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Statistics - Histograms, Bar Charts, and Pie Charts

Grade 9IGCSE

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

🔑Concepts

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Categorical vs. Continuous Data: Bar charts and pie charts are typically used for categorical or discrete data, while histograms are used for continuous data.

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Bar Charts: Bars have equal widths and gaps between them. The height represents the frequency.

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Pie Charts: Used to show proportions of a whole. Each sector's angle is proportional to the frequency of the category.

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Histograms: Unlike bar charts, there are no gaps between bars. For IGCSE, histograms often use 'Frequency Density' on the y-axis when class widths are unequal.

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Area Principle: In a histogram, the area of the bar (not just the height) represents the frequency.

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Class Width: The difference between the upper and lower boundaries of a class interval.

📐Formulae

Angle of Sector=FrequencyTotal Frequency×360∘\text{Angle of Sector} = \frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

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}

💡Examples

Problem 1:

In a survey of 60 students, 15 students chose 'Basketball' as their favorite sport. Calculate the angle for the 'Basketball' sector in a pie chart.

Solution:

90∘90^\circ

Explanation:

To find the angle, divide the frequency of the category by the total frequency and multiply by 360. 1560×360=14×360=90∘\frac{15}{60} \times 360 = \frac{1}{4} \times 360 = 90^\circ.

Problem 2:

A histogram is drawn to represent the weights of packages. For the class interval 20<w≤3020 < w \leq 30, the frequency is 45. Calculate the frequency density for this bar.

Solution:

4.54.5

Explanation:

First, find the class width: 30−20=1030 - 20 = 10. Then apply the formula: Frequency Density=FrequencyClass Width=4510=4.5\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{45}{10} = 4.5.

Problem 3:

On a histogram, a bar has a width of 5 units (from x=10x=10 to x=15x=15) and a frequency density of 1.2. What is the frequency represented by this bar?

Solution:

66

Explanation:

Frequency is calculated as the area of the bar in a histogram. Frequency=Frequency Density×Class Width=1.2×5=6\text{Frequency} = \text{Frequency Density} \times \text{Class Width} = 1.2 \times 5 = 6.