Cumulative Frequency

When data come in a grouped frequency table, you no longer know the individual values, so you can’t line them up and count to the middle. A cumulative frequency graph gets around this: it shows how many values lie below each point, so you can read off the median, the quartiles, any percentile, and how many values are above or below a cut-off. It’s the standard IB tool for summarizing large grouped data sets, and it feeds straight into a box-and-whisker diagram.

Key ideas

Cumulative frequency

The cumulative frequency of a class is the running total of the frequencies up to and including that class. It tells you how many values are less than or equal to the upper end of the class.

In IB questions, classes are written as inequalities with no gaps, like $10 \lt t \le 20$. The upper end of each class ($20$ here) is its upper class boundary.

Commute time $t$ (min)	Frequency	Upper boundary	Cumulative frequency
$0 \lt t \le 10$	$6$	$10$	$6$
$10 \lt t \le 20$	$14$	$20$	$6 + 14 = 20$
$20 \lt t \le 30$	$24$	$30$	$20 + 24 = 44$
$30 \lt t \le 40$	$20$	$40$	$64$
$40 \lt t \le 50$	$10$	$50$	$74$
$50 \lt t \le 60$	$6$	$60$	$80$

The last cumulative frequency is always the total, $n$. Here $n = 80$.

Drawing the graph

Plot each cumulative frequency against its upper class boundary, because that’s the point by which all those values have been counted. Start at the lower boundary of the first class with a cumulative frequency of $0$ (here, the point $(0, 0)$). Then join the points with a smooth curve or with straight line segments. The graph never goes down, and it often has a stretched S shape. Some books call it an ogive.

Reading values off the graph

To find the value below which a certain number of data lie, go across from the cumulative frequency axis to the graph, then down to the horizontal axis.

Statistic	Go across from
Lower quartile $Q_1$	$\dfrac{n}{4}$
Median	$\dfrac{n}{2}$
Upper quartile $Q_3$	$\dfrac{3n}{4}$
$k$th percentile	$\dfrac{k}{100} \times n$

For a large grouped data set, use $\dfrac{n}{2}$ for the median (not $\dfrac{n+1}{2}$, which is for listing individual values). Then $\text{IQR} = Q_3 - Q_1$.

To go the other way (how many values are below a given $t$), go up from $t$ to the graph, then across. The number above $t$ is $n$ minus that reading.

Values read from a graph are estimates. Different smooth curves give slightly different readings, and IB mark schemes accept a small range. On this page the points are joined with straight segments, so the readings can be checked with a little proportion (linear interpolation).

Range

The range is the largest value minus the smallest. From grouped data you don’t know these exactly, so a question will either give you the minimum and maximum, or you can only say the range is at most the largest upper boundary minus the smallest lower boundary.

From the graph to a box-and-whisker diagram

A box-and-whisker diagram needs the five-number summary: minimum, $Q_1$, median, $Q_3$, maximum. The cumulative frequency graph gives the middle three; the question supplies the minimum and maximum. Outliers (more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$, see quartiles and percentiles) are marked with a cross, and the whisker stops at the most extreme value that isn’t an outlier.

If the box and whiskers are roughly symmetric about the median, the data may be normally distributed. A clearly lopsided box plot suggests they aren’t. (See the normal distribution.)

Worked examples

Example 1: Building the table and the points

The commute times of $80$ students are in the table in Key ideas. Write down the points you would plot for the cumulative frequency graph.

Solution. Plot (upper boundary, cumulative frequency), starting at $(0, 0)$:

$$(0, 0),\ (10, 6),\ (20, 20),\ (30, 44),\ (40, 64),\ (50, 74),\ (60, 80)$$

Check: the last cumulative frequency, $80$, matches the number of students.

Cumulative frequency graph of the commute times. Going across from $20$, $40$ and $60$ gives $Q_1 = 20$, median $\approx 28.3$ and $Q_3 = 38$.

Example 2: Median, quartiles and IQR

Use the graph to estimate the median, the quartiles and the interquartile range of the commute times.

Solution. With $n = 80$:

• Median: go across from $\dfrac{80}{2} = 40$. The graph reaches $40$ at about $t = 28.3$.
• $Q_1$: go across from $\dfrac{80}{4} = 20$. This is exactly the plotted point $(20, 20)$, so $Q_1 = 20$.
• $Q_3$: go across from $\dfrac{3 \times 80}{4} = 60$. This gives $Q_3 = 38$.

$$\text{IQR} = Q_3 - Q_1 = 38 - 20 = 18 \text{ minutes}$$

Check the median by proportion: in the $20 \lt t \le 30$ class the cumulative frequency goes from $20$ to $44$, so reaching $40$ means going $20$ of the way through that class’s $24$ students:

$$\text{median} \approx 20 + \frac{40 - 20}{24} \times 10 = 28.3 \text{ min (3 s.f.)}$$

Example 3: Percentiles and counting above a value

Use the graph to estimate:

• (a) the $90$th percentile of the commute times;
• (b) the number of students whose commute is longer than $45$ minutes;
• (c) the number of students whose commute is between $15$ and $35$ minutes.

Solution.

(a) Go across from $0.90 \times 80 = 72$. That’s in the segment from $(40, 64)$ to $(50, 74)$:

$$t \approx 40 + \frac{72 - 64}{10} \times 10 = 48 \text{ minutes}$$

So $90\%$ of the students commute for $48$ minutes or less.

(b) Go up from $t = 45$: halfway between $(40, 64)$ and $(50, 74)$, the cumulative frequency is about $69$. So about $69$ students take $45$ minutes or less, and

$$80 - 69 = 11 \text{ students}$$

take longer than $45$ minutes.

(c) Up from $t = 15$: about $13$ students. Up from $t = 35$: about $54$ students. So about $54 - 13 = 41$ students commute for between $15$ and $35$ minutes.

Example 4: Drawing the box-and-whisker diagram

For the same students, the shortest commute was $3$ minutes and the longest was $58$ minutes. Draw a box-and-whisker diagram, check for outliers, and comment on whether the commute times could be normally distributed.

Solution. Five-number summary: $3$, $20$, $28.3$, $38$, $58$.

Outliers. With $\text{IQR} = 18$, $1.5 \times 18 = 27$:

$$Q_1 - 27 = -7 \qquad Q_3 + 27 = 65$$

No value is below $-7$ or above $65$, so there are no outliers. The whiskers run from $3$ to $20$ and from $38$ to $58$.

The diagram. On a number line from $0$ to $60$, draw a box from $20$ to $38$ with a line at $28.3$, and whiskers out to $3$ and $58$.

Symmetry. The two halves of the box are $28.3 - 20 = 8.3$ and $38 - 28.3 = 9.7$ minutes wide, and the whiskers are $17$ and $20$ minutes long. That’s close to symmetric, so the commute times could be roughly normally distributed (a box plot can only suggest this, not prove it).

Common mistakes

Plotting at the class midpoint. Midpoints are for estimating the mean. A cumulative frequency counts everything up to the end of the class, so plot it at the upper boundary.

Forgetting the starting point. The graph starts at the lower boundary of the first class with cumulative frequency $0$. Without it, you can’t read anything in the first class.

Reading the wrong axis. The median is a value of the variable (minutes), not a frequency. Go across from $\dfrac{n}{2}$ on the cumulative frequency axis and read the answer on the horizontal axis.

Answering “how many are above” with the graph reading. The graph gives the number below a value. For “more than $45$ minutes”, subtract the reading from the total: $80 - 69 = 11$, not $69$.

Giving the IQR as a pair of numbers. The IQR is a single number, $Q_3 - Q_1$. “From $20$ to $38$” describes the box, but the IQR is $18$.

Calling the range “60 minus 0”. That’s only the widest the range could be. Use the actual minimum and maximum if they’re given ($58 - 3 = 55$ minutes in Example 4).

Practice

1. (Warm-up) The heights of $40$ students are recorded.

Height $h$ (cm)	$150 \lt h \le 155$	$155 \lt h \le 160$	$160 \lt h \le 165$	$165 \lt h \le 170$	$170 \lt h \le 175$
Frequency	$4$	$9$	$15$	$8$	$4$

Write down the cumulative frequencies and the points you would plot.

Solution

Cumulative frequencies: $4$, $13$, $28$, $36$, $40$.

Points (upper boundary, cumulative frequency), starting at the lower boundary of the first class:

$$(150, 0),\ (155, 4),\ (160, 13),\ (165, 28),\ (170, 36),\ (175, 40)$$

2. (Warm-up) For the heights in question 1, which cumulative frequency do you go across from to find (a) the median, (b) $Q_1$, (c) the $90$th percentile? Then estimate the median, assuming the points are joined with straight lines.

Solution

(a) $\dfrac{40}{2} = 20$. (b) $\dfrac{40}{4} = 10$. (c) $0.9 \times 40 = 36$.

The median is in the $160 \lt h \le 165$ class, where the cumulative frequency goes from $13$ to $28$:

$$\text{median} \approx 160 + \frac{20 - 13}{15} \times 5 = 162 \text{ cm (3 s.f.)}$$

(More precisely $162.33\ldots$ cm.) Notice that the $90$th percentile is exactly the point $(170, 36)$, so it’s $170$ cm.

3. (Core) The masses of $200$ eggs from a farm are shown.

Mass $m$ (g)	$40 \lt m \le 45$	$45 \lt m \le 50$	$50 \lt m \le 55$	$55 \lt m \le 60$	$60 \lt m \le 65$	$65 \lt m \le 70$
Frequency	$12$	$28$	$50$	$62$	$34$	$14$

Write down the cumulative frequencies. Then, joining the points with straight lines, estimate the median and the lower quartile.

Solution

Cumulative frequencies: $12$, $40$, $90$, $152$, $186$, $200$.

Median: go across from $100$. It lies between $(55, 90)$ and $(60, 152)$:

$$\text{median} \approx 55 + \frac{100 - 90}{62} \times 5 = 55.8 \text{ g (3 s.f.)}$$

$Q_1$: go across from $50$. It lies between $(50, 40)$ and $(55, 90)$:

$$Q_1 \approx 50 + \frac{50 - 40}{50} \times 5 = 51 \text{ g}$$

4. (Core) For the eggs in question 3, estimate the upper quartile and the interquartile range.

Solution

$Q_3$: go across from $\dfrac{3 \times 200}{4} = 150$, between $(55, 90)$ and $(60, 152)$:

$$Q_3 \approx 55 + \frac{150 - 90}{62} \times 5 = 59.8 \text{ g (3 s.f.)}$$$$\text{IQR} \approx 59.84 - 51 = 8.84 \text{ g (3 s.f.)}$$

5. (Core) Eggs heavier than $62$ g are sold as “extra large”. Estimate how many of the $200$ eggs are extra large.

Solution

Go up from $m = 62$. It’s $\dfrac{2}{5}$ of the way from $(60, 152)$ to $(65, 186)$:

$$152 + \frac{2}{5} \times 34 = 165.6$$

So about $166$ eggs are $62$ g or lighter, and about $200 - 165.6 \approx 34$ eggs are extra large.

6. (Core) For the eggs, estimate (a) the $80$th percentile, and (b) the percentage of eggs lighter than $48$ g.

Solution

(a) Go across from $0.8 \times 200 = 160$, between $(60, 152)$ and $(65, 186)$:

$$60 + \frac{160 - 152}{34} \times 5 = 61.2 \text{ g (3 s.f.)}$$

(b) Go up from $48$, which is $\dfrac{3}{5}$ of the way from $(45, 12)$ to $(50, 40)$:

$$12 + \frac{3}{5} \times 28 = 28.8$$

$\dfrac{28.8}{200} = 0.144$, so about $14.4\%$ of the eggs are lighter than $48$ g.

7. (Core) The lightest egg in question 3 was $41$ g and the heaviest was $69$ g. Using $Q_1 = 51$, median $= 55.8$ and $Q_3 = 59.8$, decide whether there are any outliers and describe the box-and-whisker diagram. Is the distribution roughly symmetric?

Solution

$\text{IQR} = 59.8 - 51 = 8.8$, and $1.5 \times 8.8 = 13.2$. The fences are $51 - 13.2 = 37.8$ and $59.8 + 13.2 = 73$. Both $41$ and $69$ are inside, so there are no outliers.

The box runs from $51$ to $59.8$ with the median at $55.8$; the whiskers run to $41$ and $69$.

The halves of the box are $4.8$ and $4.0$ g wide and the whiskers are $10$ and $9.2$ g long, so the distribution is roughly symmetric (slightly longer on the left). The masses could be roughly normally distributed.

8. (Challenge) The table shows the scores of $100$ players in an online game.

Score $x$	$0 \lt x \le 10$	$10 \lt x \le 20$	$20 \lt x \le 30$	$30 \lt x \le 40$	$40 \lt x \le 50$
Frequency	$8$	$p$	$30$	$q$	$12$

Joining the points of the cumulative frequency graph with straight lines gives a median of $26$. Find $p$ and $q$.

Solution

The frequencies add to $100$: $8 + p + 30 + q + 12 = 100$, so $p + q = 50$.

The median, $26$, is in the $20 \lt x \le 30$ class. The cumulative frequency at $20$ is $8 + p$, and across that class it rises by $30$. Going $6$ of the $10$ units into the class adds $\dfrac{6}{10} \times 30 = 18$, and this must reach $\dfrac{100}{2} = 50$:

$$8 + p + 18 = 50 \quad\Rightarrow\quad p = 24$$

Then $q = 50 - 24 = 26$.

Check: cumulative frequencies $8, 32, 62, 88, 100$; median $= 20 + \dfrac{50 - 32}{30} \times 10 = 26$. ✓

9. (Challenge) At a second school, a cumulative frequency graph of $120$ students’ commute times gives $Q_1 = 15$, median $= 22$ and $Q_3 = 27$ minutes. The shortest commute is $4$ minutes, the longest is $49$ minutes, and the second longest is $41$ minutes.

• (a) Show that the longest commute is an outlier, and describe how the box-and-whisker diagram is drawn.
• (b) Compare the commute times at this school with those in Example 4 (median $28.3$, IQR $18$, range $55$).

Solution

(a) $\text{IQR} = 27 - 15 = 12$ and $1.5 \times 12 = 18$. The upper fence is $27 + 18 = 45$. Since $49 \gt 45$, the longest commute is an outlier. (The lower fence is $15 - 18 = -3$, so there are no low outliers.)

Draw the box from $15$ to $27$ with the median at $22$. The left whisker goes to $4$. The right whisker stops at $41$, the largest value that isn’t an outlier, and $49$ is marked with a cross.

(b) The median at the second school ($22$ min) is lower than in Example 4 ($28.3$ min), so its students typically have shorter commutes. Its IQR ($12$ min) is smaller than $18$ min, so the middle half of its commute times are more consistent. (Its range, $49 - 4 = 45$ min, is also smaller than $55$ min, though the range is affected by the outlier.)