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Statistics - Grouped Data and Histograms

Grade 10IGCSE

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

🔑Concepts

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Continuous vs Discrete Data: Histograms are used for continuous data where there are no gaps between intervals.

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Class Midpoint: Used as an estimate for the value of all items within a specific class interval when calculating the mean.

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Modal Class: The interval with the highest frequency density (or highest frequency if widths are equal).

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Median Class: The interval that contains the middle value (the n/2n/2-th value).

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

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Area Principle: In a histogram, the area of the bar is proportional to the frequency (Frequency = Frequency Density × Class Width).

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Class Boundaries: Ensuring there are no gaps between classes (e.g., 10-20, 20-30) to represent continuous data accurately.

📐Formulae

Estimated Mean=∑(f×x)∑f\text{Estimated Mean} = \frac{\sum (f \times x)}{\sum f} (where xx is the midpoint)

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

Midpoint=Lower Bound+Upper Bound2\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}

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

💡Examples

Problem 1:

The table shows the heights of 20 plants. Calculate an estimate for the mean height.

  • 0<h≤100 < h \le 10: Frequency 4
  • 10<h≤3010 < h \le 30: Frequency 10
  • 30<h≤4030 < h \le 40: Frequency 6

Solution:

  1. Find midpoints (xx): 5, 20, 35.
  2. Calculate f×xf \times x: (4×5)+(10×20)+(6×35)=20+200+210=430(4 \times 5) + (10 \times 20) + (6 \times 35) = 20 + 200 + 210 = 430.
  3. Sum of frequencies (∑f\sum f): 4+10+6=204 + 10 + 6 = 20.
  4. Mean = 430/20=21.5430 / 20 = 21.5 cm.

Explanation:

Since we don't know the exact heights, we use the midpoint of each class as the best estimate for the values in that group.

Problem 2:

In a histogram, a class interval 20<x≤5020 < x \le 50 has a frequency of 15. Calculate the frequency density for this bar.

Solution:

  1. Class Width = 50−20=3050 - 20 = 30.
  2. Frequency = 15.
  3. Frequency Density = 15/30=0.515 / 30 = 0.5.

Explanation:

On a histogram, the y-axis represents Frequency Density. To find the height of the bar, divide the frequency by the width of the interval on the x-axis.

Problem 3:

On a histogram, the bar for the class 0<x≤50 < x \le 5 has a height of 4 units. The bar for the class 5<x≤155 < x \le 15 has a height of 2 units. Which class has a higher frequency?

Solution:

  1. Freq(0−50-5) = 5×4=205 \times 4 = 20.
  2. Freq(5−155-15) = 10×2=2010 \times 2 = 20.
  3. Both classes have the same frequency.

Explanation:

Frequency is the area of the bar (Width ×\times Height). Even though the first bar is taller, the second bar is wider, resulting in the same total area/frequency.