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Data Handling - Histogram

Grade 8CBSE

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

🔑Concepts

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A Histogram is a specialized graphical representation of grouped data where class intervals are shown on the horizontal xx-axis and frequencies are shown on the vertical yy-axis. Visually, it consists of a set of contiguous rectangles where the area of each rectangle is proportional to the frequency of the corresponding class interval.

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Unlike a Bar Graph, there are no gaps between the bars in a Histogram. This is because Histograms represent continuous data or grouped frequency distributions where the upper limit of one class coincides with the lower limit of the next class.

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If the data on the horizontal axis does not start from zero, a 'kink' or a 'broken line' (indicated by a zig-zag symbol ≈\approx near the origin) is used. This visual device indicates that the scale along the axis does not start from zero, preventing the graph from being unnecessarily stretched.

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The width of each rectangle in a Histogram corresponds to the Class Size (hh). In standard Grade 8 problems, the class size is usually kept uniform for all intervals, resulting in bars of equal width standing side-by-side.

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The height of each rectangle represents the frequency of that particular class interval. The higher the bar, the greater the number of observations falling within that specific range.

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Class intervals are defined by a Lower Limit and an Upper Limit. In a continuous distribution like 10−2010-20 and 20−3020-30, the observation exactly equal to the upper limit (2020) is typically included in the higher class (20−3020-30) rather than the lower one.

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The total area under the histogram represents the total number of observations in the data set. Visually, you can calculate the total frequency by summing the heights of all the rectangles if the widths are equal.

📐Formulae

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

Class Size (Width)=Upper Class Limit−Lower Class Limit\text{Class Size (Width)} = \text{Upper Class Limit} - \text{Lower Class Limit}

Frequency Density=Frequency of the ClassWidth of the Class\text{Frequency Density} = \frac{\text{Frequency of the Class}}{\text{Width of the Class}}

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

💡Examples

Problem 1:

The weights (in kg) of 3030 students of a class are as follows: 40−4540-45 (5 students), 45−5045-50 (8 students), 50−5550-55 (10 students), 55−6055-60 (7 students). Construct a frequency distribution table and describe how the histogram would look.

Solution:

  1. Create the table:
    • 40−4540-45: 55
    • 45−5045-50: 88
    • 50−5550-55: 1010
    • 55−6055-60: 77
  2. Since the data starts from 4040, we draw a kink (zig-zag line) on the xx-axis between 00 and 4040.
  3. Mark weights on the xx-axis with a scale of 1 unit=5 kg1 \text{ unit} = 5 \text{ kg}.
  4. Mark frequency on the yy-axis with a scale of 1 unit=2 students1 \text{ unit} = 2 \text{ students}.
  5. Draw adjacent rectangles: the first bar from 4040 to 4545 with height 55, the second from 4545 to 5050 with height 88, and so on.

Explanation:

This example shows how to handle data that doesn't start from zero by using a kink and how to translate a frequency table into rectangle dimensions for a histogram.

Problem 2:

In a histogram representing the marks of students, the class intervals are 0−10,10−20,20−30,30−400-10, 10-20, 20-30, 30-40. The heights of the bars are 4,7,12,4, 7, 12, and 55 respectively. Find: (i) The class size (ii) The total number of students who scored 2020 or more marks.

Solution:

  1. The class size is calculated as: Upper Limit−Lower Limit=10−0=10\text{Upper Limit} - \text{Lower Limit} = 10 - 0 = 10.
  2. To find students who scored 2020 or more marks, we look at the class intervals 20−3020-30 and 30−4030-40.
  3. Frequency of 20−30=1220-30 = 12.
  4. Frequency of 30−40=530-40 = 5.
  5. Total students = 12+5=1712 + 5 = 17.

Explanation:

This demonstrates how to interpret a histogram by reading the heights of bars as frequencies and summing them for specific ranges.