9.1 Mean, Median, Mode, Frequency Table Averages, and the Effect of Outliers

Teachers routinely summarise assessment marks, attendance records, fundraising totals, and demographic figures using a single representative number called a measure of centre. Because the mean, median, and mode respond differently to unusual observations, selecting and calculating the right summary statistic is a core quantitative skill in Australian schools.

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The Three Measures of Centre: Mean, Median, and Mode

1. The Mean (Arithmetic Average)

The mean ($\bar{x}$) is the sum of all data values divided by the total number of observations ($n$):

$$\bar{x} = \frac{\sum x}{n}$$

Multiplying both sides by $n$ gives the total sum reconstruction rule:

$$\sum x = n \cdot \bar{x}$$

In LANTITE problem-solving, reconstructing the total sum is the fastest way to handle three recurring classroom scenarios:

1. Finding a required or missing score: Suppose a Year 10 student has a mean score of $74$ across $4$ science topic tests and wants to raise their mean to $76$ after the $5\text{th}$ test. The current total for $4$ tests is $4 \times 74 = 296$, and the target total for $5$ tests is $5 \times 76 = 380$. Subtracting the two totals shows the student must score $380 - 296 = 84$ on the $5\text{th}$ test.
2. Adding or removing a score: Suppose $20$ students sit a numeracy quiz and achieve a mean of $68.5$, giving an initial total of $20 \times 68.5 = 1{,}370$ marks. If an absent student sits the paper the next day and scores $89$, the updated total is $1{,}370 + 89 = 1{,}459$ across $21$ students, shifting the class mean to $\frac{1{,}459}{21} \approx 69.48$.
3. Correcting a data-entry error: Suppose a teacher records a mean of $62.0$ across $25$ students ($\text{initial sum} = 25 \times 62.0 = 1{,}550$), then discovers a student's mark of $83$ was mistyped into the spreadsheet as $38$. Remove the wrong value and add the correct value: $\text{corrected sum} = 1{,}550 - 38 + 83 = 1{,}595$. Dividing by the unchanged student count gives the true mean: $\frac{1{,}595}{25} = 63.8$.

2. The Median (Positional Middle)

The median is the middle value when observations are arranged in ascending (or descending) order. A common exam trap is presenting an unsorted list and hoping candidates simply pick the number printed in the middle of the line. Always sort the numbers first, then locate the middle position using:

$$\text{Median Position} = \frac{n + 1}{2}$$

• Odd number of values ($n$ is odd): Position $\frac{n + 1}{2}$ is a whole number corresponding to a single middle score. For $n = 9$ sorted reading scores $11, 13, 14, 15, 17, 18, 19, 20, 20$, the median is at position $\frac{9 + 1}{2} = 5\text{th}$ value, which is $17$.
• Even number of values ($n$ is even): Position $\frac{n + 1}{2}$ ends in $.5$, so the median is the arithmetic average of the two middle values at positions $\frac{n}{2}$ and $\frac{n}{2} + 1$. For $n = 8$ sorted marks $12, 14, 15, 16, 18, 19, 20, 22$, position $\frac{8 + 1}{2} = 4.5$ lies halfway between the $4\text{th}$ value ($16$) and the $5\text{th}$ value ($18$), giving $\text{Median} = \frac{16 + 18}{2} = 17$.

3. The Mode (Most Frequent Score)

The mode is the value that appears most often in the dataset:

• Unimodal: One value has the highest frequency (for example, in $4, 5, 6, 6, 6, 8, 9$, the mode is $6$).
• Bimodal or multimodal: Two or more distinct values tie for the highest frequency (for example, in $12, 14, 14, 15, 18, 18, 20$, the modes are $14$ and $18$).
• No mode: Every value appears exactly the same number of times. Be careful not to confuse "no mode" with a mode of $0$: if $18$ out of $25$ students recorded $0$ late arrivals this term, the mode is $0$.

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Calculating Mean and Median from a Frequency Table

When whole-class or year-level results are grouped into a frequency table, each distinct score $x_i$ occurs $f_i$ times. Adding all scores requires multiplying each score by its frequency first:

$$\bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$$

Critical Frequency-Table Rule: Always divide $\sum (x_i \cdot f_i)$ by the total frequency $N = \sum f_i$ (the total number of students), never by the number of rows in the table.

To find the median from a frequency table without writing out every score, add a Cumulative Frequency column by keeping a running total of $f_i$. Compute the median position $\frac{N + 1}{2}$ and find the first score row whose cumulative frequency is greater than or equal to that position.

Worked Example: Year 8 Science Quiz ($N = 25$ Students)

• Mean: $\bar{x} = \frac{201}{25} = 8.04$ marks (dividing by $5$ rows would give $40.2$, which is impossible on a $10$-mark quiz).
• Median: For $N = 25$, the middle observation is at position $\frac{25 + 1}{2} = 13\text{th}$ student. Looking down the Cumulative Frequency column, the first $8$ students scored $7$ or below, and students $9$ through $16$ all scored $8$. Because the $13\text{th}$ student sits in that row, the median is $8$.
• Mode: The highest frequency is $f_i = 8$, which corresponds to a quiz score of $8$.

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The Effect of Outliers and Skewness on Measures of Centre

An outlier is an observation that lies far above or far below the rest of the dataset:

• Effect on the Mean: Because the numerator $\sum x$ includes the exact numerical magnitude of every observation, the mean is sensitive (non-resistant) to outliers. A single very high value pulls the mean upward (positive or right skew), whereas an uncharacteristically low score—such as a student scoring $0$ after missing most of an exam—pulls the mean downward (negative or left skew).
• Effect on the Median: The median depends only on the rank order of the middle observation(s). Moving the highest value from $90$ to $900$ does not change which value sits in the middle position, so the median is resistant (robust) to outliers.
• Effect on the Mode: Changing an extreme tail observation does not alter the frequency of the most common score, so the mode normally remains unchanged.

Consider $7$ household incomes in a school catchment: $68,000, $72,000, $75,000, $75,000, $78,000, $82,000, and $450,000. The total is $900,000, giving a mean of $\frac{900{,}000}{7} \approx 128{,}571$ dollars—a figure higher than $6$ of the $7$ households! By contrast, both the median ($4\text{th}$ sorted value) and the mode are $75,000, which accurately reflect the typical household.