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.
(i) The required graphical representation is shown as follows:
(ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) Two factors may be un education and poor background.
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.
(i) The required bar graph is shown below:
(ii) We conclude that number of girls per thousand boys are maximum in scheduled tribe section whereas minimum in urban section.
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?
(i) The required bar graph is shown below:
(ii) The political party A won the maximum number of seats.
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?
(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 is a ‘frequency polygon’.
(iii) No, it is not a correct statement. The maximum number of leaves lie in the class interval 145 – 153.
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?
(i) The required histogram is shown below:
(ii) Number of lamps having life time of more than 700 hours = 74 + 62 + 48 = 184.
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.
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 more students of section A have secured good
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.
The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.
Now, the required frequency polygons are as shown below:
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.
Here, the class sizes are different. So, we calculate the adjusted frequencies corresponding to each rectangle i.e., length of the rectangle.
Adjusted frequency or length of the rectangle
Here, the minimum class size = 2 – 1 = 1
∴ We have the following table for adjusted frequencies or length of rectangles:
Now, the required histogram is shown below:
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.
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class size = 6 – 4 = 2.
Therefore, we have the following table for length of rectangles.
The required histogram is shown below:
(ii) The maximum frequency is 44, which is corresponding to the class interval 6 – 8.
∴ Maximum number of surnames lie in the class interval 6 – 8.