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Statistics - Graphical Representation of Data (Bar graphs, Histograms, Frequency Polygons)

Grade 9CBSE

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

🔑Concepts

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Bar Graphs: A bar graph is a pictorial representation of data using rectangular bars of uniform width, drawn either horizontally or vertically, with equal spacing between them. The length or height of each bar represents the frequency of the corresponding variable, making it ideal for discrete data or categorical variables.

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Histograms (Uniform Class Size): A histogram is a graphical representation of a grouped frequency distribution in the form of vertical rectangles. Unlike bar graphs, there are no gaps between the rectangles because the classes are continuous. The height of each rectangle corresponds to the frequency, and the base represents the class interval on the horizontal axis.

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Histograms (Varying Class Size): When class widths are not uniform, the area of the rectangles, rather than just the height, represents the frequency. To construct this, we calculate an adjusted frequency for each class so that the rectangles reflect the distribution accurately. Visually, some bars will appear wider than others, but their heights are scaled accordingly.

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Frequency Polygons: A frequency polygon is a line graph used to represent frequency distributions. It is formed by joining the mid-points of the upper sides of the rectangles in a histogram with straight line segments. To complete the polygon, we extend it to the x-axis by joining the mid-points of imaginary classes with zero frequency at both ends.

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Class Marks: For constructing a frequency polygon without a histogram, we use class marks. The class mark is the middle value of a class interval and is plotted on the horizontal axis against the frequency on the vertical axis. Visually, these points are then connected to show the trend and shape of the data distribution.

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The Kink (Zigzag Line): When the data on the horizontal axis does not start from zero (e.g., starting from a class like 140−150140-150), a kink or a broken line is drawn near the origin on the x-axis. This indicates that the scale is not continuous from zero to the first value shown.

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Data Continuity: For drawing histograms, the class intervals must be continuous (e.g., 10−20,20−3010-20, 20-30). If the given data is discontinuous (e.g., 10−19,20−2910-19, 20-29), it must be converted by subtracting 0.50.5 from the lower limit and adding 0.50.5 to the upper limit of each class.

📐Formulae

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

Adjusted Frequency of a Class=Frequency of the classWidth of the class×Minimum class width\text{Adjusted Frequency of a Class} = \frac{\text{Frequency of the class}}{\text{Width of the class}} \times \text{Minimum class width}

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

💡Examples

Problem 1:

Calculate the adjusted frequency for a class interval 30−4030-40 with frequency 1515, given that the minimum class width in the entire distribution is 55.

Solution:

Step 1: Identify the class width for the given interval. Width=40−30=10\text{Width} = 40 - 30 = 10. \nStep 2: Use the formula for adjusted frequency: Adjusted Frequency=FrequencyClass Width×Minimum Class Width\text{Adjusted Frequency} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width}. \nStep 3: Substitute the values: Adjusted Frequency=1510×5\text{Adjusted Frequency} = \frac{15}{10} \times 5. \nStep 4: Calculate the result: 1.5×5=7.51.5 \times 5 = 7.5.

Explanation:

In histograms with varying class widths, we adjust the height (frequency) to ensure the area of the rectangle is proportional to the actual frequency.

Problem 2:

Find the class marks for the following frequency distribution: 10−20,20−30,30−4010-20, 20-30, 30-40.

Solution:

Step 1: Use the formula Class Mark=Upper Limit+Lower Limit2\text{Class Mark} = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}. \nStep 2: For 10−2010-20: 10+202=302=15\frac{10 + 20}{2} = \frac{30}{2} = 15. \nStep 3: For 20−3020-30: 20+302=502=25\frac{20 + 30}{2} = \frac{50}{2} = 25. \nStep 4: For 30−4030-40: 30+402=702=35\frac{30 + 40}{2} = \frac{70}{2} = 35.

Explanation:

Class marks represent the central value of each class and are essential for plotting Frequency Polygons.