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Statistics - Histograms and Frequency Polygons

Grade 12IGCSE

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

🔑Concepts

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Histograms for Continuous Data: Unlike bar charts, histograms are used for continuous data and have no gaps between the bars.

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Frequency Density: In a histogram with unequal class widths, the height of the bar represents Frequency Density, not Frequency.

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Area-Frequency Relationship: The area of each bar in a histogram is proportional to the frequency of that class (Area=kimesFrequencyArea = k imes Frequency). Usually, k=1k=1.

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Class Boundaries: Adjustments made to class limits to ensure data is continuous (e.g., class 10-19 and 20-29 have a boundary at 19.5).

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Frequency Polygons: A line graph formed by connecting the midpoints of the tops of the histogram bars (or plotting frequency against midpoints).

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Interpreting Histograms: To find the number of items in a range, calculate the area of the bars or portions of bars within that range.

📐Formulae

Frequency Density=FrequencyClass Width\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}

Frequency=Frequency Density×Class Width\text{Frequency} = \text{Frequency Density} \times \text{Class Width}

Class Width=Upper Boundary−Lower Boundary\text{Class Width} = \text{Upper Boundary} - \text{Lower Boundary}

Midpoint=Lower Boundary+Upper Boundary2\text{Midpoint} = \frac{\text{Lower Boundary} + \text{Upper Boundary}}{2}

💡Examples

Problem 1:

A table shows the weights (ww) of 50 parcels. For the class 20<w≤3020 < w \leq 30, the frequency is 15. For the class 30<w≤5030 < w \leq 50, the frequency is 12. Calculate the frequency density for both classes.

Solution:

  1. For the first class (20<w≤3020 < w \leq 30): Class Width = 30−20=1030 - 20 = 10. Frequency Density = 15/10=1.515 / 10 = 1.5.
  2. For the second class (30<w≤5030 < w \leq 50): Class Width = 50−30=2050 - 30 = 20. Frequency Density = 12/20=0.612 / 20 = 0.6.

Explanation:

To find frequency density, divide frequency by the class width. Even though the second class has a higher frequency (12 vs 15 is not the case, but the width is larger), its height on the histogram will be lower (0.6) compared to the first class (1.5).

Problem 2:

In a histogram, the bar for the class 0<x≤50 < x \leq 5 has a height of 4 units. If the frequency of this class is 20, find the frequency of a class 5<x≤155 < x \leq 15 which has a height of 6 units on the same histogram.

Solution:

  1. Find the scale factor (kk): Area=Width×Height=5×4=20\text{Area} = \text{Width} \times \text{Height} = 5 \times 4 = 20. Since Area (20) = Frequency (20), k=1k=1.
  2. Calculate the area of the second bar: Width=15−5=10\text{Width} = 15 - 5 = 10. Height=6\text{Height} = 6. Area=10×6=60\text{Area} = 10 \times 6 = 60.
  3. Since k=1k=1, Frequency = 60.

Explanation:

In histograms, frequency is represented by area. By calculating the area of the known bar and comparing it to its frequency, we determine the relationship (Scale). We then apply that scale to the area of the second bar.

Problem 3:

To draw a frequency polygon for the data: 10−2010-20 (Freq: 5), 20−3020-30 (Freq: 8), 30−4030-40 (Freq: 3), what are the coordinates of the points to be plotted?

Solution:

Point 1: (10+202,5)=(15,5)(\frac{10+20}{2}, 5) = (15, 5). Point 2: (20+302,8)=(25,8)(\frac{20+30}{2}, 8) = (25, 8). Point 3: (30+402,3)=(35,3)(\frac{30+40}{2}, 3) = (35, 3).

Explanation:

A frequency polygon is plotted by using the midpoint of the class interval for the x-coordinate and the actual frequency for the y-coordinate. These points are then joined by straight lines.