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Statistics - Cumulative frequency and box plots

Grade 11A Level

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

🔑Concepts

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Cumulative frequency is a running total of frequencies. When plotted against the upper class boundary, it forms an 'S-shaped' curve called an ogive, used to estimate the median (Q2Q_2), lower quartile (Q1Q_1), and upper quartile (Q3Q_3).

Cumulative frequency curve showing how to estimate the median from the 50% mark on the y-axis.
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A Box Plot (or Box-and-Whisker Plot) provides a visual summary of a data set using five key values: the Minimum, Q1Q_1, Median, Q3Q_3, and the Maximum. The 'box' represents the Interquartile Range (IQR).

Diagram of a box plot labeling the five-number summary: Minimum, Q1, Median, Q3, and Maximum.
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The Interquartile Range (IQR) measures the spread of the middle 50% of the data. It is calculated as IQR=Q3−Q1IQR = Q_3 - Q_1. A smaller IQR indicates that the data is more consistent.

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Comparison of distributions: When comparing two box plots, look at the position of the median (average) and the width of the box (consistency/spread).

📐Formulae

PositionofMedian≈n2Position of Median \approx \frac{n}{2}

PositionofLowerQuartile(Q1)≈n4Position of Lower Quartile (Q_1) \approx \frac{n}{4}

PositionofUpperQuartile(Q3)≈3n4Position of Upper Quartile (Q_3) \approx \frac{3n}{4}

Interquartile Range (IQR) = Q_3 - Q_1

OutlierBoundaries:Lower=Q1−1.5×IQR,Upper=Q3+1.5×IQROutlier Boundaries: Lower = Q_1 - 1.5 \times IQR, Upper = Q_3 + 1.5 \times IQR

💡Examples

Problem 1:

In a survey of 80 students, their heights (h cm) were recorded. The cumulative frequency table shows: h≤150:10h \le 150: 10, h≤160:30h \le 160: 30, h≤170:65h \le 170: 65, h≤180:80h \le 180: 80. Estimate the Median and the Interquartile Range.

Solution:

  1. Total frequency (nn) = 80.
  2. Median position = 80/2=4080 / 2 = 40. Looking at the cumulative frequency, 40 falls between 160 and 170. By linear interpolation or reading a curve, Median ≈163\approx 163 cm.
  3. Q1Q_1 position = 80/4=2080 / 4 = 20. Since 20 falls in the 150<h≤160150 < h \le 160 class, Q1≈155Q_1 \approx 155 cm.
  4. Q3Q_3 position = 3/4×80=603/4 \times 80 = 60. Since 60 falls in the 160<h≤170160 < h \le 170 class, Q3≈168Q_3 \approx 168 cm.
  5. IQR=Q3−Q1=168−155=13IQR = Q_3 - Q_1 = 168 - 155 = 13 cm.

Explanation:

To estimate these values, we locate the specific rank (20th, 40th, 60th) on the cumulative frequency (y-axis), move horizontally to the curve, and then vertically down to the height (x-axis).

Problem 2:

Given the following summary for a set of test scores: Min = 20, Q1=45Q_1 = 45, Median = 55, Q3=70Q_3 = 70, Max = 95. How is this represented on a Box Plot?

Solution:

  1. Draw a horizontal scale from 20 to 100.
  2. Draw a rectangular box from 45 to 70.
  3. Draw a vertical line inside the box at 55.
  4. Draw 'whiskers' (lines) extending from the box at 45 down to 20, and from 70 up to 95.

Explanation:

The box represents the Interquartile Range (central 50%50\% of the data), the line inside shows the average (median), and the whiskers show the full extent (range) of the data set.

Problem 3:

The weights of 120 apples were recorded. The cumulative frequency graph is shown below. Use the graph to estimate the number of apples weighing more than 140 g140\text{ g}.

Cumulative frequency curve for apple weights from 80g to 180g, showing n=120.

Solution:

  1. Locate 140 g140\text{ g} on the x-axis (horizontal axis).
  2. Move vertically to the curve and then horizontally to the y-axis.
  3. The cumulative frequency at x=140x = 140 is 90.
  4. Total apples n=120n = 120.
  5. Number of apples weighing more than 140 g140\text{ g} is 120−90=30120 - 90 = 30.

Explanation:

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

Problem 4:

Two classes took the same math test. Class A's results are summarized as: Min=30, Q1=45Q_1=45, Med=60, Q3=75Q_3=75, Max=95. Class B's results are shown in the box plot. Which class performed better on average, and which class had more consistent results?

Box plot for Class B with Min=10, Q1=40, Median=50, Q3=65, Max=90.

Solution:

  1. Class A Median = 60. Class B Median = 50 (from diagram).
  2. Class A performed better on average because 60>5060 > 50.
  3. Class A IQR=75−45=30IQR = 75 - 45 = 30. Class B IQR=65−40=25IQR = 65 - 40 = 25.
  4. Class B was more consistent because its IQR (25) is smaller than Class A's IQR (30).

Explanation:

Better average performance is indicated by a higher median. Consistency is indicated by a smaller Interquartile Range (the width of the box).