Mean, median, quartiles: who finds what, how

The three indicators

Mean $\bar{x}$

$\bar{x} = \dfrac{\text{sum of values}}{\text{number of values}}$

Sensitive to extreme values (an outlier pulls it).

Median

The value that splits the ordered data into two equal halves.

• If $n$ is odd: middle value.
• If $n$ is even: mean of the two middle values.

Robust: barely affected by outliers.

Quartiles $Q_1$ and $Q_3$

• $Q_1$ = value splitting the lower quarter (25% below)
• $Q_3$ = value splitting the upper quarter (75% below)
• Interquartile range: $Q_3 - Q_1$ (spread, ignoring extremes)

Step-by-step method

Step 1: sort the data.

Step 2: count $n$.

Step 3: apply positions.

• Median:
  • $n$ odd → rank $(n+1)/2$
  • $n$ even → mean of ranks $n/2$ and $n/2 + 1$
• $Q_1$: rank $\lceil n/4 \rceil$
• $Q_3$: rank $\lceil 3n/4 \rceil$

Example

Data: 3, 7, 8, 12, 5, 6, 9, 15, 10.

Sorted: 3, 5, 6, 7, 8, 9, 10, 12, 15 (n = 9).

• Mean: 75/9 ≈ 8.33
• Median: rank 5 → 8
• $Q_1$: rank 3 → 6
• $Q_3$: rank 7 → 10
• IQR: 10 − 6 = 4

Mean vs median: which to pick?

• Mean: for symmetric data without outliers.
• Median: for skewed data or with extreme values (salaries, house prices).

Classic example: 10 employees where 9 earn $2000 and the boss earns$100,000. Mean = $11,800 (misleading), median =$2000 (representative).

Common mistakes

1. Forgetting to sort before finding the median.
2. Mixing up value and rank: if the value at rank 3 is 6, the median is NOT 3.
3. Using the mean when there are outliers.

?À toi de jouer

Data: 2, 4, 7, 9, 12. The median is:

A.7B.6.8 (the mean)C.6D.3

Valider