Lesson 9 ~25 min Unit 4 · Data & Chance +85 XP

The Median

Find the middle value of ordered data — the measure that resists extreme outliers and tells you what is truly typical.

Today's hook: The median house price in Sydney is $1.4 million. The MEAN house price is even higher because a few $10M mansions drag it up. Statisticians choose median for house prices — it tells you what’s TYPICAL, not what’s extreme.

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Think First

warm-up

Before you read on — quickly: Five employees earn $45 000, $48 000, $50 000, $52 000 and $200 000. Without calculating, which single number do you think best represents a “typical” salary in this group? Why? Try it, then check your reasoning as you go.

Record your answer in your workbook.

The Big Idea

+5 XP

The median is the middle value when data is arranged from smallest to largest. It splits the dataset into two equal halves. For an odd number of values, the median is the exact middle term. For an even number of values, the median is the mean of the two middle terms. Position formula: $\dfrac{n+1}{2}$.

Dataset: 3, 7, 9, 12, 15. Sorted already. n = 5 (odd). Position = (5+1)÷2 = 3rd. The 3rd value is 9. Three values sit below (3, 7) and three sit above (12, 15) — perfect balance. The median is a resistant statistic: unlike the mean, one extreme value won’t change it dramatically.

Median position $= \dfrac{n+1}{2}$

Sort first — always

The median of {7, 3, 9} is NOT 3 — order to {3, 7, 9} and then find the middle.

Odd n: one middle

The position formula gives a whole number. Read that value directly.

Even n: average two middles

Find positions n/2 and n/2+1 in the ordered list. Average the values there.

What You'll Master

objectives

Know

The median is the middle value of ordered data
Position formula: $(n+1) \div 2$
For even $n$: average the two middle values

Understand

Why data must be sorted before finding the median
Why median is a resistant (robust) statistic
When median is a better description than the mean

Can Do

Find the median of any dataset (odd or even $n$)
Compare mean and median and explain the difference
Identify when an outlier makes the mean misleading

Words You Need

vocabulary

MedianThe middle value when data is arranged in ascending order.

Ordered dataValues arranged from smallest to largest.

Middle valueThe value at position $(n+1)\div 2$ in ordered data.

Odd number of valuesWhen $n$ is odd there is one exact middle value.

Even number of valuesWhen $n$ is even, average the two middle values.

Resistant measureA statistic not strongly affected by outliers — the median is resistant; the mean is not.

Spot the Trap

heads-up

Wrong: NOT sorting data first. The median of {7, 3, 9, 1, 5} is not 9. You must order first: {1, 3, 5, 7, 9} → median = 5.

Right: Always sort ascending before doing anything else. Only then apply the position formula.

Wrong: For even $n$, picking just one middle value. For {4, 6, 8, 10}: median ≠ 6 and median ≠ 8.

Right: Average both middle values: median = (6+8)÷2 = 7. The median need not appear in the original data — that is fine.

Median for Odd n

+5 XP

When $n$ is odd, the position formula $\dfrac{n+1}{2}$ gives a whole number — so there is one exact middle value. Step 1: sort. Step 2: apply position formula. Step 3: read the value at that position.

Find the median of {8, 3, 15, 6, 11, 1, 9}. Sort: 1, 3, 6, 8, 9, 11, 15. Count: $n = 7$. Position = (7+1)÷2 = 4th. Count along: 1st=1, 2nd=3, 3rd=6, 4th=8. Median = 8. Three values sit below (1, 3, 6) and three above (9, 11, 15).

$n = 7$ (odd) → position $= 4$ → Median $= 8$

Sort first

Without ordering, the position formula gives the wrong value entirely.

Count carefully

Point to each value as you count: 1st, 2nd, 3rd, 4th…

Verify the balance

The same number of values should sit below and above the median.

Median for Even n

+5 XP

When $n$ is even, the position formula gives a decimal (e.g. 3.5), signalling that no single value is the middle. Find the values at positions $n/2$ and $n/2 + 1$, then average them.

Find the median of {14, 22, 18, 25, 11, 19}. Sort: 11, 14, 18, 19, 22, 25. $n = 6$. Middle positions: 3rd = 18 and 4th = 19. Median = (18 + 19) ÷ 2 = 37 ÷ 2 = 18.5. Notice 18.5 does not appear in the original data — that is perfectly fine.

$n = 6$ (even) → average positions 3 & 4 → Median $= 18.5$

Two middle values

For even $n$, positions $n/2$ and $n/2+1$ give the two values to average.

Median can be a decimal

18.5 is a valid median even though no value is 18.5. This is normal.

Don’t pick just one

Choosing only the 3rd or only the 4th value is the most common even-$n$ mistake.

When to Use Median Over Mean

+5 XP

Use the median when data is skewed or contains outliers. Use the mean when data is roughly symmetric with no extreme values. The median is a resistant statistic — it is based on position, not magnitude, so extreme values do not distort it.

Dataset: {2, 3, 4, 5, 100}. Mean = (2+3+4+5+100)÷5 = 114÷5 = 22.8. Is 22.8 typical? Four of the five values are below 6! Median: sorted, $n=5$, 3rd value = 4. The median of 4 represents the cluster of values far better.

Outlier → Mean misleading → Median stays typical

Skewed data → median

House prices, incomes, reaction times — use median when a few extreme values exist.

Symmetric data → mean

Heights, temperatures, test scores with no outliers — the mean is fair and efficient.

Always compare both

Calculate mean and median. If they differ a lot, an outlier is probably present.

Watch Me Solve It · 3 examples

Watch Me Solve It · Median of 7 values

+15 XP per step

PROBLEM

Find the median of: 23, 15, 41, 8, 37, 29, 12.

1

Sort in ascending order

8, 12, 15, 23, 29, 37, 41

Never skip this step. The unsorted list has 23 third — that is NOT the median.
2

Find the median position

$n = 7$ (odd). Position = $(7+1) \div 2 = 4$th
3

Read the value at position 4

8, 12, 15, 23, 29, 37, 41 → 4th value = 23

Check: 3 values below (8, 12, 15) and 3 above (29, 37, 41). Balanced. Median = 23.

AnswerMedian = 23

Watch Me Solve It · Median of 6 values

+15 XP per step

PROBLEM

Find the median of: 14, 22, 18, 25, 11, 19.

1

Sort in ascending order

11, 14, 18, 19, 22, 25

$n = 6$ values (even). The position formula gives 3.5 — a decimal signals we average two middle values.
2

Identify the two middle values

Positions 3 and 4: 3rd = 18, 4th = 19
3

Average the two middle values

Median = $(18 + 19) \div 2 = 37 \div 2 = \mathbf{18.5}$

18.5 does not appear in the data — that is perfectly valid for even $n$.

AnswerMedian = 18.5

Watch Me Solve It · Mean vs median with an outlier

+15 XP per step

PROBLEM

Dataset: 2, 3, 4, 5, 100. Compare the mean and median. Which better represents the typical value?

1

Calculate the mean

Mean = $(2+3+4+5+100) \div 5 = 114 \div 5 = \mathbf{22.8}$

Four of the five values are between 2 and 5. A mean of 22.8 seems very wrong for this cluster.
2

Find the median

Data already sorted. $n = 5$, position = 3rd value = 4.
3

Compare and conclude

Mean = 22.8 (distorted by outlier 100). Median = 4 (the typical value).

The outlier (100) adds 96 to the numerator but only 1 to $n$. The median is position-based and is unaffected. Report the median here.

AnswerMedian = 4 (more representative). Mean = 22.8 is distorted by the outlier.

Common Pitfalls

heads-up

Not sorting data first

The median of {7, 3, 9, 1, 5} is NOT 9 (the middle of the unsorted list). The true median, after sorting to {1, 3, 5, 7, 9}, is 5 — a completely different number.

Fix: Always write “Sorted:” as your first step. Never skip it, even if the data looks almost in order.

For even n, picking just one middle value

For {4, 6, 8, 10}, writing “median = 6” or “median = 8” is wrong. You must average both: (6+8)÷2 = 7. The correct median is 7.

Fix: Check whether $n$ is odd or even first. Even $n$ always requires averaging two values.

Confusing the median position with the median value

For $n = 7$, position = 4. Students sometimes write “median = 4”. The position (4) tells you WHERE to look in the ordered list; the value at that position is the actual median.

Fix: Always state “position = 4th, so median = [value at position 4]” to keep the two ideas separate.

Copy Into Your Books

Finding the Median

Step 1: Sort data ascending
Step 2: Position = $(n+1)\div 2$
Odd $n$: value at that position
Even $n$: average the two middle values

Key Facts

Median can be a decimal (e.g. 18.5)
Median need not appear in the original data
Median is resistant to outliers
Mean is NOT resistant

When to Use Median

Data is skewed (e.g. house prices)
Outliers are present
Mean and median differ significantly

Spot the Traps

Must sort before finding median
Even $n$ → average two middles
Position ≠ value
Check balance: equal counts each side

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · The Median

4 problems

Four drill problems to sharpen your median skills. Work each, then reveal the answer.

1 Median of: 3, 7, 12, 8, 5, 9, 4 (sort first!)

Sort: 3, 4, 5, 7, 8, 9, 12. $n=7$, position = 4th.Median = 7
2 Median of: 14, 22, 18, 25, 11, 19

Sort: 11, 14, 18, 19, 22, 25. $n=6$, positions 3 & 4: 18 and 19.Median = (18+19)÷2 = 18.5
3 If $n = 9$, the median is at which position?

Position = $(9+1)\div 2 = 5$.5th position
4 Data: 2, 3, 4, 50. Which is more representative: mean or median?

Mean = (2+3+4+50)÷4 = 14.75. Median = (3+4)÷2 = 3.5. The value 50 is an outlier.Median (3.5) is more representative

Complete in your workbook.

Quick Check · 5 questions

What is the median of: 5, 8, 3, 9, 1?

+10 XP

What is the median of: 10, 15, 20, 25?

+10 XP

Which dataset has a median of 7?

+10 XP

Why is median better than mean for reporting house prices?

+10 XP

For n = 11 values, the median is at which position?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Medium 3 MARKS

Q6. Find the median of these 9 values: 19, 5, 27, 13, 8, 21, 11, 34, 16. Show all steps including sorting.

Answer in your workbook.

Understand Easy 2 MARKS

Q7. Explain in your own words why you must sort the data into order before finding the median. What would go wrong if you didn’t?

Answer in your workbook.

Reason Hard 4 MARKS

Q8. A sports team scored: 45, 62, 38, 71, 55, 200 in 6 games. Calculate both the mean and the median. Which better represents the team’s typical performance? Justify your answer with reference to the outlier.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — 5. Sort: 1, 3, 5, 8, 9. n=5, 3rd value = 5.

2. B — 17.5. Sort: 10, 15, 20, 25. Average positions 2 & 3: (15+20)÷2 = 17.5.

3. A — {6, 7, 8}. Sorted, 2nd value = 7.

4. A — Outlier mansions pull the mean up; median is resistant.

5. C — 6th position. (11+1)÷2 = 6.

Show Your Working Model Answers

Q6 (3 marks): Sort: 5, 8, 11, 13, 16, 19, 21, 27, 34 [1]. n=9, position=(9+1)÷2=5th [1]. 5th value = 16 [1]. Median = 16.

Q7 (2 marks): Without sorting, the position formula identifies the wrong value [1]. For example, {9, 3, 7}: the 2nd value is 3 unsorted, but the median of sorted {3, 7, 9} is 7 — a completely different number [1].

Q8 (4 marks): Sort: 38, 45, 55, 62, 71, 200 [1]. Mean = (38+45+55+62+71+200)÷6 = 471÷6 = 78.5 [1]. Median: n=6, positions 3 & 4: (55+62)÷2 = 58.5 [1]. Median (58.5) is better: the score of 200 is an outlier that pulls the mean (78.5) above 5 of the 6 actual scores. The median of 58.5 better represents the team’s typical performance [1].

Stretch Challenge · +25 XP, +10 coins

The Median Inventor

Invent a dataset of 8 values where the median is exactly 10, the mean is greater than 12, and the range is at least 20. Show all calculations proving all three conditions are satisfied.

Reveal solution

Example dataset: 1, 5, 8, 10, 10, 15, 20, 25. Sorted already. n=8, positions 4 & 5: (10+10)÷2 = 10 ✓. Mean = (1+5+8+10+10+15+20+25)÷8 = 94÷8 = 11.75 — oops, just under 12. Try: 1, 5, 8, 10, 10, 15, 20, 30. Mean = 99÷8 = 12.375 > 12 ✓. Range = 30−1 = 29 ≥ 20 ✓. Many valid answers exist.

Quick Review

Sort First

Always arrange data ascending before applying the position formula.

Odd n: position $(n+1)\div 2$

Gives a whole number — one exact middle value.

Even n: average two middles

Positions $n/2$ and $n/2+1$. Average the two values there.

Resistant to outliers

Median is position-based — one extreme value barely moves it.

Compare to mean

If mean ≠ median by a lot, an outlier is present — use median.

When data is skewed

House prices, incomes, reaction times — always reach for median.

Interactive: Median Explorer

Drag values along a number line and watch how the median changes — especially when you add an extreme outlier.

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