Quartiles and Percentiles

The median splits a data set in half. Quartiles go one step further and split it into quarters, and percentiles split it into hundredths. They tell you where a value stands compared with the rest (“better than $80\%$ of the class”) and give a way to measure spread and spot outliers that isn’t fooled by extreme values. A boxplot puts all of this into one simple picture.

Key ideas

Quartiles

Quartiles divide ordered data into four parts with about the same number of values in each.

• $Q_1$ (the lower quartile) is the median of the lower half of the data.
• $Q_2$ is the median of the whole data set.
• $Q_3$ (the upper quartile) is the median of the upper half.

The method used on this site:

1. Put the data in order and find the median.
2. Split the data into a lower half and an upper half. If $n$ is odd, leave the median out of both halves. If $n$ is even, the halves are simply the first $\dfrac{n}{2}$ and the last $\dfrac{n}{2}$ values.
3. $Q_1$ is the median of the lower half, and $Q_3$ is the median of the upper half.

Graphing calculators, spreadsheets, and statistics software don’t all use this exact method. (For example, a spreadsheet’s QUARTILE function uses a different rule.) Their answers may differ slightly from yours, especially for small data sets. That’s normal: just be consistent and say which method you used.

Interquartile range

The interquartile range is the spread of the middle half of the data:

$$\text{IQR} = Q_3 - Q_1$$

Unlike the range (maximum minus minimum), the IQR is barely affected by outliers, because it ignores the lowest and highest quarters. For other measures of spread, see standard deviation.

Five-number summary

The five-number summary of a data set is

$$\text{minimum}, \quad Q_1, \quad \text{median}, \quad Q_3, \quad \text{maximum}$$

Outliers: the 1.5 × IQR rule

A value is an outlier if it is more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$. The cut-offs are called fences:

$$\text{lower fence} = Q_1 - 1.5 \times \text{IQR} \qquad \text{upper fence} = Q_3 + 1.5 \times \text{IQR}$$

Any value below the lower fence or above the upper fence is an outlier.

Boxplots

A boxplot (or box-and-whisker plot) shows the five-number summary on a number line:

• the box goes from $Q_1$ to $Q_3$, with a line at the median;
• the whiskers go from the box out to the smallest and largest values that are not outliers;
• outliers are plotted as separate dots.

The boxplot for Example 2. The whisker stops at $130$, the largest value that isn’t an outlier.

Each of the four sections (left whisker, left half of the box, right half of the box, right whisker) holds about a quarter of the data. A long section doesn’t mean more data; it means the data in that quarter are more spread out.

Percentiles and percentile rank

The $k$th percentile is the value that about $k\%$ of the data fall below. On this site, the percentile rank of a value is the percentage of the data that are below it:

$$\text{percentile rank} = \frac{\text{number of values below } x}{n} \times 100$$

rounded to a whole number. (Some textbooks also count half of the values equal to $x$; the answers come out close.)

Quartiles are special percentiles: $Q_1$ is about the 25th percentile, the median is the 50th, and $Q_3$ is about the 75th.

A percentile is not a percentage mark. Being at the 90th percentile on a test means you did better than about $90\%$ of the people who wrote it, not that you scored $90\%$.

When data follow a normal distribution, you can find percentiles from z-scores instead of from a list of values.

Worked examples

Example 1: Quartiles when n is odd

Find the quartiles and the IQR of these $9$ test scores:

$$55,\ 62,\ 68,\ 70,\ 74,\ 79,\ 81,\ 88,\ 93$$

Solution. The data are already in order. With $n = 9$, the median is the 5th value: $74$.

Leave the median out. The lower half is $55, 62, 68, 70$, so

$$Q_1 = \frac{62 + 68}{2} = 65$$

The upper half is $79, 81, 88, 93$, so

$$Q_3 = \frac{81 + 88}{2} = 84.5$$ $$\text{IQR} = 84.5 - 65 = 19.5$$

Example 2: Five-number summary, outliers, and a boxplot

Twelve students recorded their daily phone use in minutes:

$$45,\ 60,\ 72,\ 80,\ 85,\ 90,\ 95,\ 100,\ 110,\ 120,\ 130,\ 240$$

Find the five-number summary, check for outliers, and describe the boxplot.

Solution. With $n = 12$ (even), the median is the mean of the 6th and 7th values:

$$\text{median} = \frac{90 + 95}{2} = 92.5$$

The lower half is the first six values, $45, 60, 72, 80, 85, 90$, so $Q_1 = \dfrac{72 + 80}{2} = 76$.

The upper half is $95, 100, 110, 120, 130, 240$, so $Q_3 = \dfrac{110 + 120}{2} = 115$.

Five-number summary: $45,\ 76,\ 92.5,\ 115,\ 240$.

Now the fences:

$$\begin{aligned} \text{IQR} &= 115 - 76 = 39 \\ \text{lower fence} &= 76 - 1.5(39) = 76 - 58.5 = 17.5 \\ \text{upper fence} &= 115 + 1.5(39) = 115 + 58.5 = 173.5 \end{aligned}$$

No value is below $17.5$, but $240 \gt 173.5$, so $240$ is an outlier.

In the boxplot (shown above), the box runs from $76$ to $115$ with a line at $92.5$. The left whisker goes down to $45$. The right whisker stops at $130$, the largest value that isn’t an outlier, and $240$ is drawn as a separate dot.

Example 3: Percentile rank

In a class of $30$, Jun’s mark on a test was higher than the marks of $24$ classmates. Find Jun’s percentile rank and explain what it means.

Solution.

$$\text{percentile rank} = \frac{24}{30} \times 100 = 80$$

Jun is at the 80th percentile: about $80\%$ of the class scored lower than he did. This says nothing about his actual mark. On a hard test, the 80th percentile might be a $68\%$.

Example 4: Comparing two boxplots

Two Grade 12 classes wrote the same test. Their five-number summaries are:

	Min	$Q_1$	Median	$Q_3$	Max
Class A	$52$	$64$	$71$	$78$	$90$
Class B	$58$	$70$	$74$	$77$	$85$

Compare the two classes.

Solution.

• Centre: Class B’s median ($74$) is higher than Class A’s ($71$), so a typical student in B did a little better.
• Spread: Class A’s IQR is $78 - 64 = 14$, and Class B’s is $77 - 70 = 7$. The middle half of Class B is much more tightly grouped, so B’s results are more consistent.
• Extremes: Class A had the highest mark ($90$), but also the lowest ($52$).
• In Class B, about $75\%$ of students scored $70$ or more (since $Q_1 = 70$). In Class A, $70$ is between $Q_1 = 64$ and the median, so somewhere between $50\%$ and $75\%$ scored that well.

Common mistakes

Forgetting to sort the data. Quartiles, like the median, only make sense for ordered data.

Switching quartile methods partway through. With the method on this site, leave the median out when $n$ is odd. Including it is another method some books and calculators use, and it gives slightly different quartiles, so pick one method and stick to it.

Drawing the whisker all the way to an outlier. Whiskers stop at the most extreme values that are not outliers. Outliers get their own dot.

Thinking a longer section of a boxplot holds more data. Every section holds about $25\%$ of the data. A longer section means those values are more spread out.

Confusing percentile with percent. The 70th percentile means “better than about $70\%$ of people”, not “a mark of $70\%$”.

Using 1.5 × IQR from the median. The fences are measured from $Q_1$ and $Q_3$, not from the median.

Practice

1. (Warm-up) Find the five-number summary: $3, 5, 7, 8, 12, 13, 14, 18, 21$.

Solution

$n = 9$, so the median is the 5th value, $12$. Lower half: $3, 5, 7, 8$, so $Q_1 = \dfrac{5 + 7}{2} = 6$. Upper half: $13, 14, 18, 21$, so $Q_3 = \dfrac{14 + 18}{2} = 16$.

Five-number summary: $3,\ 6,\ 12,\ 16,\ 21$.

2. (Warm-up) Find $Q_1$, the median, $Q_3$, and the IQR: $22, 25, 27, 30, 31, 34, 36, 38, 41, 45$.

Solution

$n = 10$, so the median is $\dfrac{31 + 34}{2} = 32.5$.

Lower half: $22, 25, 27, 30, 31$, so $Q_1 = 27$. Upper half: $34, 36, 38, 41, 45$, so $Q_3 = 38$.

$\text{IQR} = 38 - 27 = 11$.

3. (Warm-up) A doctor says a toddler’s height is at the 85th percentile for her age. What does this mean?

Solution

About $85\%$ of toddlers her age are shorter than she is (and about $15\%$ are taller). It doesn’t mean anything is "$85\%$" about her height.

4. (Core) Use the $1.5 \times \text{IQR}$ rule to check for outliers: $12, 15, 16, 18, 19, 20, 22, 23, 25, 41$.

Solution

$n = 10$. Lower half: $12, 15, 16, 18, 19$, so $Q_1 = 16$. Upper half: $20, 22, 23, 25, 41$, so $Q_3 = 23$.

$$\begin{aligned} \text{IQR} &= 23 - 16 = 7 \\ \text{lower fence} &= 16 - 10.5 = 5.5 \\ \text{upper fence} &= 23 + 10.5 = 33.5 \end{aligned}$$

$41 \gt 33.5$, so $41$ is an outlier. No value is below $5.5$.

5. (Core) A boxplot of quiz scores has this five-number summary: $40,\ 55,\ 62,\ 70,\ 95$.

• (a) About what percentage of scores are between $55$ and $70$?
• (b) About what percentage are above $62$?
• (c) Is the maximum, $95$, an outlier?

Solution

(a) About $50\%$ (the box holds the middle half).

(b) About $50\%$ ($62$ is the median).

(c) $\text{IQR} = 70 - 55 = 15$, and the upper fence is $70 + 1.5(15) = 92.5$. Since $95 \gt 92.5$, yes, $95$ is an outlier. The boxplot should show it as a dot, with the whisker ending at the largest score that isn’t an outlier.

6. (Core) In a class of $32$ students, Leo’s mark was higher than $26$ other students’ marks. Find his percentile rank.

Solution

$$\frac{26}{32} \times 100 = 81.25$$

Leo is at about the 81st percentile.

7. (Core) Two track teams recorded their long jumps in metres.

	Min	$Q_1$	Median	$Q_3$	Max
Team X	$3.1$	$3.8$	$4.2$	$4.5$	$5.0$
Team Y	$2.9$	$3.5$	$4.0$	$4.9$	$5.6$

• (a) Which team had the longest single jump?
• (b) Which team has the higher median?
• (c) Find each team’s IQR. Which team is more consistent?

Solution

(a) Team Y ($5.6$ m).

(b) Team X ($4.2$ m compared with $4.0$ m).

(c) Team X: $4.5 - 3.8 = 0.7$ m. Team Y: $4.9 - 3.5 = 1.4$ m. Team X is more consistent: the middle half of its jumps are spread over half the distance.

8. (Challenge) The ordered data set $4, 7, x, 10, 12, 15, 19, 22$ has an IQR of $9$, where $7 \le x \le 10$. Find $x$.

Solution

$n = 8$. The upper half is $12, 15, 19, 22$, so $Q_3 = \dfrac{15 + 19}{2} = 17$.

The lower half is $4, 7, x, 10$, so $Q_1 = \dfrac{7 + x}{2}$.

$$\begin{aligned} 17 - \frac{7 + x}{2} &= 9 \\ \frac{7 + x}{2} &= 8 \\ x &= 9 \end{aligned}$$

Check: $Q_1 = \dfrac{7 + 9}{2} = 8$ and $\text{IQR} = 17 - 8 = 9$. ✓

9. (Challenge) For the data in Question 1, a classmate includes the median in both halves when finding the quartiles. What does the classmate get for $Q_1$ and $Q_3$? Why do the answers differ, and which is right?

Solution

Including the median $12$, the lower half is $3, 5, 7, 8, 12$, so $Q_1 = 7$. The upper half is $12, 13, 14, 18, 21$, so $Q_3 = 14$.

With our method, $Q_1 = 6$ and $Q_3 = 16$. The answers differ because the halves contain different values. Neither is “wrong”: statisticians use several methods, and calculators and spreadsheets pick different ones. The differences shrink as data sets get larger. What matters is using one method consistently and saying which one you used.