Discrete & Categorical Displays (Bar Charts & Pie Charts)
When displaying qualitative categories or discrete numerical values, graphs must visually reflect that categories are separate and non-continuous.
- Distinct Gaps: Gaps must be present between bars to signify that the underlying data is discrete or categorical.
- Bar Variations: Standard vertical bars, clustered/compound bars (comparing two sub-groups side-by-side), and component/stacked bars (displaying category totals with internal sub-divisions).
A circular statistical display where the circle is divided into slices proportional to the relative frequencies: $$\text{Sector Angle } \theta = \frac{f}{\sum f} \times 360^\circ$$ $$\text{Sector Percentage} = \frac{f}{\sum f} \times 100\%$$
Histograms & Frequency Density for Continuous Data
A histogram represents continuous quantitative data. Unlike a bar chart, there are no gaps between columns because continuous data flows smoothly along the horizontal axis.
In a histogram, the area of each column is proportional to the frequency: $$\text{Area} = \text{Class Width} \times \text{Height} \propto \text{Frequency}$$
- Equal Widths: The vertical axis can directly represent frequency ($f$), because all widths are identical.
- Unequal class widths: The vertical axis must represent Frequency Density: $$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$ $$\text{Frequency} = \text{Frequency Density} \times \text{Class Width} = \text{Area of Rectangle}$$
Formed by joining the mid-interval coordinates $(x_{\text{mid}}, f)$ or $(x_{\text{mid}}, \text{FD})$ of each histogram column with straight lines. The polygon is anchored down to the horizontal axis at zero frequency at the adjacent midpoints of hypothetical outer classes.
Continuous histogram with contiguous columns and overlaid frequency polygon connecting mid-interval values.
Cumulative Frequency Curves (Ogives) & Quartiles
A cumulative frequency curve (ogive) plots the running total of frequencies against the upper class boundaries of each interval.
- Plotting Rule: Plot points at $(\text{Upper Boundary}, \text{Cumulative Frequency})$. Start the curve at $(\text{Lower Boundary of first class}, 0)$.
- Median ($Q_2$): The 50th percentile. Project across from $\frac{N}{2}$ on the vertical axis to the curve, then read down to the horizontal axis.
- Lower Quartile ($Q_1$): The 25th percentile. Project across from $\frac{N}{4}$ (or $0.25N$).
- Upper Quartile ($Q_3$): The 75th percentile. Project across from $\frac{3N}{4}$ (or $0.75N$).
- Interquartile Range: $\text{IQR} = Q_3 - Q_1$.
Stem-and-Leaf Plots & Distribution Skewness
A stem-and-leaf plot organizes raw quantitative data without losing original individual scores.
- The Stem represents leading place-value digits; the Leaf represents the final digit.
- Leaves must be ordered in ascending numerical sequence without commas.
- A Key is mandatory (e.g., $4 \mid 2 = 42\text{ marks}$ or $3 \mid 5 = 3.5\text{ cm}$).
- Back-to-back stem-and-leaf plots place leaves on both sides of a central stem to compare two groups simultaneously.
- Symmetrical (Bell-shaped): Data clusters evenly around the center. $\text{Mean} \approx \text{Median} \approx \text{Mode}$.
- Positively Skewed (Right-tailed): Long tail stretches to high positive values. Bulk of data on the left. $\text{Mode} < \text{Median} < \text{Mean}$.
- Negatively Skewed (Left-tailed): Long tail stretches to small values. Bulk of data on the right. $\text{Mean} < \text{Median} < \text{Mode}$.
Worked Example 1: Unequal Class Width Histogram
The masses ($m$ kg) of 60 luggage items checked onto a flight are summarized in the table below:
| Mass ($m$ kg) | Frequency ($f$) |
|---|---|
| $0 \le m < 10$ | $10$ |
| $10 \le m < 15$ | $15$ |
| $15 \le m < 25$ | $24$ |
| $25 \le m < 35$ | $11$ |
a) Calculate the class widths and frequency densities for all four intervals.
b) State which interval is represented by the tallest column on the histogram.
Step-by-step Solution:
Part a: Widths and Frequency Densities:
Using $\text{FD} = \frac{\text{Frequency}}{\text{Class Width}}$: [M1]
- $0 \le m < 10$: Width $= 10 - 0 = 10 \implies \text{FD} = \frac{10}{10} = 1.0$. [A1]
- $10 \le m < 15$: Width $= 15 - 10 = 5 \implies \text{FD} = \frac{15}{5} = 3.0$. [A1]
- $15 \le m < 25$: Width $= 25 - 15 = 10 \implies \text{FD} = \frac{24}{10} = 2.4$. [A1]
- $25 \le m < 35$: Width $= 35 - 25 = 10 \implies \text{FD} = \frac{11}{10} = 1.1$. [A1]
Part b: Tallest column:
The height of each column in a histogram corresponds to its Frequency Density, not raw frequency.
The highest frequency density is $\text{FD} = 3.0$ for the class $10 \le m < 15$.
Therefore, $10 \le m < 15$ has the tallest column (even though $15 \le m < 25$ has more items). [Criterion A: Conceptual Mastery]
Worked Example 2: Cumulative Frequency Curve & IQR
An ogive is plotted for the test scores of $80$ students. From the graph:
The curve passes through cumulative frequency $20$ at score $44$, cumulative frequency $40$ at score $58$, and cumulative frequency $60$ at score $72$.
a) Determine the median score.
b) Calculate the Interquartile Range (IQR).
c) If the passing mark is $50$, and the cumulative frequency at score $50$ is $30$, how many students passed the test?
Step-by-step Solution:
Part a: Median ($Q_2$):
The median corresponds to $50\%$ of $80 = \frac{80}{2} = 40$.
Reading across from $cf = 40$ gives a median score of $58$. [A1]
Part b: Interquartile Range:
Lower quartile $Q_1$ corresponds to $\frac{80}{4} = 20 \implies Q_1 = 44$. [M1]
Upper quartile $Q_3$ corresponds to $\frac{3 \times 80}{4} = 60 \implies Q_3 = 72$. [M1]
$\text{IQR} = Q_3 - Q_1 = 72 - 44 = 28\text{ marks}$. [A1]
Part c: Number of passing students:
At mark $50$, $cf = 30$. This means $30$ students scored less than or equal to $50$. [M1]
Number of students who passed (scored $> 50$) $= 80 - 30 = 50\text{ students}$. [Criterion C: Clear Mathematical Interpretation]
Tiered Practice Exercises
10 problems advancing from pie chart angle calculations and stem-and-leaf diagrams to frequency density, ogive percentiles, and skewness investigations.
A survey asked 90 students to name their favorite sport: Football (35), Basketball (25), Swimming (20), Tennis (10). Calculate the sector angles required to construct an accurate pie chart.
View Solution & Markscheme
Method & Marks:
Angle formula: $\theta = \frac{f}{90} \times 360^\circ = f \times 4^\circ$. [M1]
Football: $35 \times 4^\circ = 140^\circ$. [A1]
Basketball: $25 \times 4^\circ = 100^\circ$. [A1]
Swimming: $20 \times 4^\circ = 80^\circ$. [A1]
Tennis: $10 \times 4^\circ = 40^\circ$. (Check: $140 + 100 + 80 + 40 = 360^\circ$). [A1]
Construct an ordered stem-and-leaf plot for the following 15 science quiz marks: $$23, 41, 35, 28, 44, 31, 29, 38, 47, 52, 33, 41, 36, 49, 25$$ Include a complete key.
View Solution & Markscheme
Method & Marks:
Stems are $2, 3, 4, 5$.
Stem $2$: $3, 5, 8, 9$
Stem $3$: $1, 3, 5, 6, 8$
Stem $4$: $1, 1, 4, 7, 9$
Stem $5$: $2$ [M1][A1]
Key: $3 \mid 1 = 31\text{ marks}$. [A1]
State two essential differences between a bar chart and a histogram.
View Solution & Markscheme
Method & Marks:
1. Gaps: Bar charts have gaps between columns because data is categorical or discrete; histograms have touching, contiguous columns because data is continuous. [A1]
2. Area vs. Height: In bar charts, column height represents frequency. In histograms, the area of the column represents frequency (height represents frequency density). [A1]
On a histogram, the class interval $20 \le x < 35$ is represented by a column of width $3\text{ cm}$ and height $8\text{ cm}$, representing a frequency of $24$.
a) Determine how many units of frequency are represented by $1\text{ cm}^2$ of area.
b) If another class interval $35 \le x < 40$ is drawn with a width of $1\text{ cm}$ and height of $6\text{ cm}$, what is its frequency?
View Solution & Markscheme
Method & Marks:
a) Area of first column $= 3 \times 8 = 24\text{ cm}^2$. [M1]
Area represents frequency: $24\text{ cm}^2 \equiv 24 \text{ units} \implies 1\text{ cm}^2 \equiv 1\text{ frequency unit}$. [A1]
b) Area of second column $= 1 \times 6 = 6\text{ cm}^2$.
Frequency $= 6 \times 1 = 6$. [A1]
A cumulative frequency table records the waiting times (in minutes) of 120 patients at a medical clinic:
$t < 10$: $12$; $t < 20$: $38$; $t < 30$: $76$; $t < 40$: $104$; $t < 50$: $120$.
a) Identify the coordinates plotted to construct the ogive.
b) State the cumulative frequencies used to read off the median, lower quartile, and 90th percentile.
View Solution & Markscheme
Method & Marks:
a) Points plotted: Plotted at upper class boundaries against cumulative frequency: [M1]
$(0, 0), (10, 12), (20, 38), (30, 76), (40, 104), (50, 120)$. [A1]
b) Key percentiles on vertical axis:
Median ($Q_2$): $50\%$ of $120 = 60$. [A1]
Lower quartile ($Q_1$): $25\%$ of $120 = 30$. [A1]
90th percentile: $0.90 \times 120 = 108$. [A1]
A dataset has $\text{mean} = 68$, $\text{median} = 54$, and $\text{mode} = 48$.
a) Describe the skewness of this distribution.
b) Explain what this implies about the shape of the data distribution curve and the presence of extreme values.
View Solution & Markscheme
Method & Marks:
a) Since $\text{Mode} (48) < \text{Median} (54) < \text{Mean} (68)$, the distribution is positively skewed (skewed to the right). [A1]
b) The majority of the observations are clustered toward the lower values on the left, but there is a long tail stretching to the right caused by a few exceptionally large positive scores (high outliers) that pull the mean upward. [R1]
A frequency polygon is drawn for grouped data with class intervals $10 \le x < 20$, $20 \le x < 30$, $30 \le x < 40$, and $40 \le x < 50$. What are the $x$-coordinates of the midpoints where vertices are plotted, and what two anchor points on the horizontal axis will close the polygon?
View Solution & Markscheme
Method & Marks:
Midpoints for each interval: [M1]
$x_1 = 15, x_2 = 25, x_3 = 35, x_4 = 45$. [A1]
To anchor the polygon to zero frequency, extend by one class width ($10$) on each end: [M1]
Left anchor: $15 - 10 = 5 \implies (5, 0)$.
Right anchor: $45 + 10 = 55 \implies (55, 0)$. [A1]
In a histogram representing salaries in a tech startup, the frequency density for salaries between $\$40{,}000$ and $\$60{,}000$ is $0.0025\text{ employees/dollar}$. The frequency density for executive salaries between $\$120{,}000$ and $\$200{,}000$ is $0.00015\text{ employees/dollar}$.
a) Calculate how many employees earn in each of these two brackets.
b) Calculate the ratio of the areas of these two histogram bars.
View Solution & Markscheme
Method & Marks:
a) $\text{Frequency} = \text{Class Width} \times \text{FD}$ [M1]
Bracket 1: Width $= 60{,}000 - 40{,}000 = 20{,}000$.
Employees $= 20{,}000 \times 0.0025 = 50\text{ employees}$. [A1]
Bracket 2: Width $= 200{,}000 - 120{,}000 = 80{,}000$.
Employees $= 80{,}000 \times 0.00015 = 12\text{ employees}$. [A1]
b) Since area is proportional to frequency, the ratio of areas equals the ratio of frequencies: [M1]
$\frac{\text{Area}_1}{\text{Area}_2} = \frac{50}{12} = \frac{25}{6} \approx 4.17 : 1$. [A1]
The back-to-back stem-and-leaf plot compares exam scores between Class A and Class B:
Class A | Stem | Class B
8 5 2 | 4 | 1 4 7
9 6 3 0 | 5 | 2 5 8 8
7 4 1 | 6 | 0 3 6 9
8 2 | 7 | 1 5 8
| 8 | 2 4
Key: 5 | 2 means 52 marks for Class B; 2 | 4 means 42 marks for Class A
a) Find the median score for Class A and Class B.
b) Compare the shapes and overall performance of the two classes.
View Solution & Markscheme
Method & Marks:
a) Class A has $n = 12$ students. Median is average of 6th and 7th scores:
Ordered scores: $42, 45, 48, 50, 53, 56, 59, 61, 64, 67, 72, 78$.
Median Class A $= \frac{56 + 59}{2} = 57.5$. [M1][A1]
Class B has $n = 15$ students. Median is 8th score:
Scores: $41, 44, 47, 52, 55, 58, 58, 60, 63, 66, 69, 71, 75, 78, 82, 84$.
8th score is $60 \implies \text{Median Class B} = 60$. [A1]
b) Comparison: Class B performed better overall, having a higher median ($60$ vs $57.5$) and reaching top scores in the 80s. Class A is fairly symmetric centered in the 50s-60s, while Class B extends higher. [R1]
Criterion C & D Real-World Investigation: A government report claims that "average household internet speeds increased massively over the past 5 years", citing a graph with a truncated vertical axis starting at $95\text{ Mbps}$ rather than $0\text{ Mbps}$.
a) Explain how truncating the vertical axis distorts visual perception and misleads the public.
b) Describe how using a 3D perspective pie chart distorts data interpretation.
c) Formulate two guidelines for ethical and transparent graphical communication in mathematical reports.
View Solution & Markscheme
Method & Marks:
Part a: Truncated axis distortion: By starting the vertical axis at 95 instead of 0, small relative differences (e.g. from 96 to 100 Mbps, a $4\%$ increase) appear visually magnified as if speeds quadrupled, misleading readers who judge differences by bar height rather than scale labels. [R1: Criterion C]
Part b: 3D perspective distortion: In 3D pie charts, the angle of tilt foreshortens the rear sectors and visually expands the front slice due to perspective projection, making front slices appear much larger than their actual percentage. [R1]
Part c: Ethical communication guidelines:
1. Always start quantitative ratio axes at zero or explicitly highlight a broken axis with a zigzag mark.
2. Avoid 3D charts and ensure graph areas/volumes are strictly proportional to underlying data values without decorative distortions. [A1][R1: Criterion D Reflection]
Statistical Visualizations Quick Reference
Pie Chart Sector
$$\theta = \frac{f}{\sum f} \times 360^\circ$$
Proportional sector angle.Frequency Density
$$\text{FD} = \frac{\text{Frequency}}{\text{Class Width}}$$
Height of unequal histogram bar.Histogram Area
$$\text{Area} = \text{Width} \times \text{FD} = \text{Frequency}$$
Area represents count.Ogive Coordinates
$$(\text{Upper Boundary}, \; \text{Cumulative } f)$$
Start at $(L_1, 0)$.Ogive Quartiles
$$Q_1: 0.25N, \; Q_2: 0.5N, \; Q_3: 0.75N$$
Vertical projections.Positive Skewness
$$\text{Mode} < \text{Median} < \text{Mean}$$
Long tail to the right.Top 5 Common MYP Graphing Pitfalls
- Plotting Midpoints on Cumulative Frequency Curves: Points on an ogive must be plotted at the upper class boundaries, NOT at the midpoints! Midpoints are used only for frequency polygons and estimating means.
- Drawing Gaps in Histograms: Continuous histograms must have touching, seamless bars. Leaving gaps between columns turns it into a bar chart and incorrectly implies discrete categories.
- Forgetting Frequency Density for Unequal Intervals: Plotting raw frequencies on the vertical axis when intervals have different widths inflates wider columns and distorts the distribution.
- Omitting the Key in Stem-and-Leaf Diagrams: A stem-and-leaf plot is invalid without a key. Without a key, readers cannot distinguish between $4 \mid 2 = 42$ and $4 \mid 2 = 4.2$.
- Confusing Skewness Direction: Skewness describes the direction of the long tail, not where the peak is located! A peak on the left with a tail on the right is positively skewed.
Self-Grading Mini Test
5 quiz questions on histograms, ogives, stem-and-leaf diagrams, and skewness.
In a survey of 120 students, 30 walk to school. What is the sector angle for walking on a pie chart?
Explanation: $\theta = \frac{30}{120} \times 360^\circ = \frac{1}{4} \times 360^\circ = 90^\circ$.
In a histogram with unequal classes, what does the vertical axis represent?
Explanation: When class widths are unequal, column height must represent Frequency Density ($\frac{\text{Frequency}}{\text{Width}}$) so that column area correctly represents frequency.
When constructing a cumulative frequency curve (ogive), which $x$-values should be plotted against cumulative frequency?
Explanation: Cumulative frequency represents total occurrences up to a given score, so points must be plotted at the upper boundary of each class interval.
If a salary distribution has a long tail of very high earners to the right and $\text{Mean} > \text{Median}$, the distribution is:
Explanation: A long tail toward higher values that pulls the mean above the median is defined as positively skewed (right-skewed).
The vertices of a frequency polygon are plotted at:
Explanation: A frequency polygon joins the points $(x_{\text{mid}}, f)$ located at the top midpoint of each histogram bar.