MathsStatistics › Representing and interpreting data

Representing and interpreting data

A good chart is an argument you can see. Histograms make area mean frequency, box plots put five numbers on display, and a fence built from the quartiles decides, without sentiment, which values are outliers.

Builds on Measures of location and spread.

IN THIS TOPIC

  • Construct and read histograms using frequency density, including recovering frequencies from areas.
  • Draw and compare box plots, and identify outliers with the quartile fence rule.
  • Describe skewness from quartiles or from the mean against the median.
  • Interpret cumulative frequency diagrams, and clean data before summarising it.
  • Choose or criticise a diagram for the job the question is doing.

COMMON MISCONCEPTION

In a histogram, the height of a bar tells you its frequency.

Histograms: area is frequency

A bar chart has equal-width bars, so height alone carries the information. A histogram copes with unequal class widths, and the price of that is recalculating every height as frequency density.

frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}NOT IN THE BOOKLET — LEARN IT
A histogram with unequal class widths where area gives frequency: classes of width 10, 5 and 15 with frequencies 20, 15 and 30area 20area 15area 300101530vertical axis: frequency density
FIG. 1The narrow middle class holds the fewest values yet stands tallest: area, not height, is the frequency.

Now area equals frequency, and every histogram question becomes an area question. Given a bar, multiply height by width to recover its frequency. Given a frequency, divide by the width to set the height. A histogram with a key, saying one square represents 5 observations, is the same idea with a scale factor bolted on.

WORKED EXAMPLE

Reading a histogram

In a histogram, the class 15 ≤ x < 30 has frequency 30. The class 10 ≤ x < 15 is drawn with 1.5 times the height of that bar. Find its frequency.

The first bar's density is 30/15 = 2, so the second bar's density is 1.5 × 2 = 3.

Its frequency is density times width: 3 × 5 = 15.

Check by areas. A bar a third as wide but one-and-a-half times as tall carries half the area, 15 against 30, which matches.

A frequency polygon joins the midpoints of the tops of the bars and is the usual way to lay two distributions over each other. It shows shape well and reads frequencies badly, so pick it when you are comparing and the histogram when you are counting.

Box plots and outliers

A box plot displays five numbers. The extremes, the quartiles and the median. The box spans the middle half of the data, the median line splits it, and the whiskers reach the most extreme values that are not outliers.

outlier: beyond Q1-1.5×IQR or Q3+1.5×IQR\text{outlier: beyond } Q_{1} - 1.5 \times \text{IQR} \text{ or } Q_{3} + 1.5 \times \text{IQR}NOT IN THE BOOKLET — LEARN IT
A box plot with quartiles 24, 30 and 38, whiskers to the most extreme values inside the fences, and one outlier at 62 plotted separately0102030405060fence at 5962: outliermedian 30
FIG. 2Quartiles 24 and 38 build fences at 3 and 59: the value at 62 stands beyond the fence and is plotted on its own.

The 1.5 × IQR fences are the usual rule, though a question may instead define an outlier as more than two standard deviations from the mean. Use whichever definition the question gives you, even when the other one is more familiar. An outlier gets plotted as a separate cross, and the whisker stops at the last value inside the fence, never at the fence itself.

Comparing two box plots is a two-sentence answer with a fixed shape. One sentence on location, using the medians, in context. One on spread, using the IQRs, in context. “The median mass is higher for brand A, and brand A's masses are more consistent since its IQR is smaller.” Marks vanish when the context does.

Shape and skew

A distribution is symmetric when the median sits midway between the quartiles, so Q₂ − Q₁ = Q₃ − Q₂. When the upper gap is the bigger one, the long tail points right and the data has positive skew. The reverse gives negative skew.

The mean and median tell the same story. A long right tail drags the mean above the median, so mean > median indicates positive skew. Either test earns the mark provided you quote the comparison you used, and quoting both when they disagree is a good way to start a sentence about which measure the data deserves.

Cumulative frequency

Plot cumulative frequency against the upper class boundary and join the points with a smooth increasing curve. Reading across from n/2, n/4 and 3n/4 gives the median and quartiles. Reading up from two values and subtracting counts the data in any interval. Every reading is an estimate, because the grouping already blurred the exact values, and no examiner expects three significant figures off a graph.

Cleaning the data

Real data arrives dirty, and the large data set is deliberately so. Entries are missing, rainfall reads tr, wind direction is a compass point instead of a number. Decide what to do with each before any calculation, and say what you decided. Removing a genuinely impossible value is cleaning. Removing a merely inconvenient one is bias with better manners, so an anomaly should only go when you can state a reason to doubt the measurement itself.

ASSESSMENT FOCUS

  • Every histogram mark runs through frequency density. Label the vertical axis with it, and convert between area and frequency deliberately.
  • Learn the fence rule as a sentence. One and a half IQRs beyond either quartile.
  • Compare box plots in context, one sentence for medians and one for IQRs. A comparison with no context scores the method mark at best.
  • For skew, quote the comparison. “Q₃ − Q₂ > Q₂ − Q₁, so positive skew” is a complete answer.
  • Cumulative frequency plots at upper class boundaries. Plotting at midpoints is a classic dropped mark and it is dropped every summer.

CHECK YOURSELF

A data set has Q₁ = 15 and Q₃ = 27. Using the 1.5 × IQR rule, decide whether the value 48 is an outlier.

Show a hint

Build the upper fence first.

Show the answer

IQR = 27 − 15 = 12, so the upper fence is 27 + 1.5 × 12 = 45.

48 > 45, so 48 is an outlier and would be plotted as a separate point.

In a histogram, area is frequency and the vertical axis is frequency density.

Fences stand one and a half IQRs beyond the quartiles, and whatever is beyond the fence is plotted alone.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the representing and interpreting data questions page.

CHECK YOUR PROGRESS

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  • Construct and read histograms using frequency density, including recovering frequencies from areas.
  • Draw and compare box plots, and identify outliers with the quartile fence rule.
  • Describe skewness from quartiles or from the mean against the median.
  • Interpret cumulative frequency diagrams, and clean data before summarising it.
  • Choose or criticise a diagram for the job the question is doing.

Open the full revision checklist to see every objective in the course in one place.