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Statistics - Cumulative Frequency and Box Plots

Grade 9IGCSE

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

🔑Concepts

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Cumulative frequency is the running total of frequencies. When plotted against the upper class boundaries, it forms an 'S-shaped' curve called an ogive. This curve allows us to estimate the median (Q2Q2), lower quartile (Q1Q1), and upper quartile (Q3Q3) by finding the values corresponding to n2\frac{n}{2}, n4\frac{n}{4}, and 3n4\frac{3n}{4} on the vertical axis.

Cumulative frequency curve showing the extraction of the median value.
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A Box Plot (or Box-and-Whisker Plot) provides a visual summary of the five-number summary: Minimum, Lower Quartile (Q1Q1), Median (Q2Q2), Upper Quartile (Q3Q3), and Maximum. The 'box' represents the Interquartile Range (IQRIQR), containing the middle 50%50\% of the data.

A standard box plot showing the five-number summary components.
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The Interquartile Range (IQR=Q3−Q1IQR = Q3 - Q1) is a measure of spread that is less affected by outliers than the total range. A smaller IQRIQR indicates that the data is more consistent or less varied around the median.

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Percentiles divide the data into 100 equal parts. For example, the 90th90^{th} percentile is the value below which 90%90\% of the data falls. It is found at the cumulative frequency position of 90n100\frac{90n}{100}.

📐Formulae

Cumulative Frequency Position (Median)=n2\text{Cumulative Frequency Position (Median)} = \frac{n}{2}

Lower Quartile (Q1) Position=n4\text{Lower Quartile (Q1) Position} = \frac{n}{4}

Upper Quartile (Q3) Position=3n4\text{Upper Quartile (Q3) Position} = \frac{3n}{4}

Interquartile Range (IQR)=Q3−Q1\text{Interquartile Range (IQR)} = Q3 - Q1

Range=Maximum Value−Minimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value}

💡Examples

Problem 1:

A group of 80 students took a math test. The results are: 0<x≤200 < x \le 20 (freq: 10), 20<x≤4020 < x \le 40 (freq: 20), 40<x≤6040 < x \le 60 (freq: 35), 60<x≤8060 < x \le 80 (freq: 15). Calculate the cumulative frequencies and identify the position of the median.

Solution:

  1. CF for x≤20x \le 20 is 10.
  2. CF for x≤40x \le 40 is 10+20=3010 + 20 = 30.
  3. CF for x≤60x \le 60 is 30+35=6530 + 35 = 65.
  4. CF for x≤80x \le 80 is 65+15=8065 + 15 = 80. Median Position: 802=40th\frac{80}{2} = 40^{th} value.

Explanation:

To find cumulative frequency, we keep a running total. The median position in a continuous data set of nn items is found at n/2n/2. To find the actual median score, you would locate 40 on the y-axis of a CF graph and read the corresponding x-value.

Problem 2:

From a cumulative frequency graph, the following values were found: Min = 12, Q1=25Q1 = 25, Median = 34, Q3=42Q3 = 42, Max = 58. Construct the description of the box plot.

Solution:

The box starts at 25 and ends at 42. A vertical line is drawn inside the box at 34. Whiskers extend from the box left to 12 and right to 58.

Explanation:

A box plot visually represents the five-number summary. The 'box' covers the IQR (Q1Q1 to Q3Q3), and the 'whiskers' cover the full range of the data.

Problem 3:

Compare two sets of data: Class A has a Median of 65 and IQR of 10. Class B has a Median of 60 and IQR of 20. Which class performed better and which was more consistent?

Solution:

Class A performed better on average (higher Median: 65 > 60). Class A was also more consistent (lower IQR: 10 < 20).

Explanation:

In IGCSE statistics, 'better performance' is indicated by a higher median, while 'consistency' or 'reliability' is indicated by a smaller Interquartile Range (less spread in the middle 50% of data).

Problem 4:

The cumulative frequency graph shows the heights of 120120 plants. Use the graph to estimate the number of plants with a height greater than 6060 cm.

Cumulative frequency curve for plant heights.

Solution:

  1. Locate 6060 cm on the horizontal (height) axis.
  2. Move vertically to meet the curve, then horizontally to the vertical (cumulative frequency) axis.
  3. The cumulative frequency value at 6060 cm is 9090.
  4. This means 9090 plants have a height ≤60\le 60 cm.
  5. Number of plants >60> 60 cm is 120−90=30120 - 90 = 30 plants.

Explanation:

To find 'greater than' values, subtract the cumulative frequency at that point from the total frequency (nn).

Problem 5:

Compare the distribution of test scores for two classes using the box plots provided. Which class has a higher median and which class has more spread in the middle 50%50\% of scores?

Comparative box plots for Class 1 and Class 2.

Solution:

  1. Class 1 Median (line inside box) is at 7070. Class 2 Median is at 6060. Therefore, Class 1 has a higher median score.
  2. Spread in the middle 50%50\% is represented by the length of the box (IQRIQR).
  3. Class 1 IQR=85−55=30IQR = 85 - 55 = 30. Class 2 IQR=75−35=40IQR = 75 - 35 = 40.
  4. Class 2 has a larger IQRIQR, so it has more spread in the middle 50%50\% of scores.

Explanation:

Higher median indicates better typical performance. Larger IQR indicates lower consistency in the central data.