Measures of Central Tendency

To find the mean, we use the following formula:

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

(Where $\bar{x}$ is the mean, $\sum x$ is the sum of all values, and $n$ is the number of values)

To find the median's position in an ordered list, we use:

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

Worked Example: Mean from a List

Find the mean of the following values: 3, 7, 8, 12.

Step 1: Sum the values to find $\sum x$.

• $\sum x = 3 + 7 + 8 + 12 = 30$

Step 2: Divide by the number of values ($n = 4$).

• $\bar{x} = 30 \div 4 = 7.5$

Worked Example: Median from a List

Find the median of the following values: 15, 2, 9, 11, 4.

Step 1: Arrange the data in numerical order.

• $2, 4, 9, 11, 15$

Step 2: Find the position of the median ($n = 5$).

• $\text{Position} = \frac{5 + 1}{2} = 3$

Step 3: Identify the 3rd value in your ordered list.

• $\text{Median} = 9$

Note for sets with an even $n$: If your list was $2, 4, 9, 11, 15, 20$ ($n=6$), the position would be $\frac{6+1}{2} = 3.5$. The median is the arithmetic mean of the 3rd and 4th values: $(9 + 11) \div 2 = 10$.

Worked Example: Mode from a List

Find the mode of the following values: 4, 7, 2, 7, 9, 4, 7.

Step 1: Count the frequency of each value.

• $7$ appears three times.
• $4$ appears two times.
• $2$ and $9$ appear one time each.

Step 2: Identify the value with the highest frequency.

• $\text{Mode} = 7$

Note: If the list was $4, 7, 2, 7, 9, 4$, both $4$ and $7$ would appear twice. The set would be bimodal, and the modes would be $4$ and $7$.

Estimating the Mean from Frequency Tables

Imagine trying to calculate the average age of 1,000 people by adding up a massive list of individual numbers — it would take forever. Instead, large datasets are organised into grouped frequency tables. When data is grouped into class intervals, we can only calculate an estimated mean because the exact values within each interval are unknown.

To estimate the mean, we must assume every value in a class interval is equal to the midpoint of that interval. We multiply each midpoint ($x$) by its frequency ($f$) to find the total for that row ($f \times x$). Finally, we divide the sum of these products ($\sum fx$) by the total frequency ($\sum f$).

$$\text{Mean} = \frac{\sum (x \times f)}{\sum f}$$

Worked Example: Estimated Mean from Grouped Data

Calculate an estimate for the mean from the grouped data below.

• Class $0-10$: Frequency = 3
• Class $10-20$: Frequency = 7

Step 1: Find the midpoint ($x$) for each class interval.

• Midpoint of $0-10$ is $(0 + 10) \div 2 = 5$
• Midpoint of $10-20$ is $(10 + 20) \div 2 = 15$

Step 2: Calculate $f \times x$ for each row.

• For $0-10$: $3 \times 5 = 15$
• For $10-20$: $7 \times 15 = 105$

Step 3: Calculate the total frequency ($\sum f$) and total products ($\sum fx$).

• $\sum f = 3 + 7 = 10$
• $\sum fx = 15 + 105 = 120$

Step 4: Divide $\sum fx$ by $\sum f$.

• $\text{Mean} = 120 \div 10 = 12$

Identifying the Modal Class

How do you find the most popular group when data is bundled into intervals? Instead of finding a single numerical mode, you identify the modal class. This is the specific class interval that contains the highest frequency in a grouped table.

Identifying the modal class requires no calculation; it is a direct observation of the frequency column. Be extremely careful: you must state the class interval itself as your final answer, not the frequency number.

Worked Example: Identifying the Modal Class

Identify the modal class for the following weights ($w$):

• $10 \le w < 20$ (Frequency = 9)
• $20 \le w < 30$ (Frequency = 14)

Step 1: Locate the highest number in the frequency column.

• The highest frequency is $14$.

Step 2: State the corresponding class interval.

• $\text{Modal Class} = 20 \le w < 30$

Choosing the Appropriate Measure and Handling Outliers

One extreme value can completely distort an average, making it highly unrepresentative of the overall group. An outlier is an extreme value that does not fit the general pattern of the rest of the data.

The arithmetic mean is highly sensitive to outliers because it incorporates every value; extreme numbers pull the mean heavily toward them. In contrast, the median is robust and resistant to outliers because it only relies on the middle position of the ordered data.

• Use the Mean: For broadly symmetric data that has no significant outliers.
• Use the Median: For skewed data or data containing outliers (e.g., house prices or salaries).
• Use the Mode: For qualitative data (e.g., favourite flavours) or when identifying the "most common" item for stock ordering.

Worked Example: The Effect of an Outlier

Compare the mean and median for Set A and Set B to evaluate the most appropriate measure for Set B.

• Set A: 10, 12, 13, 15, 17
• Set B: 10, 12, 13, 15, 100

Step 1: Observe the measures for Set A.

• $\text{Mean} = 13.4$, $\text{Median} = 13$

Step 2: Observe the measures for Set B (which contains the outlier 100).

• $\text{Mean} = 30$, $\text{Median} = 13$

Step 3: Evaluate the most appropriate measure for Set B.

• $30$ is not representative of Set B, as almost all values are in the low teens. The median ($13$) is the better measure because it is not affected by the extreme outlier.