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Statistics and Probability - Box-and-Whisker Plots and Cumulative Frequency Graphs

Grade 8IB

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

🔑Concepts

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A Box-and-Whisker Plot provides a visual summary of data using five key values: the Minimum, lower quartile (Q1Q_1), Median (Q2Q_2), upper quartile (Q3Q_3), and Maximum. The 'box' spans the Interquartile Range (IQRIQR), containing the middle 50%50\% of the data.

A standard box-and-whisker plot labeled with the five-number summary.
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A Cumulative Frequency Graph (Ogive) is used to estimate the median and quartiles. It is created by plotting the running total of frequencies against the upper class boundaries of data intervals.

An S-shaped cumulative frequency curve showing how to find the median at the 50% mark.
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The Interquartile Range (IQRIQR) represents the spread of the middle 50%50\% of the data, calculated as Q3−Q1Q_3 - Q_1. It is a measure of variability that is less influenced by outliers than the total range.

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Percentiles divide data into 100 equal parts. The kthk^{th} percentile is the value below which k%k\% of the data falls. For example, Q1Q_1 is the 25th25^{th} percentile and Q3Q_3 is the 75th75^{th} percentile.

📐Formulae

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

IQR=Q3−Q1IQR = Q_3 - Q_1

Position of Q1=14(n+1)\text{Position of } Q_1 = \frac{1}{4}(n+1)

Position of Q2 (Median)=12(n+1)\text{Position of } Q_2 \text{ (Median)} = \frac{1}{2}(n+1)

Position of Q3=34(n+1)\text{Position of } Q_3 = \frac{3}{4}(n+1)

💡Examples

Problem 1:

Given the following data set of test scores: 12,15,17,20,22,25,28,30,3512, 15, 17, 20, 22, 25, 28, 30, 35. Find the five-number summary and the IQRIQR.

Solution:

  1. Arrange in order (already done): 12,15,17,20,22,25,28,30,3512, 15, 17, 20, 22, 25, 28, 30, 35.
  2. Minimum: 1212, Maximum: 3535.
  3. Median (Q2Q_2): The middle value is the 5th5^{th} term, so Q2=22Q_2 = 22.
  4. Lower Quartile (Q1Q_1): The median of the lower half (12,15,17,2012, 15, 17, 20) is 15+172=16\frac{15+17}{2} = 16.
  5. Upper Quartile (Q3Q_3): The median of the upper half (25,28,30,3525, 28, 30, 35) is 28+302=29\frac{28+30}{2} = 29.
  6. IQR=Q3−Q1=29−16=13IQR = Q_3 - Q_1 = 29 - 16 = 13.

Explanation:

The five-number summary provides the bounds for the box-and-whisker plot. The IQRIQR tells us the spread of the middle half of the students' scores is 1313 marks.

Problem 2:

A group of 8080 students took a math quiz. The cumulative frequency graph shows that the 20th20^{th} student (at the 25th25^{th} percentile) scored 4545 marks and the 60th60^{th} student (at the 75th75^{th} percentile) scored 7575 marks. Calculate the IQRIQR for these marks.

Solution:

From the data provided:

  • Q1Q_1 (25th percentile) = 4545
  • Q3Q_3 (75th percentile) = 7575
  • IQR=Q3−Q1IQR = Q_3 - Q_1
  • IQR=75−45=30IQR = 75 - 45 = 30

Explanation:

In a cumulative frequency graph, percentiles are used to find quartiles. The IQRIQR is the difference between the 75th75^{th} and 25th25^{th} percentiles.

Problem 3:

Calculate the cumulative frequencies for the following frequency table: Score RangeFrequency0−10510−201220−308\begin{array}{|l|c|} \hline \text{Score Range} & \text{Frequency} \\ \hline 0-10 & 5 \\ 10-20 & 12 \\ 20-30 & 8 \\ \hline \end{array}

Solution:

To find cumulative frequency (CFCF):

  • For 0−100-10: CF=5CF = 5
  • For 10−2010-20: CF=5+12=17CF = 5 + 12 = 17
  • For 20−3020-30: CF=17+8=25CF = 17 + 8 = 25

Final Table: Score Upper BoundFrequencyCumulative Frequency105520121730825\begin{array}{|l|c|r|} \hline \text{Score Upper Bound} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 10 & 5 & 5 \\ 20 & 12 & 17 \\ 30 & 8 & 25 \\ \hline \end{array}

Explanation:

Cumulative frequency is calculated by summing the frequencies up to the current interval. These values are plotted against the upper bounds (10,20,3010, 20, 30) to draw the graph.

Problem 4:

A researcher records the weights of 120 apples in grams. The cumulative frequency graph of the data is shown. Use the graph to estimate the number of apples that weigh more than 160160 grams.

Cumulative frequency curve for apple weights showing a value of 90 at 160 grams.

Solution:

  1. Find the value 160160 on the x-axis (Weight).
  2. Move vertically to hit the curve, then horizontally to the y-axis (Cumulative Frequency).
  3. The cumulative frequency at x=160x = 160 is 9090.
  4. This means 9090 apples weigh 160160g or less.
  5. To find those weighing more than 160160g: Total - Cumulative Frequency = 120−90=30120 - 90 = 30.

Answer: 3030 apples.

Explanation:

To find the count of items 'above' a certain value, you subtract the cumulative frequency at that point from the total frequency (nn).

Problem 5:

Given a data set with a Minimum of 55, Q1Q_1 of 1212, Median of 1818, Q3Q_3 of 2525, and Maximum of 4040. Draw the box-and-whisker plot and determine the Interquartile Range (IQRIQR).

Box-and-whisker plot showing Min 5, Q1 12, Med 18, Q3 25, and Max 40.

Solution:

  1. Draw a number line from 00 to 4545.
  2. Draw a rectangle (the box) from x=12x = 12 to x=25x = 25.
  3. Draw a vertical line inside the box at x=18x = 18 (Median).
  4. Extend whiskers from 1212 down to 55 and from 2525 up to 4040.
  5. Calculate IQRIQR: IQR=Q3−Q1=25−12=13IQR = Q_3 - Q_1 = 25 - 12 = 13

Answer: IQR=13IQR = 13.

Explanation:

The box covers the range between the quartiles, and the whiskers connect the box to the extremes of the data.