S1.2 - Representation and summary of data

A statistical display should make the distribution's centre, spread, shape and unusual values easier to compare. Histograms, stem-and-leaf diagrams and box plots preserve different information, so interpret the feature each display actually encodes.

For classes $10\le x<15$ with frequency $20$ and $15\le x<25$ with frequency $30$, the densities are $20/5=4$ and $30/10=3$. The first bar is taller even though the second class contains more observations. To estimate a frequency in part of a class, use the corresponding fraction of its bar area, assuming values are evenly spread within that class.

Compare distributions in context: a larger median means a larger typical value, while a larger IQR means the central half is more variable. Quote the statistics or visible features that support the comparison; do not write only that one box or bar is 'bigger'.

A measure of location describes where a distribution is centred. The mean uses every value, the median divides ordered data into two halves, and the mode is the most frequent value or category.

For grouped continuous data, use each class midpoint as its representative value, so the resulting mean is an estimate. Locate a grouped median from cumulative frequency and use linear interpolation within its class:

Here $L$ is the lower class boundary, $C$ the cumulative frequency before the class, $f$ the class frequency, $w$ its width and $pN$ the target position. For example, if $N=80$ and the median class $40\le x<50$ has $C=30$ and $f=25$, then $Q_2=40+(40-30)10/25=44$. Use the position convention specified or established by the data.

Coding reduces arithmetic. If $y=(x-a)/b$, then $x=a+by$ and$$\bar x=a+b\bar y.$$For $n=20$, $a=50$, $b=5$ and $\sum y=12$, the original mean is $50+5(12/20)=53$. Decode the final value and restore its unit.

Dispersion measures how widely observations vary. Compare spread only after checking that the groups and units are meaningful; a smaller measure indicates greater consistency, not necessarily a better outcome.

For $2,4,4,6$, the mean is $4$ and$$\operatorname{Var}(X)=\frac{2^2+4^2+4^2+6^2}{4}-4^2=2.$$Hence $sigma=\sqrt2\approx1.41$ in the data's units, while the range is $6-2=4$. Variance is not reported in the original unit because its deviations were squared.

For a linear coding $X=a+bY$,$$\operatorname{Var}(X)=b^2\operatorname{Var}(Y),\qquad \sigma_X=|b|\sigma_Y.$$Adding a constant changes location but not spread; multiplying by $b$ multiplies range, IQR and standard deviation by $|b|$. For grouped data, midpoint-based variance and standard deviation are estimates.

Skewness describes asymmetry in a distribution. Name the direction and support it with the graph or statistic requested; the long or more spread-out tail gives the direction of skew.

If a numerical measure is specified, substitute consistently and use its sign. Common forms include$$\frac{Q_3-2Q_2+Q_1}{Q_3-Q_1},\qquad \frac{3(\bar x-Q_2)}{\sigma},\qquad \frac{\bar x-\text{mode}}{\sigma}.$$A positive result indicates positive skew and a negative result negative skew; a value near zero indicates little skew for that measure.

An outlier rule is supplied when required. For the common $1.5\times\mathrm{IQR}$ rule, calculate both fences:$$Q_1-1.5\mathrm{IQR},\qquad Q_3+1.5\mathrm{IQR}.$$Values strictly beyond the fences are outliers and are plotted separately on a box plot; whiskers extend to the most extreme non-outliers under this convention.

If $Q_1=12$ and $Q_3=20$, then $mathrm{IQR}=8$ and the fences are $0$ and $32$. A value of $35$ is an upper outlier, whereas $32$ lies on the fence and is not beyond it. The calculation identifies unusual position; context is still needed to interpret it.