}\mathrm{term}\\ =52\\ \text{Here},\text{the frequency of 52 is heighest i.e., 3.}\\ \text{So, the mode is 52.}\end{array}$

Q.23 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

Ans.

$\begin{array}{l}\mathrm{The}\text{given data are:}\\ \text{29},\text{32},\text{48},\text{5}0,\mathrm{x},\mathrm{x}+\text{2},\text{72},\text{78},\text{84},\text{95}\\ \text{Number of data}=10\\ \mathrm{Median}=63\\ \mathrm{Median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{term}+{(\frac{\mathrm{n}}{2}+1)}^{\mathrm{th}}\mathrm{term}}{2}\\ =\frac{{\left(\frac{10}{2}\right)}^{\mathrm{th}}\mathrm{term}+{(\frac{10}{2}+1)}^{\mathrm{th}}\mathrm{term}}{2}\\ 63=\frac{5^{\mathrm{th}}\mathrm{term}+6^{\mathrm{th}}\mathrm{term}}{2}\\ 63=\frac{5^{\mathrm{th}}\mathrm{term}+6^{\mathrm{th}}\mathrm{term}}{2}\\ 63=\frac{\mathrm{x}+\mathrm{x}+2}{2}\\ 63=\frac{2\mathrm{x}+2}{2}\\ \Rightarrow 63=\mathrm{x}+1\\ \Rightarrow \mathrm{x}=63-1=62\end{array}$

Q.24 Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Ans.

$\begin{array}{l}Ascending\text{order of data is:}\\ \text{14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28}\\ \text{Here 14 is repeated more time than any other data}\text{.}\\ \text{So, mode of data is 14}\text{.}\end{array}$

Q.25 Find the mean salary of 60 workers of a factory from the following table:

Ans.

$\begin{array}{l}\mathrm{Mean}=\frac{{\displaystyle \sum \mathrm{fx}}}{{\displaystyle \sum \mathrm{f}}}\\ =\frac{305000}{60}\\ =5083.33\end{array}$

Q.26 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.

Ans.

(i) Here is an example showing that mean is an appropriate measure of central tendency. The heights of five friends: 110.8 cm, 110.9 cm, 111.1 cm, 111.2 cm and 111.4 cm.

\begin{array}{l}\text{Then},\text{Mean height}=\frac{\begin{array}{l}\text{11}0.\text{8 cm}+\text{11}0.\text{9 cm}+\text{111}.\text{1 cm}+\text{111}.\text{2 cm}\\ +\text{111}.\text{4 cm}\end{array}}{5}\\ =\frac{555.4}{5}\\ =111.08cm\end{array}

We see that mean height is approximately 111 cm which is closer to each given data.
So, we can say that mean is an appropriate measure of central tendency.
(ii) Example of median:

$\begin{array}{l}\mathrm{The}\text{temperatures in °C of 6 different cities are given below:}\\ \text{26°,\hspace{0.17em}\hspace{0.17em}28°,\hspace{0.17em}\hspace{0.17em}29°,\hspace{0.17em}\hspace{0.17em}32°,\hspace{0.17em}\hspace{0.17em}35°,\hspace{0.17em}\hspace{0.17em}41°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mean}=\frac{\text{26°}+\text{\hspace{0.17em}\hspace{0.17em}28°}+\text{\hspace{0.17em}\hspace{0.17em}29°}+\text{\hspace{0.17em}\hspace{0.17em}32°}+\text{\hspace{0.17em}35°}+\text{\hspace{0.17em}\hspace{0.17em}41°}}{6}\\ =\frac{191\mathrm{°}}{6}\\ =31.8\mathrm{°}\\ \text{Here, n}=6\left(\mathrm{even}\right)\\ \mathrm{So},\text{median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\text{term}+{(\frac{\mathrm{n}}{2}+1)}^{\mathrm{th}}\text{term}}{2}\\ =\frac{{\left(\frac{6}{2}\right)}^{\mathrm{th}}\text{term}+{(\frac{6}{2}+1)}^{\mathrm{th}}\text{term}}{2}\\ =\frac{3^{\mathrm{rd}}\text{term}+4^{\mathrm{th}}\text{term}}{2}\\ =\frac{\text{29°}+\text{\hspace{0.17em}\hspace{0.17em}32°}}{2}\\ =\frac{\text{61°}}{2}\\ =30.5\mathrm{°}\\ \mathrm{We}\text{see that median is better representation than mean.}\\ \text{Because, Meadian does not affect the heighest temperature.}\end{array}$