Split the data in half, then split each half again.
The data
3, 7, 8, 12, 13, 14, 18, 21, 22, 25
Find Q1.
With ten values the median falls between the 5th and 6th (13.5), splitting the list into two halves of five. Q1 is the median of the lower five — 3, 7, 8, 12, 13 — which is 8. B averages 8 and 12, interpolating where a clean middle value already exists.
This list is already in order. If it were not, that would be step zero.
n = 10, so the middle is between the 5th and 6th
an even count
(13 + 14) ÷ 2
average the two
median = 13.5
also called Q2
five below, five above
lower half: 3, 7, 8, 12, 13
five values
Q1 = 8
the middle one
upper half: 14, 18, 21, 22, 25
the other five
Q3 = 21
its middle
The median itself is left out of both halves. For 7 values the median is the 4th, and each half has the remaining three.
IQR = Q3 − Q1
the spread of the middle half
= 21 − 8 = 13
where half the data lives
25 − 3 = 22
one wild value ruins it
13
ignores both extremes
The IQR is the resistant measure of spread, just as the median is the resistant measure of centre.
1.5 × IQR = 19.5
the fence distance
lower fence = 8 − 19.5 = −11.5
Q1 minus it
upper fence = 21 + 19.5 = 40.5
Q3 plus it
nothing outside → no outliers
3 and 25 are both inside
Add a value of 60 and the upper fence still sits at 40.5 — 60 would be flagged as an outlier while the quartiles barely move.
| Part of the plot | Holds |
|---|---|
| left whisker | the bottom 25% |
| the box | the middle 50% |
| right whisker | the top 25% |
| line in the box | the median |
Each quarter holds the same number of values. A stretched whisker means those values are spread out, not that there are more of them.
9 values: median = 5th value = 13
leave it out of both halves
lower half 4, 6, 9, 10: Q1 = 6 + 92 = 7.5
four values, so average the middle two
upper half 15, 16, 20, 24: Q3 = 16 + 202 = 18
IQR = 18 − 7.5 = 10.5
Unsorted, with a repeat, halves of four. Find Q1, Q3 and the IQR of 7, 3, 9, 3, 12, 5, 10, 6.
sorted: 3, 3, 5, 6, 7, 9, 10, 12
both 3s stay in
lower half 3, 3, 5, 6: Q1 = (3 + 5) ÷ 2 = 4
even half: average the middle two
upper half 7, 9, 10, 12: Q3 = (9 + 10) ÷ 2 = 9.5
the median 6.5 splits the halves
IQR = 9.5 − 4 = 5.5
What is different: sort first and keep every copy of a repeated value. With 8 values each half has 4, so the quartiles themselves are averages.
Writing the 3 once leaves seven values: 3, 5, 6, 7, 9, 10, 12, with Q1 = 5 and Q3 = 10. Every value counts, however many times it appears.
Backwards, with an unknown value: 2, 5, y, 9, 12, 14, 17, 21 are in order and Q1 = 6. The lower half is 2, 5, y, 9, so Q1 = (5 + y) ÷ 2 = 6, giving y = 7. Check the order: 5 ≤ 7 ≤ 9 ✓.
A low value under test. Is 2 an outlier in 2, 18, 20, 21, 24, 26, 30, 31, 33?
median = 24; Q1 = (18 + 20) ÷ 2 = 19, Q3 = (30 + 31) ÷ 2 = 30.5
9 values: leave 24 out of the halves
IQR = 11.5, 1.5 × IQR = 17.25
the fence distance
lower fence = 19 − 17.25 = 1.75
the side that matters for 2
2 > 1.75: not an outlier
it sits just inside
What is different: the suspect value is at the bottom, so the lower fence decides. Looking far from the rest is not enough; only the fence test counts.
× 3: every gap triples, so the IQR becomes 18
Q1 and Q3 both triple
− 4: both quartiles move down 4
the gap between them does not change
new IQR = 18
| Change to every value | Quartiles | IQR |
|---|---|---|
| add or subtract c | move by c | unchanged |
| multiply by k (k > 0) | × k | × k |
14 takes the 4 off the gap. Shifting every value moves Q1 and Q3 together, so their difference stays the same.
Expect Q1 and Q3 of a short list (odd and even counts), a box plot to read, the 1.5 × IQR outlier fences, or the effect of a change to every value. Sort first, and leave the median out of both halves when the count is odd. Ten-second check: Q1 ≤ median ≤ Q3, and about half the data lie between Q1 and Q3.
The data
12, 15, 18, 20, 22, 25, 28, 31, 45
Is 45 an outlier?
Q1 = 16.5 and Q3 = 29.5, so the IQR is 13 and the upper fence is Q3 + 1.5 × IQR = 29.5 + 19.5 = 49. Since 45 < 49 it is not an outlier. A judges by eye — the 1.5 × IQR rule exists precisely so that "looks far away" is not the test.
