Exercise 12.1 Solutions (Class 9)

(i) Represent the information given above graphically.
A bar graph is the most suitable representation for this data because we are comparing the values of discrete categories (the different causes).
Construction Steps:
1. Draw two perpendicular axes. On the horizontal axis (x-axis), represent the ‘Causes’.
2. On the vertical axis (y-axis), represent the ‘Female fatality rate (%)’ with an appropriate scale (e.g., 1 cm = 5%).
3. Draw rectangular bars of equal width for each cause, leaving equal spacing between them. The height of each bar will correspond to its percentage value.

(ii) Which condition is the major cause of women’s ill health and death worldwide?
By observing the data table, the highest percentage is 31.8%, which corresponds to ‘Reproductive health conditions’.

Answer: Reproductive health conditions.

(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.
Two major factors contributing to high fatality rates from reproductive health conditions include:
1. Lack of Medical Facilities: Inadequate access to hospitals, skilled doctors, and proper medical care, especially during childbirth, in many parts of the world.
2. Lack of Awareness and Education: Insufficient knowledge about health, hygiene, disease prevention, and family planning can lead to preventable complications.

(i) Represent the information above by a bar graph.
A bar graph is appropriate to compare the number of girls across different discrete sections of society.
Construction Steps:
1. On the x-axis, represent the ‘Section’.
2. On the y-axis, represent the ‘Number of girls per thousand boys’. A scale starting from 900 can be used to better highlight the differences.
3. Draw bars of equal width for each section, with the height corresponding to its value.

(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Conclusions:
1. The girl-to-boy ratio is highest in the Scheduled Tribe (ST) section (970).
2. The girl-to-boy ratio is lowest in the Urban section (910).
3. The number of girls per thousand boys is less than 1000 in all sections, which points towards a gender imbalance in society.

(i) Draw a bar graph to represent the polling results.
A bar graph is suitable for comparing the seats won by each distinct political party.
Construction Steps:
1. Represent ‘Political Party’ on the x-axis.
2. Represent ‘Seats Won’ on the y-axis.
3. Draw bars of equal width for each party with height corresponding to the seats won.

(ii) Which political party won the maximum number of seats?
By observing the table, the maximum value is 75.

Answer: Political Party A won the maximum number of seats.

(i) Draw a histogram to represent the given data.
The given class intervals are discontinuous. To draw a histogram, we must make them continuous. We find the gap between consecutive classes (e.g., 127 – 126 = 1), take half of it (0.5), and adjust the limits.
Adjusted Continuous Class Intervals:

Now, a histogram can be drawn using the continuous class intervals on the x-axis and the number of leaves on the y-axis. The bars will be adjacent to each other.

(ii) Is there any other suitable graphical representation for the same data?
Yes, a Frequency Polygon is another suitable representation for continuous grouped data. It can be drawn by plotting the class marks against the frequencies and joining the points.

(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
No, this conclusion is incorrect. The class interval 145-153 (or 144.5-153.5) has the maximum frequency (12). This means the maximum number of leaves have lengths that lie *within this range*, not that they are all exactly 153 mm long.

(i) Represent the given information with the help of a histogram.
The class intervals are continuous and have equal width. A histogram is drawn with ‘Life time (in hours)’ on the x-axis and ‘Number of lamps’ on the y-axis. The bars will be adjacent.

(ii) How many lamps have a life time of more than 700 hours?
We sum the frequencies for the classes 700-800, 800-900, and 900-1000.
Number of lamps = 74 + 62 + 48 = 184.

Answer: 184 lamps have a lifetime of more than 700 hours.

To draw frequency polygons, we need the class marks (mid-points).

Construction: Plot the points (Class Mark, Frequency) for both sections on the same graph using different colors or line styles. Join the points for each section with straight lines to form two frequency polygons.

Comparison: From the graph, we can observe that Section B has a higher peak (more students scoring in the 10-20 range), while Section A has more students scoring in the 20-30 and 30-40 ranges. This suggests that while Section B had more students in one particular low-scoring bracket, Section A’s performance was generally better and more consistent in the higher mark ranges.

The data has discontinuous class intervals. First, we make them continuous and find the class marks.

(Table completed for all rows). Now, plot the points (Class Mark, Frequency) for both teams on the same graph and join them with line segments to form the frequency polygons.

The class intervals are of varying widths. The heights of the rectangles in the histogram must be adjusted. The adjusted height (or frequency density) is calculated as (Frequency / Class Width) × (Minimum Class Width).

Now, a histogram is drawn where the x-axis represents the ‘Age’ and the y-axis represents the ‘Proportion of children per 1 year interval’ (the adjusted height). The width of each bar corresponds to its original class width.

(i) Draw a histogram.
The class intervals have varying widths, so we must adjust the heights.

(Note: Adjusted height is scaled by the minimum class width, 2, for easier graphing). A histogram is drawn using these adjusted heights.

(ii) Write the class interval in which the maximum number of surnames lie.
Looking at the original frequency table, the highest frequency is 44.

Answer: The class interval with the maximum number of surnames is 6 – 8.