S1 2.1 - Reading statistical diagrams

S1 2.1 - Reading statistical diagrams

Read histograms, stem-and-leaf diagrams and box plots, including displays with unequal intervals and two groups. Use their centre and spread to make comparisons in context, and understand the information each display keeps or loses.

Histograms with unequal intervals

Can the tallest histogram bar contain fewer observations than a shorter bar? Yes: frequency depends on the bar's area, so its width matters too.

A histogram represents grouped numerical data with adjoining bars. The horizontal axis shows class boundaries. For continuous data, an interval such as 10≤t<1510\le t<15 includes 1010 but excludes 1515, so adjacent classes do not overlap.

Frequency density

When the vertical axis is frequency density,

frequency density=frequencyclass width,frequency=class width×frequency density.\text{frequency density}=\frac{\text{frequency}}{\text{class width}},\qquad \text{frequency}=\text{class width}\times\text{frequency density}.

Thus bar area represents frequency. This formula is not printed in the formula booklet: learn it.

The diagram shows waiting times. Its 1010–1515 bar is tallest, but its narrow width means it need not have the greatest frequency.

[DIAGRAM: asset_name: Unequal histogram intervals; asset_slug: unequal-histogram-intervals; caption: The tallest bar has area 20; the wider final bar has area 30.; description: Waiting-time histogram with intervals zero to ten, ten to fifteen and fifteen to thirty minutes, frequency densities two, four and two. Each bar is labelled with its frequency twenty, twenty and thirty.]
The tallest bar has area 20

To construct the display, calculate each class width and density, use one horizontal scale and one vertical scale, then draw adjoining rectangles. If heights are measured in centimetres rather than density units, use a known bar to find the common scale. For example, density 44 drawn as 8 cm8\,\mathrm{cm} means density 22 is drawn as 4 cm4\,\mathrm{cm}.

Worked example 1: area gives a class frequency

Question. A waiting-time histogram has density 22 on 15≤t<3015\le t<30, where tt is measured in minutes. Find the number of observations in that class. Compare it with a bar of density 44 on 10≤t<1510\le t<15.

Plan. Multiply each density by its own class width.

f15–30=(30−15)×2=30f10–15=(15−10)×4=20\begin{aligned} f_{15\text{–}30}&=(30-15)\times2=30\\ f_{10\text{–}15}&=(15-10)\times4=20 \end{aligned}

Answer. The wider, shorter bar represents 3030 observations, compared with 2020 in the taller bar.

Check. The widths are in the ratio 3:13:1 and the heights in the ratio 1:21:2, so the areas are in the ratio 3:23:2, matching 30:2030:20.

Change only the interval requested, from a whole class to part of it. Grouped data do not reveal the individual waiting times, so we must now make an estimate.

Worked example 2: part of a bar

Question. For the same histogram, estimate the number of waiting times in 20≤t<3020\le t<30.

Plan. Assume observations are spread uniformly within the 1515–3030 class, so the required fraction of its width gives the same fraction of its frequency.

estimated count=30−2030−15×30=20.\text{estimated count}=\frac{30-20}{30-15}\times30=20.

Answer. An estimated 2020 waiting times. The word “estimated” matters because the exact positions within the class are unknown.

Check. The required interval covers two thirds of the class, so it should contain two thirds of 3030. Equivalently, its area is 10×2=2010\times2=20.

If several classes are involved, add their whole or partial areas. A decimal estimated count may be rounded to a sensible whole number when the task asks how many people or objects.

Watch out

Do not compare frequencies by bar height when widths differ. Use areas. The modal class is the class with greatest frequency density; the class with greatest total frequency can be different.

Use the supplied intervals and densities independently of the earlier diagram.

Stem-and-leaf and back-to-back displays

A stem-and-leaf diagram retains the individual observations while arranging them in order.

The stem gives the leading digit or digits and the leaf gives the final digit. A key fixes the place value and units: 3∣1=31 s3\mid1=31\,\mathrm{s} is a different scale from 3∣1=3.1 s3\mid1=3.1\,\mathrm{s}. Repeated observations need repeated leaves, and an empty stem may be kept so that a gap in the data is visible.

A back-to-back stem-and-leaf diagram places two groups either side of shared stems. On the left, leaves increase as you move away from the stem towards the left; on the right, they increase towards the right. Read a leaf together with its stem, rather than reading the printed row as a multi-digit number. A two-sided key such as 1∣3∣11\mid3\mid1 means 31 s31\,\mathrm{s} in either group: 33 is the central stem and each 11 is a leaf.

[DIAGRAM: asset_name: Back-to-back response times; asset_slug: back-to-back-response-times; caption: Each leaf represents one response time; the central stem fixes the tens digit.; description: Back-to-back stem-and-leaf display. Group A has twenty-four, twenty-eight, thirty-one, thirty-one, thirty-six, forty-two and forty-seven seconds. Group B has twenty-two, twenty-five, twenty-nine, thirty-three, thirty-eight, forty-one and forty-four seconds. Stems are two, three and four. The key one bar three bar one means thirty-one seconds in either group, with three explicitly identified as the central stem.]
Each leaf represents one response time

To construct one, choose stems that preserve the required precision, place each leaf beside its stem, order both sides and include a key. The shared stems allow direct comparison without hiding the raw values.

Worked example 3: reading one side

Question. Group A's back-to-back rows are 8 4∣28\ 4\mid2, 6 1 1∣36\ 1\ 1\mid3, and 7 2∣47\ 2\mid4. The key 1∣3∣11\mid3\mid1 means 31 s31\,\mathrm{s} in either group, with 33 the central stem. Find its median and range.

Plan. Read from the lowest stem upwards, taking the leaves nearest the stem first on the left side.

24, 28, 31, 31, 36, 42, 47.24,\ 28,\ 31,\ 31,\ 36,\ 42,\ 47.

There are 77 values, so the median is the fourth; the range uses the largest and smallest.

median=31 s,range=47−24=23 s.\text{median}=31\,\mathrm{s},\qquad \text{range}=47-24=23\,\mathrm{s}.

Answer. Median 31 s31\,\mathrm{s}, range 23 s23\,\mathrm{s}.

Check. Three values lie on each side of the fourth position. The two leaves 1,11,1 on stem 33 represent two observations, not one.

Change only the side read: the key and stems stay the same, but the right-hand leaves are read outwards towards the right.

Worked example 4: reading the other side

Question. Group B's rows are 2∣2 5 92\mid2\ 5\ 9, 3∣3 83\mid3\ 8, and 4∣1 44\mid1\ 4, with the same key. Find its median and range and compare the medians with Group A.

Plan. Recover the seven values, then use the middle value and extremes.

22, 25, 29, 33, 38, 41, 44.22,\ 25,\ 29,\ 33,\ 38,\ 41,\ 44. median=33 s,range=44−22=22 s.\text{median}=33\,\mathrm{s},\qquad \text{range}=44-22=22\,\mathrm{s}.

Answer. Group B's median is 33 s33\,\mathrm{s}, 2 s2\,\mathrm{s} higher than Group A's, so its typical response time is longer by this measure. Its range is 22 s22\,\mathrm{s}.

Check. The fourth value is 3333, and both endpoints appear in the diagram. The median comparison concerns typical times, not every individual's time.

The diagram preserves exact values, but it can become crowded for a very large dataset.

Watch out

A leaf is one observation. Never combine leaves 1 1 61\ 1\ 6 into 116116, discard repeats, or assume a decimal scale without reading the key.

Read the new decimal key carefully.

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