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How Quantities Combine: Understanding Data - Visualising and Interpreting Data

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Data can be organized and presented visually using Bar Graphs, where the height of bars represents the frequency of discrete categories. The bars are of equal width with uniform spacing between them.

A standard bar graph showing four bars of varying heights representing frequencies of different categories.
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Histograms are used for continuous data grouped into class intervals. Unlike bar graphs, there are no gaps between the rectangles, and the area of each rectangle is proportional to its frequency. For equal class widths, height is proportional to frequency.

A histogram with continuous rectangular bars showing distribution across class intervals.
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A Frequency Polygon is a line graph formed by joining the mid-points (class marks) of the upper sides of the rectangles in a histogram. It can also be drawn independently by plotting class marks against frequencies.

A frequency polygon formed by connecting mid-points of class intervals with straight lines.
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When class widths are unequal in a histogram, the heights of the rectangles must be adjusted so that the area (not just height) is proportional to the frequency. The adjusted frequency is calculated relative to the minimum class width.

Histogram bars with different widths, showing how frequency density affects height.

📐Formulae

Class Mark=Upper Limit+Lower Limit2\text{Class Mark} = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}

Adjusted Frequency=FrequencyClass Width×Minimum Class Width\text{Adjusted Frequency} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width}

Range=Maximum Value−Minimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value}

💡Examples

Problem 1:

Given the class interval 20−3020 - 30 with a frequency of 1515, and the minimum class width in the entire data set being 55, calculate the length (height) of the rectangle for a histogram with unequal class widths.

Diagram showing a histogram bar representing class 20-30 with an adjusted height of 7.5 units.

Solution:

Class Width=30−20=10\text{Class Width} = 30 - 20 = 10 Minimum Class Width=5\text{Minimum Class Width} = 5 Frequency=15\text{Frequency} = 15 Length of Rectangle=1510×5=1.5×5=7.5\text{Length of Rectangle} = \frac{15}{10} \times 5 = 1.5 \times 5 = 7.5

Explanation:

To maintain area proportionality in a histogram with varying widths, we calculate adjusted frequency using the ratio of the current frequency to its width, scaled by the minimum width found in the distribution.

Problem 2:

Represent the following data using a frequency polygon: Class marks 5,15,25,355, 15, 25, 35 with corresponding frequencies 4,10,8,64, 10, 8, 6.

A frequency polygon plotted through points representing class marks 5, 15, 25, and 35.

Solution:

  1. Plot points (5,4),(15,10),(25,8),(35,6)(5, 4), (15, 10), (25, 8), (35, 6).
  2. To close the polygon, assume classes before and after with zero frequency: (−5,0)(-5, 0) and (45,0)(45, 0).
  3. Connect points with straight lines.

Explanation:

Points are plotted using the class mark as the xx-coordinate and the frequency as the yy-coordinate. The polygon is 'closed' by connecting it to the horizontal axis at the midpoints of the preceding and succeeding imaginary classes.