Measures of Central Tendency: Where Is the Middle?

"The average salary at this company is 50,600 euros." Sounds reasonable — until you realize most employees earn under 45,000 and a single manager’s salary pulls the number up. One word can hide the truth: "average."

Mean and median both claim to describe the "middle" of a dataset. But they answer fundamentally different questions — and choosing the wrong one can mislead you about what is "typical." In this article, you will learn what a dataset is, how the mean works as a balance point, and why the median often gives a more honest answer.

The Dataset — From Single Value to Collection

In tabular data, a feature corresponds to a column and a data point to a row. In Python, you can represent a dataset as a list (one feature) or as a list of dictionaries (multiple features).

# One-dimensional: single feature
salaries = [32, 35, 38, 40, 42, 45, 48, 55, 120]

# Multi-dimensional: multiple features per data point
employees = [
    {"name": "Anna",  "salary": 32, "dept": "Support"},
    {"name": "Ben",   "salary": 120, "dept": "Management"},
    ...
]

For the rest of this article, we will work with nine salaries in thousands of euros: [32, 35, 38, 40, 42, 45, 48, 55, 120]. Eight employees cluster between 32k and 55k. One outlier sits at 120k. This asymmetry is exactly what makes the choice between mean and median matter.

The Mean — The Balance Point

The seesaw analogy breaks at one point: a physical seesaw has limits (it tips over), while a dataset can contain arbitrarily extreme values that shift the mean without any natural boundary. This break point highlights exactly the mean’s vulnerability.

salaries = [32, 35, 38, 40, 42, 45, 48, 55, 120]
mean = sum(salaries) / len(salaries)  # 50.555...

# Or using the standard library:
import statistics
mean = statistics.mean(salaries)      # 50.555...

The Median — The Outlier-Proof Center

The analogy breaks at scale: with millions of data points, you cannot "line them up" — you need sorting algorithms, which cost O(n log n) time. Here your knowledge from Path I.A about sorting and complexity pays off.

import statistics

salaries = [32, 35, 38, 40, 42, 45, 48, 55, 120]
mean   = statistics.mean(salaries)    # ≈ 50.6
median = statistics.median(salaries)  # 42.0

# The gap reveals the skew:
print(f"Mean: {mean:.1f}, Median: {median}")
# Mean > Median → right-skewed (outlier pulls mean up)

# Replace CEO salary with 1,000 (= 1 million €):
salaries_extreme = [32, 35, 38, 40, 42, 45, 48, 55, 1_000]
print(statistics.mean(salaries_extreme))    # ≈ 148.3
print(statistics.median(salaries_extreme))  # 42 — unchanged!

Interactive: Compute Central Tendency

You have learned about mean, median, and mode. Enter your own data points and observe live how the three measures change. Try the outlier dataset — and see how a single extreme value shifts the mean while the median stays stable.

Takeaways

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