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How do you summarise, display and interpret data, and how do you measure how two variables are related?

Measures of location and spread, diagrams for single and bivariate data, outliers and cleaning, correlation and the equation of a regression line, and interpolation versus extrapolation.

A focused answer to the Edexcel A-Level Mathematics data presentation and interpretation content, covering measures of location and spread, histograms and box plots, outliers and cleaning, correlation, the regression line, and interpolation versus extrapolation.

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  1. What this dot point is asking
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What this dot point is asking

Edexcel wants you to calculate and interpret measures of location (mean, median, mode) and spread (range, interquartile range, variance, standard deviation), draw and read histograms, box plots and cumulative frequency diagrams, identify and clean outliers, describe correlation, find and use the equation of a regression line, and know the difference between interpolation and extrapolation.

The answer

Location and spread

Outliers

Diagrams for data

Correlation and regression

Correlation describes the strength and direction of a linear relationship between two variables. A scatter diagram with points clustering tightly along a rising line shows strong positive correlation; points scattered around a falling line show negative correlation; a formless cloud shows little or none. The regression line of yy on xx has the form y=a+bxy = a + bx, where bb is the gradient (the change in yy per unit change in xx) and aa is the intercept (the predicted yy when x=0x = 0). It is used to predict yy from xx.

Examples in context

Try this

Q1. For data with x=50\sum x = 50, x2=310\sum x^2 = 310 and n=10n = 10, find the mean and standard deviation. [3 marks]

  • Cue. Mean =5= 5; variance =3101025=6= \tfrac{310}{10} - 25 = 6, so standard deviation =62.45= \sqrt{6} \approx 2.45.

Q2. A box plot has Q1=12Q_1 = 12 and Q3=20Q_3 = 20. Determine the outlier boundaries using 1.5×IQR1.5 \times \text{IQR}. [3 marks]

  • Cue. IQR =8= 8; boundaries are 1212=012 - 12 = 0 and 20+12=3220 + 12 = 32.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20195 marksThe summary statistics for a sample of n=8n = 8 values are x=96\sum x = 96 and x2=1240\sum x^2 = 1240. Calculate the mean and the standard deviation, and determine whether a value of 3030 would be an outlier using the two-standard-deviation rule.
Show worked answer →

Mean xˉ=xn=968=12\bar{x} = \dfrac{\sum x}{n} = \dfrac{96}{8} = 12 (M1, A1).

Variance =x2nxˉ2=12408122=155144=11= \dfrac{\sum x^2}{n} - \bar{x}^2 = \dfrac{1240}{8} - 12^2 = 155 - 144 = 11, so standard deviation σ=113.317\sigma = \sqrt{11} \approx 3.317 (M1, A1).

The upper boundary is xˉ+2σ=12+6.63=18.63\bar{x} + 2\sigma = 12 + 6.63 = 18.63. Since 30>18.6330 > 18.63, the value 3030 is an outlier (A1).

Markers reward the mean, the variance formula, the standard deviation, and the outlier comparison.

Edexcel 20224 marksA regression line of yy on xx is y=4.5+1.8xy = 4.5 + 1.8x, fitted from data with xx ranging from 22 to 1515. Estimate yy when x=10x = 10 and when x=25x = 25, and comment on the reliability of each estimate.
Show worked answer →

For x=10x = 10: y=4.5+1.8×10=4.5+18=22.5y = 4.5 + 1.8 \times 10 = 4.5 + 18 = 22.5 (M1, A1).

For x=25x = 25: y=4.5+1.8×25=4.5+45=49.5y = 4.5 + 1.8 \times 25 = 4.5 + 45 = 49.5 (A1).

The first uses x=10x = 10, inside the data range [2,15][2, 15], so it is interpolation and reliable. The second uses x=25x = 25, outside the range, so it is extrapolation and unreliable (A1).

Markers reward both substitutions and a correct interpolation/extrapolation comment.

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