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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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A histogram is a graphical representation of a frequency distribution in the form of rectangles. If class intervals are continuous, rectangles are adjacent. The width of each rectangle equals the class size hh, and the area is proportional to the frequency. For unequal class intervals, height is proportional to Frequency Density.

Histogram showing adjacent rectangles for continuous class intervals
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A Frequency Polygon is formed by joining the midpoints of the tops of the rectangles in a histogram with straight lines. To complete the polygon at both ends, it is extended to the class marks of the imaginary classes with zero frequency preceding the first class and following the last class.

Frequency polygon formed by connecting midpoints of class intervals
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An Ogive (Cumulative Frequency Curve) is a graph representing cumulative frequency. In a 'Less Than' ogive, points are plotted with Upper Class Limits as xx-coordinates and Cumulative Frequencies as yy-coordinates. The curve is always non-decreasing and takes an S-shape.

Less Than Ogive curve rising from left to right
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For discontinuous class intervals (e.g., 10−19,20−2910-19, 20-29), an adjustment factor dd must be calculated to make them continuous (9.5−19.5,19.5−29.59.5-19.5, 19.5-29.5) before drawing a histogram or ogive. This ensures no gaps exist between the boundaries of consecutive classes.

Visualizing the gap between discontinuous classes and the adjustment point
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A 'Kink' or Zig-Zag line is used on the horizontal axis (xx-axis) if the first class interval does not start from zero. This indicates that the scale has been broken to skip the empty region between the origin and the first data point.

Zig-zag line on x-axis representing a scale break

📐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.

Problem 3:

Construct a histogram for the following data representing weights of students: 40−4540-45 kg: 4 students, 45−5045-50 kg: 12 students, 50−5550-55 kg: 8 students, 55−6055-60 kg: 6 students.

Histogram for student weights with heights 4, 12, 8, and 6.

Solution:

  1. Plot Weight on the x-axis starting from 40 (use a kink/zigzag line if necessary).
  2. Plot Number of Students on the y-axis.
  3. Draw rectangles with heights corresponding to the frequencies. Rectangle heights: 40−4540-45 is 4, 45−5045-50 is 12, 50−5550-55 is 8, 55−6055-60 is 6.

Explanation:

Since the class intervals are continuous and have the same width, the area of each rectangle is proportional to its frequency.

Problem 4:

Represent the following data using a frequency polygon: Class 0−100-10 (Freq: 5), 10−2010-20 (Freq: 15), 20−3020-30 (Freq: 10), 30−4030-40 (Freq: 5).

Frequency polygon starting at -5 and ending at 45.

Solution:

  1. Find class marks: 5,15,25,355, 15, 25, 35.
  2. Points to plot: (5,5),(15,15),(25,10),(35,5)(5, 5), (15, 15), (25, 10), (35, 5).
  3. Include imaginary classes: (−5,0)(-5, 0) and (45,0)(45, 0).
  4. Connect points with straight lines.

Explanation:

The frequency polygon is a closed figure, so we connect the ends to the horizontal axis at the mid-points of the preceding and succeeding empty classes.

Graphical Representation (Histograms, Frequency Polygon, Ogive) Class 9 Notes & Examples