NCERT Solutions for Class 9 Maths Chapter 14 Statistics

 

NCERT solutions for class 9 maths chapter 14 Statistics: Statistics is one of the important topics under class 9 NCERT syllabus. You must have seen information like the temperature of cities, cricket batsman or bowler ranking, results of particular exams either in news, tv, online sites or magazines. All the mentioned examples are represented through a table or graphs. CBSE NCERT solutions for class 9 maths chapter 14 Statistics is covering the solutions for this particular topic in detail. In statistics, the numbers or facts are collected for a purpose and the collection is called data. From these collected data you can understand and conclude some results. For example, if we have data of temperature of 40 cities for a particular day, we can infer that which city has a higher temperature, which city has lower temperature, what is the average temperature of 40 cities, which value of temperature is most repeated, what is the middle value of temperature if we arranged the temperature in either increasing or decreasing order etc. In this, you are arranging the collected data and then  analyze it and can infer certain things. In solutions of NCERT for class 9 maths chapter 14 Statistics the answers to the questions of mode, average, median, etc are covered. Statistics is the part of maths in which you learn to get some results based on the collected data. In this chapter, there are a total of 4 exercises consist of total of 35 questions. NCERT solutions for class 9 maths chapter 14 Statistics is covering every question is a step by step manner so that you do not lose a single mark in any question. Apart from this particular chapter NCERT solutions are available for other chapters as well.

NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.1

Q1 Give five examples of data that you can collect from your day-to-day life.

Answer:

Five examples of data that we can collect in our daily life are

(i) Number of students in a class.

(ii) The number of books in a library.

(iii) Toys sold on a particular day at a shop.

(iv) People who voted for a particular candidate.

(v) Runs scored by a batsman on each ball in a particular evening.

Q2 Classify the data in Q.1 above as primary or secondary data.

Answer:

(i) The number of students in a class.

(ii) The number of books in a library.

(iii) Toys sold on a particular day at a shop.

(iv) People who voted for a particular candidate.

(v) Runs scored by a batsman on each ball on a particular evening.

All of the data in Q.1 is primary data.

NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.2

Q1 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?

Answer:

The representation of the given data in the form of a frequency distribution table is as follows.

From the table we can see that O is the most common and AB is the rarest blood group.

Q2 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?

Answer:

As the minimum and maximum distances of an engineer from his place of work is 2 and 32 respectively the class intervals with class size 5 would be the following.

0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35

The representation of the given data in the form of a grouped frequency distribution table is as follows

Frequencies of the class intervals 5 - 10 and 10 - 15 are maximum and equal to 11 each and frequencies of the class intervals 20 - 25 and 125 - 30 are minimum and equal to 1 each.

Q7 (i) The value of \pi upto \50 decimal places is given below:

3.14159265358979323846264338327950288419716939937510

Make a frequency distribution of the digits from 0 to 9 after the decimal point.

Answer:

The representation of the given data in the form of a frequency distribution table is as follows.

Q4 (i) 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

Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]

Answer:

As we can see from the given table that the data is discontinous and the difference between the upper limit of a class and the lower limit of the next class is 1 and therefore we change both of them by a value 1/2.

e.g 127 - 135 would become 126.5 - 235.5

The modified table therefore is

The representation of the above data through a histogram is as follows

Q4 (iii) 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

Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Answer:

No it is certainly not correct to conclude that the maximum number of leaves are 153 mm long because the given data does not tell us about the exact length of the leaves. It only tells us about the range in which their lengths lie. We can only conclude that the maximum number of leaves (12) have their lengths in the region 145 - 153.

Q5 (ii) The following table gives the life times of 400 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

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

Answer:

Lamps having life time in the range 700 - 800 = 74

Lamps having life time in the range 800 - 900 = 62

Lamps having life time in the range 900 - 1000 = 48

Lamps having a life time of more than 700 hours = 74 + 62 + 48 = 184.

Q6 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

Answer:

To make the frequency polygon we first modify the table as follows

\\Class\ marks= \frac{Upper\ limit\ of\ class\ interval+Lower\ limit\ of\ class\ interval}{2}

To make the frequency polygon we mark the marks on the x-axis and the number of students on the y-axis.

The representation of the given information in the form of frequency polygon is as follows.

From the frequency polygon we can see that the performance of section A is better.

Q7 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

2

5

7-12

1

6

13-18

8

2

19-24

9

10

25-30

4

5

31-36

5

6

37-42

6

3

43-48

10

4

49-54

6

8

55-60

2

10

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

Answer:

The given data is not continous we therefore modify the limits of the class intervals as well to make the class intervals continous.

To make the frequency polygon we first modify the table as follows

\\Class\ marks= \frac{Upper\ limit\ of\ class\ interval+Lower\ limit\ of\ class\ interval}{2}

To make the frequency polygon we mark the number of balls on the x-axis and the runs scored on the y-axis.

The representation of the given information in the form of frequency polygon is as follows.

Q8 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

5

2-3

3

3-5

6

5-7

12

7-10

9

10-15

10

15-17

4

Draw a histogram to represent the data above.

Answer:

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width

\\Weighted\ frequency=\frac{Minimum\ class\ size}{Class\ size\ of\ the\ interval}\times Frequency\\ Length\ of\ the\ rectangle=\frac{Minimum\ width}{Width\ of\ the\ rectangle}\times Frequency

Minimum class size = 2 - 1 = 1

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

Q9 (i) 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

6

4-6

30

6-8

44

8-12

16

12-20

4

Answer:

Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width

\\Weighted\ frequency=\frac{Minimum\ class\ size}{Class\ size\ of\ the\ interval}\times Frequency\\ Length\ of\ the\ rectangle=\frac{Minimum\ width}{Width\ of\ the\ rectangle}\times Frequency

Minimum class size = 6 - 4 = 2

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

NCERT solutions for class 9 maths chapter 14 Statistics Excercise: 14.4

Q1 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.

Answer:

Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}

Number of observations, n = 10

\\\bar{X}=\frac{\sum_{i=1}^{n=10}x_{i}}{n}\\ \bar{X}=\frac{ 2+ 3+ 4+ 5+ 0+ 1+ 3+ 3+ 4+ 3}{10}\\ \bar{X}=\frac{28}{10}\\ \bar{X}=2.8

Mean is 2.8

To find the median we have to arrange the given data in ascending order as follows:

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

n = 10 (even)

\\Median=\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}\ term}{2}\\ Median=\frac{(\frac{10}{2})^{th}\ term+(\frac{10}{2}+1)^{th}\ term}{2}\\ Median=\frac{(5)^{th}\ term+(6)^{th}\ term}{2}\\ Median=\frac{3+3}{2}\\ Median=3

In the given data 3 occurs the maximum number of times (4)

Therefore, Mode = 3

Q2 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.

Answer:

Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}

Number of observations, n = 15

\\\bar{X}=\frac{\sum_{i=1}^{n=15}x_{i}}{n}\\ \bar{X}=\frac{41+ 39+ 48+ 52+ 46+ 62+ 54+ 40+ 96+ 52+ 98+ 40+ 42+ 52+ 60}{15}\\ \bar{X}=\frac{822}{15}\\ \bar{X}=54.8

Mean is 54.8

To find the median we have to arrange the given data in ascending order as follows:

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

n = 15 (odd)

\\Median=(\frac{n+1}{2})^{th}\ term\\ Median=(\frac{15+1}{2})^{th}\ term\\ Median=8^{th}\ term\\ Median=52

In the given data 52 occurs the maximum number of times ()

Therefore, Mode = 52

Q4 Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Answer:

In the given data 14 is occuring the maximum number of times (4)

Mode of the given data is therefore 14.

Q5 Find the mean salary of 60 workers of 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

Answer:

Salary ( in Rs)(xi)

Number of workers(fi)

fixi

3000

16

48000

4000

12

48000

5000

10

50000

6000

8

48000

7000

6

42000

8000

4

32000

9000

3

27000

10000

1

10000

Total

\sum_{i=1}^{8}f_{i}=60

\sum_{i=1}^{8}f_{i}x_{i}=305000

 

The mean of the above data is given by

\\\bar{X}=\frac{\sum_{i=1}^{8}f_{i}x_{i}}{\sum_{i=1}^{8}f_{i}}\\ \bar{X}=\frac{305000}{60}\\ \bar{X}=5083.33

The mean salary of the workers working in the factory is Rs 5083.33

Q6 (i) Give one example of a situation in which the mean is an appropriate measure of central tendency.

Answer:

The mean is an appropriate measure of central tendency in case the observations are close to each other. An example of such a case is height of the students in a class.

Statistics Excercise: 14.4

Q6 (ii) Give one example of a situation in which the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Answer:

The mean is not an appropriate measure of central tendency in case the observations are not close to each other. An example of such a case is prices of the toys in a toy shop.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 14 Statistics?

  • Go through some tables and understand that representation of data.

  • Take a look through some examples to understand the calculation of mean, median, mode.

  • Memorize the formulas to calculate mean, median, mode.

  • Implement the application of all the concepts and formulas on the practice problems.

  • While practicing you can use NCERT solutions for class 9 maths chapter 14 Statistics.

Keep working hard and happy learning!

 

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