Students can Download Class 9 Maths Chapter 14 Statistics Ex 14.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 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) Scale: x axis ➝ 1 horizontal axis = 1 cm.

y axis ➝ 5% = 1 cm.

(ii) Reproductive health conditions.

(iii) Nerve and skin.

Question 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.

(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) Scale: x axis ➝ 1 horizontal = 1 cm.

y axis ➝ 100 girls = 1 cm.

(ii) After verification we came to know that Strength of Girls is more in ST section and it is less in Urban section.

Question 3.

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 ?

(ii) A Political Party won the maximum number of seats.

Solution:

(i) Scale: x axis ➝ 1 horizontal(Party) = 1 cm.

y axis ➝ 10 Parties = 1 cm.

(ii) ‘A’ political Party won the maximum number of seats.

Question 4.

The length of 40 leaves of a plant are measured correct 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 are 153 mm long ? Why ?

Solution:

(i) Scale: x axis ➝ 1 Class interval = 1 cm.

y axis ➝ 2 leaves = 1 cm.

(ii) Frequency Polygon.

(iii) No. Because from 145 to 153, length of leaves is 153.

Question 5.

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 ?

Solution:

(i) Scale: x axis ➝ 1 Class interval = 1 cm.

y axis ➝ 10 Bulbs = 1 cm.

(ii) 184 lamps have a life time of more than 700 hours.

Question 6.

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.

Solution:

(i) Scale: x axis ➝ 5 marks = 1 cm.

y axis ➝ 2 frequency = 1 cm.

Let us take class màrks on X-axis and frequencies on Y-axis. To plot frequency polygon of Section-A, we plot the points (5, 3), (15, 9). (25, 17), (35. 12), (45, 9) and join these points by line segments. To plot frequency polygon of Section-B, we plot the points (5, 5), (15,.19), (25, 15),(35. 1O)(45, 1)on the same scale and join these poil”lts by dotted line segments.

Section B students scored less marks in great numbers.

Question 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 the teams on the same graph by frequency polygons.

(Hint: First make the class intervals continuous.)

Solution:

Class interval is continuous,

1 – 6, 7 – 12 Here the difference is 1.

∴ \(\frac{1}{2}\) = 0.5 is lower limit is subtracted,

0.5 is taken upper limit, it becomes

0.5 – 6.5

6.5 – 12.5.

Let us represent class marks on X-axis and frequencies on Y-axis on a suitable scale. To get the frequency polygon of team A, plot the points (3.5,2), (9.5,1), (15.5,8). (21.5,9), (27.5,4), (33.5,5), (39.5,6), (45.5,10), (5 1.5,6) and (57.5,2) and join the points by the line segments.

To get the frequency polygon of team B, plot the points (3.5,5), (9.5,6), (15.5,2), (21.5, 10), (27.5,5). (33.5,6), (39.5,3), (45.5,4), (51 .5,8) and (57.5.10) and join the points by dotcd line segnients.

Question 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:

Scale: x-axis ➝ 1 year = 1 cm.

y – axis ➝ 1 child = 1 cm.

Let us represent the class-intervals along X-axis and corresponding adjusted frequencies on Y-axis on a suitable scale.

Now, draw rectangles with the class-intervals as bases and the corresponding adjusted frequencies as the heights.

Therefore, the required histogram is as given below:

Question 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) In this question, the class sizes are different. So, calculate the adjusted frequency for each class by using the formula:

Here, the minimum class size 6 – 4 = 2.

Therefore, the adjusted frequency table is as under

Let us represent the class-intervals along X-axis and corresponding adjusted frequency on Y-axis on suitable scale.

Now draw rectangles with the class-intervals as bases and the corresponding adjusted frequency the heights. The required histogram is as under:

(ii) Class interval which has maximum number of surnames is: 6 – 8.