# NCERT Solutions for class 9 Maths Chapter 14- Statistics Exercise 14.3 (2023)

### Exercise 14.3

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 %)

 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.

Ans: The graph of the information presented above can be produced as follows by depicting causes on the x-axis and family fatality rate on the y-axis, and selecting an acceptable scale (1 unit = 5% for the y axis).

All the rectangle bars are of the same width and have equal spacing between them.

ii. Which condition is the major cause of women’s ill health and death worldwide?

Ans: Reproductive health issues are the leading cause of women's illness and mortality globally, affecting 31.8% of women.

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: The factors are as follows:

a. Lack of medical facilities

b. Lack of correct knowledge of treatment

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/STBackward districtsNon – backward districtsRuralUrban 940970920950920930910

i. Represent the information above by a bar graph.

Ans: The graph of the information presented above may be built by choosing an appropriate scale (1 unit = 100 girls for the y-axis) and representing section (variable) on the x-axis and number of girls per thousand boys on the y-axis.

Here, all the rectangle bars are of the same length and have equal spacing in between them.

ii. In the classroom discuss what conclusions can be arrived at from the graph.

Ans: The largest number of females per thousand boys (i.e., 970) is found in ST, while the lowest number of girls per thousand boys (i.e., 910) is found in urban areas.

In addition, the number of females per thousand boys is higher in rural regions than in cities, in backward districts than in non-backward districts, and in SC and ST districts than in non-SC/ST districts.

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.

Ans:

Here, all the rectangle bars are of the same length and have equal spacing in between them.

ii. Which political party won the maximum number of seats?

Ans: From the above graph it is clear that Political party ‘A’ won the maximum number of seats.

4. 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 117.5-126.5 3 126.5-135.5 5 135.5-144.5 9 144.5-153.5 12 135.5-162.5 5 162.5-171.5 4 171.5-180.5 2

i. Draw a histogram to represent the given data.

Ans: The length of leaves is represented in a discontinuous class interval with a difference of $1$ between them, as can be seen. To make the class intervals continuous, $\dfrac{1}{2} = 0.5$ must be added to each upper class limit and $0.5$ must be subtracted from the lower class limits.

 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 12 135.5-162.5 5 162.5-171.5 4 171.5-180.5 2

The above histogram may be built using the length of leaves on the x-axis and the number of leaves on the y-axis.

On the y-axis, one unit symbolises two leaves.

ii. Is there any other suitable graphical representation for the same data?

Ans: Frequency polygon is another good graphical representation of this data.

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

Ans: No, because the maximum number of leaves (i.e.$12$) has a length of $144.5{\text{mm}}$ to $153.5{\text{mm}}$ It is not necessary for all of them to be $153{\text{mm}}$long.

5. The following table gives the life times of neon lamps:

 Length (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.

Ans: The histogram of the given data may be produced by plotting the life duration (in hours) of neon lamps on the x-axis and the number of lamps on the y-axis. Here,1

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

ii. How many lamps have a lifetime of more than $700$ hours?

Ans: It may be deduced that the number of neon lamps with a lifetime more than $700$is equal to the sum of the numbers of neon lamps with lifetimes of $700,800$and $900$. As a result, there are $184$ neon bulbs with a lifetime of more than $700$ hours $(74 + 62 + 48 = 184)$.

6. The following table gives the distribution of students of two sections according to the mark 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: We can find the class marks of the given class intervals by using the following formula.

${\text{Class mark = }}\dfrac{{{\text{Upper class limit + Lower class limit}}}}{2}$

 Section A Section B Marks Class Marks Frequency Marks Class Marks 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 polygon can be constructed as follows, with class markings on the x-axis and frequency on the y-axis, and an appropriate scale $(1{\text{ unit = 3 for the y - axis}})$.

It can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.

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 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.)

Ans: As it can be seen data is not continuous, and the difference in upper limit and

lower limit is 1, so to make class interval continuous 0.5 needed to be added in each

limit.

Class Mark=$(\frac{\text{Upper Limit + Lower Limit} }{\text{2}})$

 No. 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

A frequency polygon can be created by plotting class grades on the x-axis and running times on the y-axis.

8. A random survey of the number of children of various age groups playing in 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.

Ans:

 Age (in years) Frequency (Number of Children) Width of Class Length of Rectangle 1-2 5 1 $\dfrac{{5 \times 1}}{1} = 5$ 2-3 3 1 $\dfrac{{3 \times 1}}{1} = 3$ 3-5 6 2 $\dfrac{{6 \times 1}}{2} = 3$ 5-7 12 2 $\dfrac{{12 \times 1}}{2} = 6$ 7-10 9 3 $\dfrac{{9 \times 1}}{3} = 3$ 10-15 10 5 $\dfrac{{10 \times 1}}{5} = 2$ 15-17 4 2 $\dfrac{{4 \times 1}}{2} = 2$

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 1-44-66-88-1212-20 63044164

i. Draw a histogram to depict the given information.

Ans:

 Number of Letters Frequency (Number of Surnames) Width of Class Length of Rectangle 1-4 6 3 $\dfrac{{6 \times 2}}{3} = 4$ 4-6 30 2 $\dfrac{{30 \times 2}}{2} = 30$ 6-8 44 2 $\dfrac{{44 \times 2}}{2} = 44$ 8-12 16 4 $\dfrac{{16 \times 2}}{4} = 8$ 12-20 4 8 $\dfrac{{4 \times 2}}{8} = 1$

The histogram can be generated using the number of letters on the x-axis and the fraction of the number of surnames per 2 letters interval on the y-axis, as well as an acceptable scale (1 unit = 4 students for the y axis).

ii. Write the class interval in which the maximum number of surnames lie.

Ans: The maximum number of surnames in the class interval is 6-8 since it contains 44 surnames, which is the maximum for this data.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

### NCERT Solution Class 9 Maths of Chapter 14 All Exercises

 Chapter 14 - Statistics Exercises in PDF Format Exercise 14.1 2 Questions & Solutions (2 Short Answers) Exercise 14.2 9 Questions & Solutions (9 Long Answers) Exercise 14.3 9 Questions & Solutions (6 Short Answers, 3 Long Answers) Exercise 14.4 6 Questions & Solutions (2 Short Answers, 4 Long Answers)

## NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.3 Free PDF

The Exercise 14.3 Class 9 NCERT Solutions are also available in PDF format on the official website of Vedantu. It is free of cost to download so all the students can avail of it. Students can store either a physical copy or a soft copy for future use. It helps the students for combined studies and can practice in their free time. Also, they can use it during the time of examinations.

### Chapter 14 Maths Class 9 Exercise 14.3

Exercise 14.3 Class 9 NCERT Solutions has 9 sums. Each sum has been explained below.

### Exercise 14.3 Class 9 NCERT Solutions Question 1

The first question of exercise 14.3 is about a survey of causes for women's death. A set of three questions were given to solve using the data. It is very easy for students to understand and answer. It is a short answer type of question. Students can show it on the graph by taking values on the x-axis and y-axis.

### 14.3 Maths Class 9 Question 2

The next question of Ex 14.3 Class 9 is data about the percentage of girls and boys in various castes. Based on the tabular data, students need to represent in a bar graph and need to write conclusions by finding cumulative frequency. Then it is necessary to find the median by putting values.

### Chapter 14 Exercise 14.3 Question 3

The third question of 14.3 Class 9 has given the data about political parties and their seats. Students are asked to represent the data using a bar graph. And the next question is to find the majority of seats the party has won and which party it is? This can be found using the mean value. Students can understand and substitute the formula of the mean using the upper limit and lower limit.

### Chapter 14.3 Maths Class 9 Question 4

The next question of Exercise 14.3 Chapter 14 Class 9 is a long answer type question. Three questions were given under the data. By referring to NCERT Solutions Class 9 Maths pdf, students can easily convert the data into continuous and answer the remaining in detail.

### Class 9 Maths Chapter 14 Exercise 14.3 Question 5

This question is simple and straightforward. Students can answer it by understanding the formula itself. By practising previous papers in the PDF, students can solve these kinds of sums within minutes.

### Ex 14.3 Class 9 Question 6

The sixth question of ex 14.3 Maths Class 9 is about the representation of polygon. Students need to find cumulative frequency first. Then finding the median is helpful to move further. Now calculate the class mark and prepare a table. From that table, students need to find the mode using the formula.

### Ex 14.3 Class 9 Solutions Question 7

It is a similar question to represent the data using a polygon. As the given data is continuous, it is easy to find the cumulative frequency, and thereby they can represent the data easily. Even though these sums are easy to solve, students should be attentive while copying data. Otherwise, the answer will be wrong.

### Class 9 Maths Chapter 14 Exercise 14.3 Question 8

It is also a short answer type question for which the data of children who are playing in the park was given. As continuous data with equal intervals were provided, it is easy to calculate the values and represent them in the histogram.

### Chapter 14 Exercise 14.3 Question 9

The last question of Exercise 14.3 Class 9 NCERT Solutions is the data taken from the telephone directory. Two questions were followed by the data for the students to answer. The first is to represent the data into a histogram and then finding the greatest interval.

For solving all these questions, Class 9 Maths Chapter 14 Exercise 14.3 will help the students to get practised and understand the logic.

### Key Takeaways of NCERT Solutions Class 9 Maths Chapter 14 Exercise 14.3 Free PDF

As the pdf provides several benefits, let see some of the takeaways are as follows:

• Students can store either soft copy or hard copy.

• The material has both solved and unsolved examples.

• The updated versions were uploaded after thorough verification.

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