CBSE Class 9 Mathematics Graphical Distribution of Data Assignment Set 01

Question. The rainfall recorded in a city are as follows, in various years:

Depict this information in the form of bar graph.
Answer: To represent the recorded rainfall data on a bar graph, we plot the years along the horizontal axis (X-axis) and the rainfall (in cm) along the vertical axis (Y-axis). Let the scale on the vertical axis be \( 1 \text{ cm} = 1 \text{ unit} \) of rainfall. We construct vertical rectangular bars of equal width with uniform spacing between them.
In simple words: Draw a straight column for each year. The height of the columns represents the quantity of rainfall in centimeters.
Exam Tip: Be sure to keep the width of the bars and the spaces between them absolutely uniform to secure full marks.

Question. Draw a histogram for the following data:

Answer: The class intervals given are discontinuous. To construct a histogram, we first convert them into continuous class intervals by subtracting \( 0.5 \) from the lower limits and adding \( 0.5 \) to the upper limits. The adjusted intervals and frequencies are as follows:

In simple words: First, modify the class limits so there are no empty gaps between intervals. Next, draw rectangles directly touching each other, where the height represents the frequency.
Exam Tip: Make sure to draw a kink (zigzag line) on the X-axis because the intervals start from 24.5 instead of 0.

Question. Draw the histogram of the following data:

Answer: Since the class intervals have unequal widths, we must adjust the frequencies before drawing the histogram. The minimum class width is \( 10 \). We compute the adjusted frequency (or height of the rectangle) using the formula:
\[ \text{Adjusted Frequency} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width} \]
The calculations are shown in the table below:

In simple words: When interval sizes vary, we calculate a revised frequency height for each column. This keeps the total area of the blocks properly matched to the data.
Exam Tip: For unequal class widths, always adjust the frequencies based on the minimum class width before drawing the heights of the rectangles.

Question. The marks scored by the students in an examination are given in the form of a frequency distribution table as below:

Represent this data in the form of a frequency polygon after drawing the histogram.
Answer: To draw the frequency polygon, we first plot the histogram for the given classes. Then, we connect the mid-points of the top edges of the rectangles. To close the polygon, we extend it to the mid-points of the imaginary class intervals at both ends (class \( -10 - 0 \) with mid-point \( -5 \), and class \( 60 - 70 \) with mid-point \( 65 \)), both having zero frequency.
The mid-points (class marks) are:

In simple words: Draw the continuous bars first. Next, link the top center of each bar with straight lines, and extend both ends to the floor line to complete the shape.
Exam Tip: Never forget to close the frequency polygon at both ends by connecting it to the horizontal axis at the preceding and succeeding imaginary class marks.

Question. Draw a histogram and a frequency polygon of the marks secured by eighty students of a school in the final examination out of 400:

Answer: We are given the class marks (mid-points). The consecutive class marks have a difference of \( 10 \), so the class width is \( 10 \). We construct the continuous class intervals around these mid-points using \( \text{Class Mark} \pm 5 \). The converted values are:

To close the polygon, we include imaginary mid-points \( 295 \) and \( 365 \) with zero frequency at the ends.
In simple words: Convert the single midpoints back into normal intervals of width 10 first. Draw the histogram rectangles, and then connect their top center points together.
Exam Tip: Be extra careful to identify if class marks are given in descending order; always rearrange them in ascending order before performing calculations.

Question. Represent the following information in the form of a frequency curve:

Answer: The data is presented in a cumulative format. We first find the non-cumulative frequencies for each class interval and determine their mid-points:

A frequency curve is obtained by joining the plotted mid-points of class intervals with a smooth, freehand curve. We close the curve by extending it to the horizontal axis at the preceding mid-point \( -5 \) and succeeding mid-point \( 55 \) with zero frequency.
In simple words: Convert the cumulative data to direct interval frequencies first. Plot the midpoints on a grid, then use a smooth freehand line to trace the path without straight angles.
Exam Tip: A frequency curve must be drawn smooth and freehand, whereas a frequency polygon is drawn using straight line segments with a ruler.

Question. Construct an ogive for the following data:

Answer: We construct a "Less Than" cumulative frequency distribution to draw the ogive:

We plot the points \( (2, 0) \), \( (5, 3) \), \( (8, 4) \), \( (11, 9) \), \( (14, 17) \), \( (17, 19) \), and \( (20, 20) \) and connect them smoothly.
In simple words: Create a running total of the frequencies. Point by point, plot the upper boundary of each class with its cumulative total, then join them with an S-shaped curve.
Exam Tip: An ogive represents cumulative frequencies. Plot the cumulative values against the upper limits of the classes, not the class marks.

Question. Draw an ogive and a frequency polygon for the following data of 300 patients treated in a hospital on a certain day:

Answer: We rearrange the classes in ascending order and construct both tables for the frequency polygon and ogive (cumulative frequency).

1. Less than Ogive Curve: we plot the points \( (10, 0) \), \( (20, 40) \), \( (30, 70) \), \( (40, 115) \), \( (50, 150) \), \( (60, 220) \), and \( (70, 300) \).
2. Frequency Polygon: we plot class marks with their standard frequency values. To close the ends, we use preceding mid-point \( 5 \) and succeeding mid-point \( 75 \) (both at zero frequency). In simple words: First, rewrite the categories from youngest to oldest. Construct the ogive using accumulated totals, and construct the polygon using the midpoints of each age group.
Exam Tip: Always verify that the sum of the individual class frequencies equals the final cumulative frequency value (300 in this case) to avoid errors.

Question. The following are the scores of two groups of a certain class in a test given below. Construct a frequency polygon for each of these two groups on the same axes.

Answer: We plot the class marks on the X-axis and the frequencies on the Y-axis. To draw the polygons, we also include the preceding class mark \( 30 \) and succeeding class mark \( 54 \), both with zero frequency.

In simple words: Draw two different lines on the same chart. Use a solid line for Group A and a dashed line for Group B, plotting the midpoints against their test scores.
Exam Tip: Include a clear legend (color key) on the graph when plotting multiple frequency polygons together, as examiners award specific marks for this.

Question. Construct an ogive for the following distribution:

Answer: The given distribution is of the "More Than" type. To construct the ogive, we plot the lower limits on the X-axis and the cumulative frequencies on the Y-axis. The coordinates to plot are: \( (0, 100) \), \( (10, 96) \), \( (20, 78) \), \( (30, 68) \), \( (40, 26) \), and \( (50, 6) \). We extend the curve to the next limit \( (60, 0) \) to complete it on the axis.
In simple words: Unlike the "less than" ogive, this graph goes downwards. It starts high at the top left and moves lower to the bottom right because fewer students get higher marks.
Exam Tip: A "more than" ogive always slopes downwards from top-left to bottom-right, whereas a "less than" ogive slopes upwards from bottom-left to top-right.