Histograms, the mode and ogives · Statistics & correlation

General Maths drew histograms with equal class widths and read the median and quartiles from a cumulative frequency curve (see statistics and cumulative frequency). Further Maths adds unequal widths, the mode found from the histogram, and more readings from the ogive.

Histograms with unequal widths

In a histogram the area of each bar stands for its frequency. When the classes have different widths, the height must be the frequency density:

Frequency densityHeight = frequency ÷ class width, so area = frequency

Use class boundaries for the widths: the class 23–28 runs from 22.5 to 28.5, a width of 6.

Frequency densitySwitch the bar heights; tap a bar

f = 18class 20–30, width 101.8height = f ÷ width

Each bar's height is its frequency density, so its area (width × height) is its frequency: 10 × 1.8 = 18. The widest class, 30–50, now looks as sparse as it really is.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q13

Age (years)	17–19	20–22	23–28	29–34	35–43
Number of patients	6	9	12	18	18

The table shows the frequency distribution of the ages of patients in a clinic.

Draw a histogram for the distribution.

Find, correct to two decimal places, the mean age of the patients.

Widths and heights
- Widths: ${3,\ 3,\ 6,\ 6,\ 9}$ .
- Heights: ${\frac63 = 2}$ , ${\frac93 = 3}$ , ${\frac{12}{6} = 2}$ , ${\frac{18}{6} = 3}$ , ${\frac{18}{9} = 2}$ .
- Draw each bar over its class boundaries with these heights.
Think first. The widths are 3, 3, 6, 6, 9. Divide each frequency by its width.
The mean
- The products ${fx}$ : ${108,\ 189,\ 306,\ 567,\ 702}$ .
- Add them: ${\sum fx = 1872}$ .
- ${\bar x = \frac{1872}{63} \approx 29.71}$ years.
Think first. Class marks 18, 21, 25.5, 31.5, 39.

More: histograms

The mode from a histogram

On the tallest bar, draw a line from each top corner to the top corner of the neighbouring bar on the opposite side. Where the lines cross, read down to the axis: that is the mode.

Reading the modeThe crossed lines lean towards the taller neighbour

By calculation, the same construction gives $\text{mode} = L + \dfrac{D_1}{D_1 + D_2} \times c$ , where $L$ is the lower class boundary of the modal class, $c$ its width, $D_1$ the rise from the class before and $D_2$ the fall to the class after.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q6

Age (years)	20–24	25–29	30–34	35–39	40–44	45–49	50–54	55–59
Number of workers	22	24	30	38	36	30	18	12

The table shows the age distribution of workers in a factory.

Using a graphical method, find the modal age of the workers.

The modal class
- The modal class is 35–39, with 38 workers. Its boundaries are ${34.5}$ and ${39.5}$ , so ${c = 5}$ .
Think first. Which class has the highest frequency?
The differences
- ${D_1 = 38 - 30 = 8}$ and ${D_2 = 38 - 36 = 2}$ .
Think first. Compare 38 with its neighbours, 30 and 36.
The mode
- ${34.5 + \frac{8}{8 + 2} \times 5 = 34.5 + 4 = 38.5}$ years. The graph gives the same.

More: the mode

Readings from an ogive

Plot cumulative frequency against upper class boundaries and join with a smooth curve. Then read across and down:

Reading the curveAcross from ¼N, ½N and ¾N, then down

Median at $\frac12N$ , quartiles at $\frac14N$ and $\frac34N$ . The semi-interquartile range is $\frac12(Q_3 - Q_1)$ .
A percentile or decile works the same way: the 60th percentile (6th decile) is at $0.6N$ .
“How many scored between 32 and 74”: read up from 32 and from 74 and subtract.
“The pass mark if 18% failed”: read across from $0.18N$ .

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q12

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Number of students	5	5	10	18	23	23	9	4	2	1

The table shows the marks obtained by students in an examination.

Construct a cumulative frequency table for the distribution.

Draw an ogive for the distribution.

Use the ogive to determine the: (i) median mark; (ii) semi-interquartile range.

If a student is selected at random, what is the probability that he obtained at least 60 marks?

The cumulative frequencies
- Upper boundaries ${9.5, 19.5, \ldots, 99.5}$ with cumulative frequencies ${5, 10, 20, 38, 61, 84, 93, 97, 99, 100}$ .
The median
- 50 lies between 38 (at 39.5) and 61 (at 49.5).
- ${39.5 + \frac{50 - 38}{23} \times 10 \approx 44.7}$ .
Think first. N = 100, so read at 50. Which two points is it between?
The quartiles
- ${Q_1 = 29.5 + \frac{25 - 20}{18} \times 10 \approx 32.3}$ .
- ${Q_3 = 49.5 + \frac{75 - 61}{23} \times 10 \approx 55.6}$ .
- Semi-interquartile range ${= \frac12(55.6 - 32.3) \approx 11.7}$ .
Think first. Read at 25 and 75.
At least 60 marks
- ${9 + 4 + 2 + 1 = 16}$ students, so the probability is ${\frac{16}{100} = 0.16}$ .
Think first. How many scored 60 or more?

More: ogives

Your turn

WAEC 2017 · Paper 2 · Q6

Statistics & correlation

Height (cm)	36–40	41–45	46–50	51–55	56–60
Frequency	3	9	21	12	5

The table shows the heights, in cm, of some seedlings in a certain garden.

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot the cumulative frequencies $3, 12, 33, 45, 50$ against the upper class boundaries $40.5, \ldots, 60.5$ , starting from $(35.5, 0)$ , and join them with a smooth S-shaped curve. For (b): reading across from 12.5 and 37.5 gives $Q_1 \approx 45.6$ and $Q_3 \approx 52.4$ . Readings from a hand-drawn curve vary a little; examiners accept a small range.
(b)
Using the curve, find the semi-interquartile range.

Try it on a graph

The ogive with the quartile readings.

Worked solution (try it first)

(a)

Plot the cumulative frequencies $3, 12, 33, 45, 50$ at the upper boundaries $40.5, 45.5, 50.5, 55.5, 60.5$ , starting from $(35.5, 0)$ , and join with a smooth curve.

(b)

$N = 50$ .
Read across from 12.5 and 37.5: $Q_1 \approx 45.6$ and $Q_3 \approx 52.4$ .
Semi-interquartile range $= \frac12(52.4 - 45.6) \approx 3.4$ .

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