NCERT Solutions for Class 9 Mathematics Chapter 14 – Statistics

Mathematics helps in developing various skill sets of the students. It tests students’ knowledge based on numbers, statistics, graphs, objects and figures. Students must have a clear understanding of the concept of Mathematics and should be able to interpret it.

Data study is a significant part of engineering studies. Therefore, students must have the right knowledge about the handling of data. This is what is covered in this chapter. Students will learn topics like handling of data, interpretation of data, presentation of data, collection of data, measures of central tendency and graphical representation of data in this chapter.

Noting the importance of the chapter, every detail covered in Class 9 Mathematics Chapter 14 has been included in the NCERT Solutions for Class 9 Mathematics Chapter 14. It has the detailed solution to each exercise and every question and test yourself easily from the tip to the toe.  The key points are highlighted in a proper format.Also, It has certain examples which will help students to connect to their statistical studies in the future directly. Thus, the NCERT Solutions have been an essential factor in the students’ preparation .

Extramarks’ website has all these NCERT Solutions. Apart from the NCERT Solutions for Class 9 Mathematics Chapter 14, you can also find all the NCERT-related revision notes, NCERT Exemplar, NCERT-based sample papers and mock tests on the website. Students can avail this material  from the Extramarks’ website anytime to leverage their performance and excel in the examinations.

Key Topics Covered In NCERT Solutions forClass 9 Mathematics Chapter 14

Chapter 14 Class 9 Mathematics is about Statistics. Statistics is the study of collecting information about a large data set. Understanding statistics can be really helpful for day to day life . As a result, it has become one of the introductory chapters of Class 9th Mathematics. In the chapter, you will learn how to manage such data efficiently without making many mistakes.

It would help you in the future classes for group studies and for doing any research work. It will also lay the foundation for various graphical interpretations in Mathematics. To be good at this chapter, it is necessary that students learn to handle data accurately, which can be achieved with lots of practice. For this reason, we have included many examples in NCERT Solutions for Class 9 Mathematics Chapter 14, which is available on the Extramarks’ website.

NCERT Solutions for Class 9 Mathematics Chapter 14 requires students to use their critical thinking ability and apply a wide range of formulas they have learnt.

Introduction

In this chapter, we will learn about important data, its collection, and its presentation.

There is a lot of data available these days in the world, but every data is not so informative, so we have to extract the essential, meaningful and informative data. The study of extracting meaningful information from such data is called statistics.

In this, we will study data collection, organisation, analysis and interpretation.

Collection of data extramarks

In this part of the chapter, you will find activities related to data collection and learn how to collect those data using that activity. Post that, they learn the difference between primary and secondary data as well.

To get more details about the collection of the data and its activities, visit the Extramarks’ website and refer to the NCERT Solutions for Class 9 Mathematics Chapter 14.

Presentation of data

In the above part of the chapter, we studied the data collection. Now we will discuss how to represent the data in an easy-to-understand way and get its main features at a glance.

Students of ‘raw data’, ‘arranged data’, and ‘range’, etc. in this section.

To get more details on how to represent the data in a meaningful and easy way. You can visit our Extramarks’ website and refer to the NCERT Solutions for Class 9 Mathematics Chapter 14.

Graphical representation of data

In this part of the chapter, we will learn how to present the data by putting less effort and in a structured way to understand the data and the graphical form of the data in a better way. Some graphs available are given below:

• Bar graph:

It is constructed in the form of rectangular vertical columns, representing the comparison among the different categories.

• Histogram:

It represents a bar graph in rectangular vertical columns, representing the data in intervals.

• Frequency polygons:

It represents a polygon that shows the data’s frequency

Students will learn about the graphical representations from this topic. To clarify their concepts, students can practice the questions provided in the NCERT Solutions for Class 9 Mathematics Chapter 14, available on the Extramarks’ website.

Measures of central tendency

In this chapter, we have already studied how to represent the data in frequency distributions table, bar graphs, histograms and frequency polygons. Now the question lies do we need to remember all the data. We can remember some essential features of the data representation by using measures of the central tendency.

To learn more about measures of central tendency, visit the Extramarks’ website and refer to NCERT Solutions for Class 9 Mathematics Chapter 14.

Summary

In This chapter, we have learnt about representing the informative data in a very simple way, remembering the data quickly and learnt ways to convert the essential informative data in a very organised way.

The points we learned from this chapter are as follows:

• Collection of data

We went through some examples of the data collection.

• Presentation of data

We learnt to present the data in a managed way.

• Graphical representation of data

We used some graphs to represent the data. They are:

• Bar graph
• Histogram
• Frequency polygons
• Measures of central tendency

We learnt how we could represent any data very easily using the measures of central tendency.

Students can learn more regarding this chapter from the NCERT Solutions for Class 9 Mathematics Chapter 14 available on the Extramarks’ website.

NCERT Solutions for Class 9 Mathematics Chapter 14 Exercise & Solutions.

In this online world, students can easily find question banks for a certain chapter from various sources, but it is difficult to get the right solutions to all the exercises by experts. Hence, we focus on exercises and its solution in our NCERT Solutions for Class 9 Mathematics Chapter 14, available on the Extramarks’ website. This would help students to get answers to every question  at one place. This way they can  save time during their preparation.

The solutions are entirely designed as per examination pattern and the tendency of the paper setters. Thus, ensuring that students do not lack in any aspect of their preparation and pass their examinations with flying colours.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 14:

• Chapter 14: Exercise 14.1
• Chapter 14: Exercise 14.2
• Chapter 14: Exercise 14.3
• Chapter 14: Exercise 14.4

Along with Class 9 Mathematics Solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

• NCERT Solutions Class 1,
• NCERT Solutions Class 2,
• NCERT Solutions Class 3,
• NCERT Solutions Class 4,
• NCERT Solutions Class 5,
• NCERT Solutions Class 6,
• NCERT Solutions Class 7,
• NCERT Solutions Class 8,
• NCERT Solutions Class 9,
• NCERT Solutions Class 10,
• NCERT Solutions Class 11.
• NCERT Solutions Class 12.

NCERT Exemplar Class 9 Mathematics

Accuracy during examinations is necessary to score well. Students can acquire accuracy with a lot of practice. They can practise if they have the right sets of questions. NCERT Exemplar Class 9th Mathematics is the perfect resource for all these topics given above.

One can find questions from all the core topics and fundamental concepts of the NCERT textbook in the NCERT Exemplar. It is a useful guide for all the aspirants preparing for competitive exams like JEE, NEET etc. All levels of questions including   easy, medium and challenging are covered in the book, making it extremely helpful for all levels of students.

Experts suggest all the students should include NCERT Exemplar in their study material due to its quality content and multiple questions framed on a single topic with different difficulty levels. The challenging questions in the book aid students in using their analytical abilities. As a result, students develop open and broad analytical thinking skills.

Key Features for NCERT Solutions for Class 9 Mathematics Chapter 14

Students can excel in their examinations only if they have the right skill set to solve their papers. Hence, NCERT Solutions for Class 9 Mathematics Chapter 14 helps build the required skill set. The key features are as follows:

• Students will develop confidence while doing calculations once they refer to NCERT Solutions for Class 9 Mathematics Chapter 14.
• They will learn the art of time management and be able to complete their paper on time.
• After completing the NCERT Solutions for Class 9 Mathematics Chapter 14, they can deal with the statistical portion of mathematics quickly and efficiently.

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:

 Blood Group Number of students A 9 B 6 AB 3 O 12 Total 30

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:

 Class Interval Tally mark Number of engineers 0-5 $\overline{)||||}$ 5 5-10 $\overline{)||||}\text{\hspace{0.17em}}\overline{)||||}\text{\hspace{0.17em}}|$ 11 10-15 $\overline{)||||}\text{\hspace{0.17em}}\overline{)||||}\text{\hspace{0.17em}}|$ 11 15-20 $\overline{)||||}\text{\hspace{0.17em}}||||$ 9 20-25 $|$ 1 25-30 $|$ 1 30-35 $||$ 2 Total 40

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:

 Class interval Tally mark Frequency 84-86 $|$ 1 86-88 $|$ 1 88-90 $||$ 2 90-92 $||$ 2 92-94 $\overline{)||||}||$ 7 94-96 $\overline{)||||}|$ 6 96-98 $\overline{)||||}||$ 7 98-100 $||||$ 4 Total 30

(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:

 Class intervals Tally mark Number of students(f) 150 – 155 $\overline{)||||}\overline{)||||}||$ 12 155 – 160 $\overline{)||||}\text{\hspace{0.17em}}||||$ 9 160 – 165 $\overline{)||||}\overline{)||||}||||$ 14 165- 170 $\overline{)||||}\overline{)||||}$ 10 170 – 175 $\overline{)||||}$ 5 Total 50

(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:

 Class interval Tally mark Number of days 0.00 – 0.04 $||||$ 4 0.04 – 0.08 $\overline{)||||}||||$ 9 0.08 – 0.12 $\overline{)||||}||||$ 9 0.12 – 0.16 $||$ 2 0.16 – 0.20 $||||$ 4 0.20 – 0.24 $||$ 2 Total 30

(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:

 Number of heads Tally mark Frequency 0 $\overline{)||||}|$ 6 1 $\overline{)||||}\overline{)||||}$ 10 2 $\overline{)||||}||||$ 9 3 $||||$ 5 Total 30

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:

 Digits 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 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:

 Class interval Frequency 0 – 5 10 5 – 10 13 10 – 15 5 15 – 20 2 Total 30

(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:

 Class interval Frequency 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

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 %):

 S.No. Causes Female fatality rate (%) 1. 2. 3. 4. 5. 6. Reproductive health conditions Neuropsychiatric conditions Injuries Cardiovascular conditions Respiratory conditions Other causes 31.8 25.4 12.4 4.3 4.1 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.

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.

 Section Number of girls per thousand boys Schedule Cast (SC) Schedule 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.

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:

 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?

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:

 Length (in mm) Number of leaves 118 – 126 127 – 135 136 – 144 145 – 153 154 – 162 163 – 171 172 – 180 3 5 9 12 5 4 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 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:

 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 3 5 9 12 5 4 2 Total 40

(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:

 Life time (in hours) Number of lamps 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900 900 – 1000 14 56 60 86 74 62 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?

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:

 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.

Ans.

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

I
n this way, frequency table will be as given below:

 Section A Section B Marks Class Mark Frequency Marks Class Mark 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

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:

 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.

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:

 Number of balls Midpoint 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 1

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:

 Age (in years) Number of children 1 – 2 2 – 3 3 – 5 5 – 7 7 – 10 10 – 15 15 – 17 5 3 6 12 9 10 4

Draw a histogram to represent the data above.

Ans.

The modified frequency table is given below:

 Age (in years) Number of children Width of interval Length of rectangle 1 – 2 5 1 $\frac{5}{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}5$ 2 – 3 3 1 $\frac{3}{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3$ 3 – 5 6 2 $\frac{6}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3$ 5 – 7 12 2 $\frac{12}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}6$ 7 – 10 9 3 $\frac{9}{3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3$ 10 – 15 10 5 $\frac{10}{5}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}2$ 15 – 17 4 2 $\frac{4}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1\text{\hspace{0.17em}}=\text{\hspace{0.17em}}2$

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:

 Number of letters Number of surnames 1 – 4 4 – 6 6 – 8 8 – 12 12 – 20 6 30 44 16 4

(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:

 Number of letters Number of surnames Width of interval Length of rectangle 1 – 4 6 3 $\frac{6}{3}×2=4$ 4 – 6 30 2 $\frac{30}{2}×2=30$ 6 – 8 44 2 $\frac{44}{2}×2=44$ 8 – 12 16 4 $\frac{16}{4}×2=8$ 12 – 20 4 8 $\frac{4}{8}×2=1$

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}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2+3+4+5+0+1+3+3+4+3}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{28}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=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\text{\hspace{0.17em}}\left(even\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}Median=\frac{{\left(\frac{n}{2}\right)}^{th}term+{\left(\frac{n}{2}+1\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{\left(\frac{10}{2}\right)}^{th}term+{\left(\frac{10}{2}+1\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{\left(5\right)}^{th}term+{\left(6\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3+3}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=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}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\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{\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}\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}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{4}0+\text{42}+\text{52}+\text{6}0\end{array}}{15}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{822}{15}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=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}\\ \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}}={\left(\frac{15+1}{2}\right)}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\\ \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}}={8}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\\ \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}}=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,\text{}\mathrm{x},\text{}\mathrm{x}+\text{2},\text{72},\text{78},\text{84},\text{95}\\ \text{Number of data}=10\\ \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}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{Median}=63\\ \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}}\mathrm{Median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \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}\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} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{10}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{10}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}63=\frac{{5}^{\mathrm{th}}\mathrm{term}+{6}^{\mathrm{th}}\mathrm{term}}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{{5}^{\mathrm{th}}\mathrm{term}+{6}^{\mathrm{th}}\mathrm{term}}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{\mathrm{x}+\mathrm{x}+2}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{2\mathrm{x}+2}{2}\\ ⇒\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}63=\mathrm{x}+1\\ ⇒\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\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:

 Salary (in Rs) Number of workers 3000 4000 5000 6000 7000 8000 9000 10000 16 12 10 8 6 4 3 1 Total 60

Ans.

 Salary(x) Number of workers(f) fx 3000 4000 5000 6000 7000 8000 9000 10000 16 12 10 8 6 4 3 1 48000 48000 50000 48000 42000 32000 27000 10000 Total 60 305000

$\begin{array}{l}\mathrm{Mean}=\frac{\sum \mathrm{fx}}{\sum \mathrm{f}}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{305000}{60}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5083.33\end{array}$

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.

$\begin{array}{l}\text{Then},\text{Mean height}=\frac{\begin{array}{l}\text{11}0.\text{8 cm}+\text{11}0.\text{9 cm}+\text{111}.\text{1 cm}+\text{111}.\text{2 cm}\\ \text{}+\text{111}.\text{4 cm}\end{array}}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{555.4}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=111.08\text{\hspace{0.17em}}cm\end{array}$

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}\\ \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}\hspace{0.17em} \hspace{0.17em}}=\frac{191\mathrm{°}}{6}\\ \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}\hspace{0.17em}\hspace{0.17em}}=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}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\text{term}}{2}\\ \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}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{6}{2}\right)}^{\mathrm{th}}\text{term}+{\left(\frac{6}{2}+1\right)}^{\mathrm{th}}\text{term}}{2}\\ \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} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{3}^{\mathrm{rd}}\text{term}+{4}^{\mathrm{th}}\text{term}}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{29°}+\text{\hspace{0.17em}\hspace{0.17em}32°}}{2}\\ \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} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{61°}}{2}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=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}$

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