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Statistics - Graphical Representation (Histograms, Frequency Polygon, Ogive)

Grade 9ICSE

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

🔑Concepts

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Graphical representation of statistical data involves translating numerical frequency distributions into visual forms like Histograms, Frequency Polygons, and Ogives to easily interpret trends and central tendencies. These visual tools help identify the mode and median of a dataset quickly.

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A Histogram is a set of adjacent rectangles where the base represents the class interval on the xx-axis and the height represents the frequency on the yy-axis. If the data is in inclusive form (e.g., 10−19,20−2910-19, 20-29), it must be converted to exclusive form (e.g., 9.5−19.5,19.5−29.59.5-19.5, 19.5-29.5) so that the rectangular bars touch each other without any gaps. If the first class does not start at zero, a 'kink' or zig-zag line is drawn on the xx-axis near the origin.

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A Frequency Polygon is a line graph used to represent the frequency distribution of a dataset. It is constructed by joining the mid-points of the top edges of the rectangles in a histogram with straight line segments. Visually, it looks like a mountain peak or a series of connected slopes. To complete the polygon and enclose the area, the line is extended to the xx-axis at the mid-points of the imaginary class intervals before the first class and after the last class.

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Construction of a Frequency Polygon without a Histogram involves calculating the Class Mark for each interval. These class marks are plotted on the xx-axis and the corresponding frequencies on the yy-axis. The points are then connected using a ruler to form a continuous jagged line that starts and ends on the horizontal axis.

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The Cumulative Frequency Curve or Ogive is a smooth free-hand curve representing cumulative frequency. In a 'Less Than' Ogive, we plot the upper limits of class intervals against their corresponding cumulative frequencies. The curve starts from the lower limit of the first class on the xx-axis and rises continuously toward the right, typically forming an 'S' shape. This curve is particularly useful for finding the median and quartiles of the data.

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The Class Mark (or Mid-value) is the central value of a class interval, representing the entire group for calculations and plotting polygons. It is visually located exactly in the middle of a class interval on the horizontal scale of a graph.

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Adjustment Factor calculation is necessary when dealing with discontinuous (inclusive) classes. By subtracting half the difference between two consecutive classes from the lower limit and adding it to the upper limit, we create a continuous boundary. This ensures that the bars of a histogram are perfectly adjacent and that the Ogive is plotted accurately.

📐Formulae

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

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

Adjustment Factor(d)=12(Lower limit of one class−Upper limit of the previous class)\text{Adjustment Factor} (d) = \frac{1}{2}(\text{Lower limit of one class} - \text{Upper limit of the previous class})

Adjusted Lower Limit=Lower Limit−d\text{Adjusted Lower Limit} = \text{Lower Limit} - d

Adjusted Upper Limit=Upper Limit+d\text{Adjusted Upper Limit} = \text{Upper Limit} + d

Frequency Density (for unequal classes)=Frequency of the classWidth of the class\text{Frequency Density (for unequal classes)} = \frac{\text{Frequency of the class}}{\text{Width of the class}}

💡Examples

Problem 1:

Given the following frequency distribution, calculate the class marks and draw a frequency polygon without using a histogram: Class Intervals: 10−20,20−30,30−40,40−5010-20, 20-30, 30-40, 40-50 Frequencies: 5,12,8,45, 12, 8, 4

Solution:

Step 1: Calculate the Class Marks for each interval. For 10−2010-20: x1=10+202=15x_1 = \frac{10+20}{2} = 15 For 20−3020-30: x2=20+302=25x_2 = \frac{20+30}{2} = 25 For 30−4030-40: x3=30+402=35x_3 = \frac{30+40}{2} = 35 For 40−5040-50: x4=40+502=45x_4 = \frac{40+50}{2} = 45

Step 2: Identify the points to plot as (x,f)(x, f): P1(15,5),P2(25,12),P3(35,8),P4(45,4)P_1(15, 5), P_2(25, 12), P_3(35, 8), P_4(45, 4)

Step 3: To close the polygon, find mid-points of preceding and succeeding classes: Preceding: 0−100-10 mid-point is 55. Point P0(5,0)P_0(5, 0) Succeeding: 50−6050-60 mid-point is 5555. Point P5(55,0)P_5(55, 0)

Step 4: Plot points P0P_0 to P5P_5 on a graph and connect them with straight lines.

Explanation:

To draw a frequency polygon without a histogram, the class marks are treated as the xx-coordinates. Closing the polygon by extending it to the xx-axis ensures the total area under the polygon remains equivalent to the area of the corresponding histogram.

Problem 2:

Construct a 'Less Than' Ogive for the following data: Marks: 0−10,10−20,20−30,30−400-10, 10-20, 20-30, 30-40 Frequency: 3,7,10,53, 7, 10, 5

Solution:

Step 1: Construct the Cumulative Frequency (CF) table. Marks <10< 10: CF=3CF = 3 Marks <20< 20: CF=3+7=10CF = 3 + 7 = 10 Marks <30< 30: CF=10+10=20CF = 10 + 10 = 20 Marks <40< 40: CF=20+5=25CF = 20 + 5 = 25

Step 2: Identify the coordinates (UpperLimit,CF)(Upper Limit, CF) to plot: (10,3),(20,10),(30,20),(40,25)(10, 3), (20, 10), (30, 20), (40, 25)

Step 3: Also include the point where CF is 0 at the lower limit of the first class: (0,0)(0, 0).

Step 4: Plot these points on a graph where the xx-axis is 'Marks' and the yy-axis is 'Cumulative Frequency'. Connect the points with a smooth, free-hand curve.

Explanation:

An Ogive represents the running total of frequencies. By plotting the upper limit against the cumulative frequency, we show how many observations fall below a certain value. Using a smooth curve instead of straight lines distinguishes the Ogive from a frequency polygon.