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Statistics and Probability - Graphical Representation of Data (Bar Graphs, Histograms, Pie Charts)

Grade 8IB

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

🔑Concepts

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Bar Graphs: These are used for comparing discrete categories. The height or length of each bar represents the frequency of that category. Bars should have equal width and should be separated by equal gaps.

A standard bar graph showing four categories A, B, C, and D with varying frequencies.
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Histograms: Unlike bar graphs, histograms are used for continuous data grouped into intervals. There are no gaps between the bars because the data is continuous. The area of each bar is proportional to the frequency.

A histogram showing continuous data intervals from 10 to 90 with no gaps between bars.
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Pie Charts: A circular representation where the circle is divided into sectors. Each sector represents a proportion of the whole. The central angle of a sector is calculated by dividing the category's frequency by the total frequency and multiplying by 360∘360^{\circ}.

A pie chart showing a 25 percent sector with a 90 degree central angle.
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Data Interpretation: Visual representations allow us to identify the 'Mode' (the most frequent category shown by the highest bar or largest sector) and 'Trends' (increasing or decreasing patterns in data over intervals).

📐Formulae

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

Percentage of Total=FrequencyTotal Frequency×100%\text{Percentage of Total} = \frac{\text{Frequency}}{\text{Total Frequency}} \times 100\%

Relative Frequency=FrequencyTotal Frequency\text{Relative Frequency} = \frac{\text{Frequency}}{\text{Total Frequency}}

Frequency Density (for advanced histograms)=FrequencyClass Width\text{Frequency Density (for advanced histograms)} = \frac{\text{Frequency}}{\text{Class Width}}

💡Examples

Problem 1:

A group of 4040 students was surveyed about their favorite sports. 1010 students chose Football, 1414 chose Basketball, 88 chose Swimming, and 88 chose Tennis. Calculate the central angles required to represent this data in a pie chart.

Solution:

  1. Identify the total frequency: 10+14+8+8=4010 + 14 + 8 + 8 = 40.
  2. Calculate the angle for Football: 1040×360∘=0.25×360∘=90∘\frac{10}{40} \times 360^{\circ} = 0.25 \times 360^{\circ} = 90^{\circ}.
  3. Calculate the angle for Basketball: 1440×360∘=0.35×360∘=126∘\frac{14}{40} \times 360^{\circ} = 0.35 \times 360^{\circ} = 126^{\circ}.
  4. Calculate the angle for Swimming: 840×360∘=0.20×360∘=72∘\frac{8}{40} \times 360^{\circ} = 0.20 \times 360^{\circ} = 72^{\circ}.
  5. Calculate the angle for Tennis: 840×360∘=0.20×360∘=72∘\frac{8}{40} \times 360^{\circ} = 0.20 \times 360^{\circ} = 72^{\circ}.
  6. Verification: 90∘+126∘+72∘+72∘=360∘90^{\circ} + 126^{\circ} + 72^{\circ} + 72^{\circ} = 360^{\circ}.

Explanation:

Each frequency is converted into a fraction of the total population (4040). This fraction is then multiplied by 360∘360^{\circ} to find the portion of the circle the category occupies.

Problem 2:

The heights (in cm) of 1515 seedlings are recorded: 2,5,7,12,14,15,18,21,23,25,26,31,33,34,382, 5, 7, 12, 14, 15, 18, 21, 23, 25, 26, 31, 33, 34, 38. Create a frequency table with class intervals of width 1010 and describe the resulting histogram.

Solution:

  1. Group the data into intervals:
    • 0≤h<100 \le h < 10: 2,5,72, 5, 7 (Frequency = 33)
    • 10≤h<2010 \le h < 20: 12,14,15,1812, 14, 15, 18 (Frequency = 44)
    • 20≤h<3020 \le h < 30: 21,23,25,2621, 23, 25, 26 (Frequency = 44)
    • 30≤h<4030 \le h < 40: 31,33,34,3831, 33, 34, 38 (Frequency = 44)
  2. The x-axis will be labeled with the boundaries 0,10,20,30,400, 10, 20, 30, 40.
  3. The y-axis will show frequencies from 00 to 55.
  4. Four bars will be drawn with heights 3,4,4,43, 4, 4, 4 respectively, touching each other to show continuity.

Explanation:

Since height is continuous data, a histogram is used. Data is sorted into 'bins' or intervals. The equal width of the intervals (1010) ensures that the height of each bar accurately represents the frequency of seedlings in that height range.

Problem 3:

A school library tracks the number of books borrowed over four days: Monday: 2020, Tuesday: 4545, Wednesday: 3030, Thursday: 1515. Represent this data using a bar graph.

Bar graph showing library books borrowed: Mon(20), Tue(45), Wed(30), Thu(15).

Solution:

  1. Label the x-axis with the Days (Monday, Tuesday, Wednesday, Thursday).
  2. Label the y-axis with the Number of Books (frequency) using a scale of 1010 units.
  3. Draw bars for each day: Monday =20= 20, Tuesday =45= 45, Wednesday =30= 30, Thursday =15= 15.

Explanation:

A bar graph is chosen here because the days are discrete categories. The heights clearly show that Tuesday was the busiest day for borrowing books.

Problem 4:

In a survey of 6060 people, 3030 preferred Vanilla, 2020 preferred Chocolate, and 1010 preferred Strawberry. Calculate the sector angles for a pie chart and represent the data.

Pie chart representing Vanilla (50%), Chocolate (33.3%), and Strawberry (16.7%).

Solution:

Total =60= 60 Vanilla angle: 3060×360∘=180∘\frac{30}{60} \times 360^{\circ} = 180^{\circ} Chocolate angle: 2060×360∘=120∘\frac{20}{60} \times 360^{\circ} = 120^{\circ} Strawberry angle: 1060×360∘=60∘\frac{10}{60} \times 360^{\circ} = 60^{\circ} Sum of angles: 180∘+120∘+60∘=360∘180^{\circ} + 120^{\circ} + 60^{\circ} = 360^{\circ}

Explanation:

Since Vanilla is exactly half of the total (30/6030/60), it occupies a semi-circle (180∘180^{\circ}). The remaining half is split between Chocolate and Strawberry in a 2:12:1 ratio.