8.3 Histograms, Cumulative Frequency, Stem-and-Leaf Plots, and Dot Plots

When educators analyse the distribution of a single numerical variable—such as test marks, reading ages, or fitness scores—they choose between displays that preserve every individual score (dot plots and stem-and-leaf plots) and displays that summarise scores into class intervals (histograms and cumulative frequency curves).

Comparing Univariate Data Displays

---

Dot Plots: Mode, Range, and Exact Median Position

In a dot plot, each dot plotted above a horizontal number line represents one observation (unless a key states otherwise, such as $1\text{ dot} = 2\text{ students}$). To analyse a dot plot systematically:

1. Sample size ($N$): Count the total number of dots across the entire plot.
2. Mode: Identify the horizontal scale value that has the tallest vertical column of dots. Do not report the number of dots in that column as the mode! For instance, if the value $4\text{ books}$ has $6\text{ dots}$ above it (the tallest stack), the mode is $4\text{ books}$, not $6$.
3. Range: Subtract the lowest populated scale value from the highest populated scale value ($\text{Max} - \text{Min}$). If the printed axis runs from $0$ to $10$ but the lowest dot is at $2$ and the highest dot is at $8$, the range is $8 - 2 = 6$ (never $10 - 0 = 10$).
4. Median: Locate the middle position $\frac{N + 1}{2}$ and count dots in ascending order from left to right:
   • If $N = 19$, the median is the $\frac{19 + 1}{2} = 10\text{th}$ dot.
   • If $N = 20$, the median lies halfway between the $10\text{th}$ and $11\text{th}$ dots.
5. Clusters, gaps, and outliers: A cluster is a narrow interval where many dots bunch together; a gap is an interior range of scale values with zero dots; an outlier is an isolated dot separated from the main body of data by a wide gap.

---

Histograms vs. Cumulative Frequency Tables and Ogives

A histogram groups continuous numerical measurements into adjacent class intervals of equal width (such as $50 \le x < 60$, $60 \le x < 70$, $70 \le x < 80$). Because exact individual scores are merged into bins, you cannot read the exact sample mean or exact individual minimum from a histogram alone, but you can:

• Identify the modal class (the interval with the highest bar).
• Sum bar heights across multiple intervals to find how many students scored "at least $70$" ($x \ge 70$) or "less than $60$" ($x < 60$), and divide by $N$ to convert to a relative frequency percentage.
• Locate the class interval containing the median by accumulating bar frequencies from the lowest interval upward until you pass the $\frac{N + 1}{2}$ position.

Worked Example 1: Summing Histogram Intervals and Finding the Median Class

Suppose a histogram of Year 10 science exam scores ($N = 80$ students) uses five intervals: $40 \le x < 50$ ($6$ students), $50 \le x < 60$ ($14$ students), $60 \le x < 70$ ($28$ students), $70 \le x < 80$ ($20$ students), and $80 \le x < 90$ ($12$ students):

• Percentage scoring at least $70$ ($x \ge 70$): Sum the top two bars ($20 + 12 = 32$ students) and divide by $80$: $\frac{32}{80} \times 100\% = 40\%$.
• Class interval containing the median: With $N = 80$, the median lies between the $40\text{th}$ and $41\text{st}$ ordered scores. The first two bars accumulate $6 + 14 = 20$ students, and adding the third bar ($60 \le x < 70$) brings the running total to $20 + 28 = 48$ students. Since positions $21$ through $48$ all lie inside $60 \le x < 70$, the median falls in the $60 \le x < 70$ interval.

Cumulative Frequency Tables and Ogives

A cumulative frequency table (or its graphical curve, called an ogive) records the running total $CF(x)$ of observations that fall at or below each upper boundary $x$. The final cumulative frequency always equals the total sample size $N$.

Suppose $N = 200$ Year 9 students sit a reading test with the following cumulative frequencies:

• Score $\le 50$: $CF(50) = 24$
• Score $\le 60$: $CF(60) = 58$
• Score $\le 70$: $CF(70) = 130$
• Score $\le 80$: $CF(80) = 176$
• Score $\le 90$: $CF(90) = 200$

From these running totals, we apply three fundamental rules:

1. Count within an interval $(a, b]$: Subtract consecutive cumulative totals:

$$\text{Students scoring } 60 < x \le 70 = CF(70) - CF(60) = 130 - 58 = 72\text{ students}$$

1. Count strictly above a threshold $b$: Subtract $CF(b)$ from the total cohort $N$:

$$\text{Students scoring } x > 70 = N - CF(70) = 200 - 130 = 70\text{ students } (35\%)$$

1. Median and Interquartile Range ($\text{IQR}$) from an Ogive:
   • Lower Quartile ($Q_1$, $25\text{th}$ percentile): Read across from $0.25N = 50$ on the vertical cumulative frequency axis to the curve, then down to the score axis.
   • Median ($Q_2$, $50\text{th}$ percentile): Read across from $0.50N = 100$ on the vertical axis.
   • Upper Quartile ($Q_3$, $75\text{th}$ percentile): Read across from $0.75N = 150$ on the vertical axis.
   • Interquartile Range ($\text{IQR}$): Calculate $\text{IQR} = Q_3 - Q_1$, which measures the spread of the middle $50\%$ of scores.

---

Stem-and-Leaf Plots and the Back-to-Back Left-Side Trap

A stem-and-leaf plot splits each number into a leading stem (such as tens or ones) and a single-digit trailing leaf (such as ones or tenths). Always inspect the Key/Legend first: under Key: 6 | 4 = 64, the entry represents $64$, whereas under Key: 6 | 4 = 6.4, it represents $6.4$.

In a back-to-back stem-and-leaf plot, two groups share a central column of stems—with Group A plotted to the left of the stem and Group B plotted to the right of the stem. In an ordered plot, leaves always increase moving outward away from the central stem:

• Right side (Group B): Leaves increase from left to right (e.g., | 6 | 1 3 8 represents $61, 63, 68$).
• Left side (Group A): Leaves increase from right to left moving outward from the central stem! Thus the row 9 5 2 | 6 | represents $62, 65, 69$ in ascending order—where 2 (nearest the stem 6) is the smallest value in that row ($62$) and 9 (farthest to the left) is the largest value in that row ($69$).

When counting to the $\frac{N+1}{2}\text{th}$ leaf to find the median of the left-hand group, start at the smallest stem and count leaves from the central stem outward to the left within each row.