Mathematics 9709 · Probability & Statistics 1

The median & choosing an average

The median is the value in the middle when you line everyone up.
It cares about order, not size.
One extreme value can drag the mean, but it barely moves the median.
That robustness is why median + IQR are the usual choice for skewed data.

Position first, then the value there

First sort the data.
For a discrete ordered list, the median sits at position $\tfrac{n+1}{2}$ .
Odd $n$ : one middle value. Even $n$ : average the two middle values.
From a cumulative frequency graph, enter at $\tfrac n2$ , go across to the curve, then down.
Use $n/2$ on the graph because it is continuous; do not use $(n+1)/2$ there.

Median + IQR from a stem-and-leaf:

Count $n=\sum f$ .
Median position $\tfrac{n+1}{2}$ → read that ordered value.
$Q_1$ = middle of the lower half, $Q_3$ = middle of the upper half.
On the left side of a back-to-back, read leaves right-to-left.
$\text{IQR}=Q_3-Q_1$ ; label median, $Q_1$ , $Q_3$ clearly.

Worked example

Start simple: median + IQR from an ordered list (odd n)

11 ordered times: $31,35,38,42,44,47,51,53,58,62,70$ . Median at the $\tfrac{12}{2}=6$ th → $47$ ; $Q_1$ at the 3rd $=38$ , $Q_3$ at the 9th $=58$ , so $\text{IQR}=20$ . (9709/52/O/N/23 Q4)

Answer

\text{median }47,\ \text{IQR }20

Worked example

Median from a frequency table

$x=40,41,42,43,44$ with $f=11,23,19,8,4$ . $n=65$ , so position $\tfrac{66}{2}=33$ rd. Counting up: 40 covers 1–11, 41 covers 12–34 — the 33rd falls in 41.

Answer

\text{median}=41

Not 42 (the middle x-value): the frequencies decide where the 33rd reading lands.

Estimate the median from a cumulative frequency graph

300 museum artefacts. Enter the CF axis at n/2 = 150, go across to the curve, then down: the median mass is about 2.6 kg.

Worked example

Read the median off the curve (use n/2)

The masses of 300 artefacts are shown by the cumulative frequency curve. For a continuous graph you enter at $\tfrac n2=\tfrac{300}{2}=150$ (not $(n+1)/2$ ), draw across to the curve and drop down.

Answer

\text{median}\approx2.6\text{ kg}

Using $n/2$ (not $(n+1)/2$ ) is what makes counting up from the bottom and down from the top land on the same value.

Common mistake

Reading the left side of a back-to-back diagram left-to-right — its leaves run the other way, so the quartiles come out scrambled. And using

(n+1)/2

on a continuous CF graph (use

n/2

), or

n/2

on an ordered discrete list.

Which average — and reading the skew

Match the average to the job.
The mean chases the tail, so skewed data often needs the median.
The median stays more stable because it only uses position.
Drag the outlier and watch the mean run after it while the median holds:

outlier

mean = 22.0·median = 20·mean > median → positive skew

The question is about…	Use
An outlier / skew	median (robust) — and say WHY
Symmetric data, or further algebra	mean
'Most popular' / qualitative	mode
The spread	IQR (robust) or range — not an average
Comparing two sets' centre	compare medians
Comparing consistency	compare IQRs

Positive skew: the long right tail drags the mean past the median past the mode. Negative skew is the mirror image (mean < median < mode).

Skew or an outlier? → median, and say why (the extreme value). Symmetric or further calc? → mean. Spread question? → IQR or range, not an average.

Examiner note

Identify the median (write “median =” or “

Q_2

=”), don't leave a bare number. State both quartiles and the subtraction for the IQR. For the judgement mark, name the outlier / skew in context — “because of the values” scores zero.

Your turn— tap to reveal the worked answer (outlier robustness)

House prices ($000s): 220, 236, 242, 350, 480, 3500. Compare mean and median.

Median (even $n$ ) $=\tfrac{242+350}{2}=296$ ; mean $=\tfrac{5028}{6}=838$ . The lone mansion drags the mean to a price nobody paid.

Answer

\text{median }\$296{,}000\ \text{vs mean }\$838{,}000

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The median & choosing an average

The median is the value in the middle when you line everyone up.
It cares about order, not size.
One extreme value can drag the mean, but it barely moves the median.
That robustness is why median + IQR are the usual choice for skewed data.

Position first, then the value there

First sort the data.
For a discrete ordered list, the median sits at position $\tfrac{n+1}{2}$ .
Odd $n$ : one middle value. Even $n$ : average the two middle values.
From a cumulative frequency graph, enter at $\tfrac n2$ , go across to the curve, then down.
Use $n/2$ on the graph because it is continuous; do not use $(n+1)/2$ there.

Median + IQR from a stem-and-leaf:

Count $n=\sum f$ .
Median position $\tfrac{n+1}{2}$ → read that ordered value.
$Q_1$ = middle of the lower half, $Q_3$ = middle of the upper half.
On the left side of a back-to-back, read leaves right-to-left.
$\text{IQR}=Q_3-Q_1$ ; label median, $Q_1$ , $Q_3$ clearly.

Worked example

Start simple: median + IQR from an ordered list (odd n)

11 ordered times: $31,35,38,42,44,47,51,53,58,62,70$ . Median at the $\tfrac{12}{2}=6$ th → $47$ ; $Q_1$ at the 3rd $=38$ , $Q_3$ at the 9th $=58$ , so $\text{IQR}=20$ . (9709/52/O/N/23 Q4)

Answer

\text{median }47,\ \text{IQR }20

Worked example

Median from a frequency table

$x=40,41,42,43,44$ with $f=11,23,19,8,4$ . $n=65$ , so position $\tfrac{66}{2}=33$ rd. Counting up: 40 covers 1–11, 41 covers 12–34 — the 33rd falls in 41.

Answer

\text{median}=41

Not 42 (the middle x-value): the frequencies decide where the 33rd reading lands.

Estimate the median from a cumulative frequency graph

300 museum artefacts. Enter the CF axis at n/2 = 150, go across to the curve, then down: the median mass is about 2.6 kg.

Worked example

Read the median off the curve (use n/2)

The masses of 300 artefacts are shown by the cumulative frequency curve. For a continuous graph you enter at $\tfrac n2=\tfrac{300}{2}=150$ (not $(n+1)/2$ ), draw across to the curve and drop down.

Answer

\text{median}\approx2.6\text{ kg}

Using $n/2$ (not $(n+1)/2$ ) is what makes counting up from the bottom and down from the top land on the same value.

Common mistake

Reading the left side of a back-to-back diagram left-to-right — its leaves run the other way, so the quartiles come out scrambled. And using

(n+1)/2

on a continuous CF graph (use

n/2

), or

n/2

on an ordered discrete list.

Which average — and reading the skew

Match the average to the job.
The mean chases the tail, so skewed data often needs the median.
The median stays more stable because it only uses position.
Drag the outlier and watch the mean run after it while the median holds:

outlier

mean = 22.0·median = 20·mean > median → positive skew

The question is about…	Use
An outlier / skew	median (robust) — and say WHY
Symmetric data, or further algebra	mean
'Most popular' / qualitative	mode
The spread	IQR (robust) or range — not an average
Comparing two sets' centre	compare medians
Comparing consistency	compare IQRs

Positive skew: the long right tail drags the mean past the median past the mode. Negative skew is the mirror image (mean < median < mode).

Skew or an outlier? → median, and say why (the extreme value). Symmetric or further calc? → mean. Spread question? → IQR or range, not an average.

Examiner note

Identify the median (write “median =” or “

Q_2

=”), don't leave a bare number. State both quartiles and the subtraction for the IQR. For the judgement mark, name the outlier / skew in context — “because of the values” scores zero.

Your turn— tap to reveal the worked answer (outlier robustness)

House prices ($000s): 220, 236, 242, 350, 480, 3500. Compare mean and median.

Median (even $n$ ) $=\tfrac{242+350}{2}=296$ ; mean $=\tfrac{5028}{6}=838$ . The lone mansion drags the mean to a price nobody paid.

Answer

\text{median }\$296{,}000\ \text{vs mean }\$838{,}000