GCSE · Statistics

Cumulative Frequency, Box Plots and Histograms Higher

Three Higher-tier diagrams for big sets of grouped data. A cumulative frequency graph lets you read off the median and quartiles. A box plot squeezes a whole distribution into five numbers. A histogram with unequal class widths shows the shape of the data honestly, by making area, not height, stand for frequency.

GCSE subject content (Statistics, Higher): construct and interpret diagrams for grouped discrete data and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs; interpret and compare distributions including box plots, quartiles and inter-quartile range

Before you start. You need grouped frequency tables, class intervals such as 40 < t ≤ 50 and running totals: see grouped data and the estimated mean and the KS3 page frequency tables. You also need the median and range from averages and range, and reading values off a graph.

1 · Cumulative frequency tables Higher

80 runners took part in a fun run. The table groups their finishing times. Cumulative frequency is the running total of the frequencies: how many runners had finished by the end of each class. "Cumulative" just means "added up as you go".

For example, 5 runners took 30 to 40 minutes and 10 took 40 to 50 minutes, so 5 + 10 = 15 runners had finished within 50 minutes. Fill in the cumulative frequency column yourself.

Check: the last cumulative frequency must equal the total frequency (here 80). Each cumulative frequency answers "how many values are less than or equal to the upper end of this class?" That's why the next step plots it at the upper end.

2 · Drawing a cumulative frequency graph Higher

Plot each cumulative frequency against the upper class boundary (the end of the class), because that's the point by which that many runners had finished. Also plot a starting point at the lower end of the first class with cumulative frequency 0 (nobody finished in under 30 minutes). Then join the points with a smooth curve, or with straight lines: both are accepted.

Tap the grid to plot the 7 points (tap a point again to remove it).

Common mistakes: plotting at the midpoints (that's for a frequency polygon), plotting the frequencies instead of the cumulative frequencies, and forgetting the starting point. The graph always climbs (or stays level): it can never go down, because a running total can't shrink. Its typical stretched S-shape is called an ogive.

3 · Reading the median, quartiles and more Higher

Lined up in time order, the runners split into four equal quarters. The lower quartile (LQ) is the time a quarter of the way along, the median is halfway, and the upper quartile (UQ) is three quarters of the way. With n = 80 runners:

LQ: go across from cumulative frequency n ÷ 4 = 20 to the curve, then down to read the time.
Median: from n ÷ 2 = 40. UQ: from 3n ÷ 4 = 60.
Inter-quartile range IQR = UQ − LQ: the spread of the middle half of the data. Unlike the range it ignores the extreme values at each end.
A percentile works the same way: the 90th percentile is read from 90% of 80 = 72.
"How many took more than 75 minutes?" Go up from 75 to the curve and across: about 66 had finished by then, so 80 − 66 = 14 took longer.

(For grouped data on a graph use n ÷ 2, not (n + 1) ÷ 2: with large n the difference is invisible.) Drag the horizontal line up and down and the vertical line left and right, or use the buttons.

smooth curve (untick for straight lines) show all three quartiles

Across from cf

Up from time

More than that time

4 · Box plots Higher

A box plot (box-and-whisker diagram) draws five numbers on a scale: the minimum, LQ, median, UQ and maximum. The box runs from LQ to UQ with a line at the median, so the box holds the middle half of the data and its length is the IQR. The whiskers run out to the minimum and maximum.

From a short list of raw data, put it in order and find the median first. Then the LQ is the median of the lower half and the UQ the median of the upper half (leave out the median itself when n is odd). With 11 values that's the 3rd, 6th and 9th: the positions (n + 1) ÷ 4, (n + 1) ÷ 2 and 3(n + 1) ÷ 4. Drag the dots and watch the box plot follow.

Notice: drag the largest value far to the right. The whisker stretches and the range jumps, but the box (and so the median and IQR) hardly moves. That's why the median and IQR are good at ignoring extreme values. Each of the four sections of a box plot holds about a quarter of the data, so a long whisker or a long half-box means the data is spread out there, not that there is more of it.

5 · Build a box plot from the cumulative frequency graph Higher

A cumulative frequency graph gives the LQ, median and UQ, but not the exact minimum and maximum, so questions tell you those. Here: the fastest runner took 33 minutes and the slowest 88 minutes. Use your readings from section 3 and drag the five handles into place.

6 · Histograms: why frequency density? Higher

A bar chart for grouped data works only while every class has the same width. Here are the times 100 customers spent in a shop. Pick unequal widths with height = frequency: the 30 to 60 minute bar, which lumps three 10-minute classes together, towers over its neighbours even though customers there are thinly spread. Your eye reads area, and a wide bar has a lot of area.

The fix is to make area represent frequency. Since area = width × height, the height must be

frequency density = frequency ÷ class width

Frequency density is "frequency per unit": here, customers per minute of time. A diagram drawn this way is a histogram. Switch to frequency density and the shape matches what the equal-width charts show.

classes height =

Drawing a histogram: add a class width column and a frequency density column to the table; label the vertical axis "frequency density" (never "frequency"); draw bars with no gaps, from boundary to boundary. With equal widths, frequency and frequency density give the same shape, which is why ordinary bar charts are fine there.

7 · Reading frequencies back from a histogram Higher

Turn it round: frequency = frequency density × class width = the area of the bar. To estimate how many values lie in part of a class, assume the values are spread evenly across the class and take that part of the area. Drag the two shaded-range handles (or type the ends).

from to

Try it

label bar heights answer

Exam traps. Reading the height as the frequency (multiply by the width). Using the whole class when the question asks about part of it. And when the axis has no numbers: use one bar whose frequency you know to find how many people one square (or 1 cm²) of area stands for, then scale.

Worked examples

Median, quartiles and IQR from a cumulative frequency graph

The fun-run times: 30 < t ≤ 40: 5; 40 < t ≤ 50: 10; 50 < t ≤ 60: 25; 60 < t ≤ 70: 20; 70 < t ≤ 80: 12; 80 < t ≤ 90: 8. The points are joined with straight lines.

Time up to (min)	30	40	50	60	70	80	90
Cum. freq.	0	5	15	40	60	72	80

Positions: n = 80, so LQ at 80 ÷ 4 = 20, median at 80 ÷ 2 = 40, UQ at 3 × 80 ÷ 4 = 60.

Median: cf 40 is reached exactly at t = 60, so median = 60 min. UQ: cf 60 is reached at t = 70, so UQ = 70 min.

LQ: cf 20 is between (50, 15) and (60, 40). It is 5 of the 25 runners into that class, so 5 ÷ 25 = 1/5 of the way across: 50 + 10 × 1/5 = 52 min. (A smooth curve gives about 52 too.)

IQR = UQ − LQ = 70 − 52 = 18 min.

"How many more than…" and a percentile

Same graph. (a) Estimate how many runners took more than 75 minutes. (b) Estimate the 90th percentile.

(a) Up from t = 75 to the graph: halfway between (70, 60) and (80, 72), so cf = 60 + 12 ÷ 2 = 66. That many took 75 minutes or less.

More than 75 minutes: 80 − 66 = 14 runners. (Giving 66 is the classic slip: that's how many took less.)

(b) 90% of 80 = 0.9 × 80 = 72. Across from cf 72: the graph reaches 72 at t = 80, so the 90th percentile is 80 minutes.

Box plot from raw data

The scores of 11 players: 26, 15, 38, 21, 17, 24, 12, 29, 18, 31, 23. Draw a box plot.

Order: 12, 15, 17, 18, 21, 23, 24, 26, 29, 31, 38.

Median: (11 + 1) ÷ 2 = 6th value = 23.

LQ: lower half 12, 15, 17, 18, 21 → middle = 17 (the 3rd value). UQ: upper half 24, 26, 29, 31, 38 → middle = 29 (the 9th value).

Min 12, max 38. Draw a scale, the box from 17 to 29 with a line at 23, whiskers to 12 and 38. IQR = 29 − 17 = 12; range = 38 − 12 = 26.

Drawing a histogram with unequal widths

Shop times t (min): 0 < t ≤ 10: 10; 10 < t ≤ 15: 14; 15 < t ≤ 20: 18; 20 < t ≤ 30: 28; 30 < t ≤ 60: 30.

Class	f	Width	FD = f ÷ w
0–10	10	10	1
10–15	14	5	2.8
15–20	18	5	3.6
20–30	28	10	2.8
30–60	30	30	1

Widths: upper − lower for each class.

Frequency density: 10 ÷ 10 = 1; 14 ÷ 5 = 2.8; 18 ÷ 5 = 3.6; 28 ÷ 10 = 2.8; 30 ÷ 30 = 1.

Draw: horizontal axis time 0 to 60, vertical axis "frequency density" 0 to 4. Each bar spans its class with height = FD, no gaps. The tallest bar is 15–20, the class where customers are most tightly packed, even though 30–60 has the most customers.

Estimating within part of a class

Using that histogram, estimate how many customers spent between 25 and 40 minutes.

Split at the class boundary: 25 to 30 lies in the 20–30 class; 30 to 40 lies in the 30–60 class.

25 to 30: area = FD × width = 2.8 × 5 = 14.

30 to 40: area = 1 × 10 = 10.

Total: 14 + 10 = 24 customers (an estimate: it assumes customers are spread evenly within each class).

Exam-style: a histogram with no scale

On a histogram the bar for 20 < t ≤ 30 is 2.8 cm tall and represents 28 people. The bar for 30 < t ≤ 60 is 1 cm tall. Each 1 cm on the horizontal axis is 1 minute. How many people does the 30 < t ≤ 60 bar represent?

Area of the known bar: 10 × 2.8 = 28 cm², representing 28 people, so 1 cm² = 1 person.

Area of the other bar: 30 × 1 = 30 cm².

Answer: 30 × 1 = 30 people. (Comparing heights, 1 cm against 2.8 cm, would wrongly suggest only 10.)

Quick check

What you've learned

Cumulative frequency is the running total of the frequencies. Plot it at the upper class boundary, start at (lowest boundary, 0) and join with a smooth curve or straight lines.
From the graph read the LQ at n ÷ 4, the median at n ÷ 2 and the UQ at 3n ÷ 4. IQR = UQ − LQ measures the spread of the middle half and ignores extremes.
For "more than x", read the cf at x and subtract it from the total. Percentiles are read at that percentage of n.
A box plot shows min, LQ, median, UQ and max on a scale. From raw data, the quartiles are the medians of the lower and upper halves.
In a histogram, area represents frequency: frequency density = frequency ÷ class width, and frequency = frequency density × class width.
To estimate within part of a class, take the matching fraction of that bar's area (assuming values are evenly spread).