NCERT solutions for class 9 Maths
Statistics
Chapter 14
Exercise 14.3
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) Graphical representation
(ii) Reproductive health conditions is the major cause of women’s ill health and death worldwide .
(iii) Two factors responsible.
- Lack of medical facilities.
- Lack of correct knowledge of treatment.
Ex 14.3 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) The bar graph :
(ii) We conclude that number of girls per thousand boys are maximum in scheduled tribe section whereas minimum in urban section.
Ex 14.3 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?
Solution:
(i) The required bar graph is shown below:
(ii) The political party A won the maximum number of seats.
Ex 14.3 Question 4.
The length of 40 leaves of a plant measured correct to one millimetre 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 number of leaves 153 mm long and Why?
Solution:
(i) The given frequency distribution table is not continuous. Therefore, first we have to modify it to be continuous distribution.
Thus, the modified frequency distribution table is:
Now, the required histogram of the frequency distribution is shown below :
(ii) Yes, other suitable graphical representation of the data is ‘frequency polygon’.
(iii) No, it is not a correct. The maximum number of leaves lie in the class interval 145 – 153.So, it is not necessary that all have their lengths as 153.
Ex 14.3 Question 5.
The following table gives the lifetimes of 400 neon lamps
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more 700 h?
Solution:
(i) The required histogram is shown below:
(ii) Number of lamps having life time of more than 700 hours = 74 + 62 + 48 = 184.
Ex 14.3 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:
To draw a frequency polygon, we mark the class marks along x-axis. Therefore, the modified table is:
So, the two frequency polygons are as shown below:
From the above frequency polygon, we can see that the performence of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.
Ex 14.3 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.
Solution:
It can be observed that the class intervals of given data are not continuous. Therefore, we first modify the distribution as continuous.
Now, the required frequency polygons are as shown below:
Ex 14.3 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:
Here, it can be observed that the data has class intervals of varying width. The propertion of children per 1 year interval can be calculated as follows:
Now, the required histogram is shown below:
Ex 14.3 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) Since, the given frequency of class intervals of distribution are unequal, and the minimum class size = 6 – 4 = 2.
The required histogram is shown below:
(ii) The class interval in which the maximum number of surnames lies is 6 – 8 as it has 44 surnames in it. i.e. the maximum for this data.
