Standard Deviation, Quartiles & Percentiles • Topic 2 of 4

Quartiles & the Interquartile Range

Quartiles cut a sorted data set into four equal-count parts. The second quartile (Q2) is just the median. The first quartile (Q1) is the median of the lower half, and the third quartile (Q3) is the median of the upper half. The interquartile range, IQR = Q3 − Q1, captures the spread of the middle 50% of the data and, crucially, ignores the extremes — which makes it a robust companion to the range in exactly the way the median is a robust companion to the mean. On the GRE the reliable method is: sort, find the median, then find the median of each half. A common convention question is whether the overall median is included in the halves; for GRE purposes, when the count is odd, leave the middle value out of both halves. The IQR is what a boxplot draws as its central box, so this topic and boxplots are two views of the same idea.

A boxplot of the five-number summary with the interquartile range markedBoxplot: five-number summaryIQR = Q3 - Q1 = 8min 2Q1 5median 9Q3 13max 16

✅ Solved examples

1. Find Q1, Q2 and Q3 for 2, 4, 6, 8, 10, 12, 14, 16.
Sorted, 8 values. Median Q2 = average of 4th and 5th = (8+10)/2 = 9. Lower half {2,4,6,8}: Q1 = (4+6)/2 = 5. Upper half {10,12,14,16}: Q3 = (12+14)/2 = 13.
2. Find the IQR of the set above.
IQR = Q3 − Q1 = 13 − 5 = 8.
3. Find Q1 and Q3 for 5, 7, 9, 11, 13 (odd count).
Median Q2 = 9 (the 3rd value); exclude it. Lower half {5, 7}: Q1 = (5+7)/2 = 6. Upper half {11, 13}: Q3 = (11+13)/2 = 12.
4. Quantitative Comparison. For the set 1, 3, 3, 5, 7, 9, 11 — Quantity A: the IQR. Quantity B: the range minus 6. Which is greater?
Sorted, 7 values, median = 5 (4th), excluded. Lower half {1,3,3}: Q1 = 3. Upper half {9,11} plus 7 → {7,9,11}: Q3 = 9. IQR = 9 − 3 = 6. Range = 11 − 1 = 10, so range − 6 = 4. Quantity A (6) > Quantity B (4). Answer: A is greater.

✏️ Practice — try these, take hints as needed

1. Find the median (Q2) of 3, 6, 9, 12, 15, 18.
Even count of 6.
Average the 3rd and 4th values.
(9 + 12)/2.
10.5
2. Find Q1 for 10, 20, 30, 40, 50, 60, 70, 80.
Median splits into halves {10,20,30,40} and {50,60,70,80}.
Q1 = median of the lower half.
(20 + 30)/2.
25
3. A data set has Q1 = 18 and Q3 = 42. Find its IQR.
IQR = Q3 − Q1.
42 − 18.
Just subtract.
24
4. For 4, 8, 12, 16, 20 (odd count), find Q3.
Median = 12, the 3rd value; exclude it.
Upper half is {16, 20}.
Average them.
18
5. Which is more affected by an extreme high outlier: the IQR or the range?
The range uses the maximum directly.
The IQR uses only the middle 50%.
An outlier sits outside Q1–Q3.
The range (the IQR is robust to outliers)

📝 Topic test — 8 questions

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