Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
In a histogram the AREA of a bar is the frequency, so the height is the frequency density: \(\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}\). On a cumulative frequency graph, read across from a position (such as \(\tfrac{n}{2}\) for the median) and down to the value.
Plotting frequency (not frequency density) on the vertical axis when the class widths differ. It makes wide classes look far too important and scores no marks for the heights.
Forgetting to reverse the left-hand leaves. In the row \(8\ 6\ 5\ 2\ 0 \mid 3\), the smallest Class A value is 30 (the leaf next to the stem), not 38.
Plotting cumulative frequencies at the class midpoints. A cumulative frequency counts everything UP TO the end of a class, so it belongs at the upper boundary.
The mean \(\bar{x} = \dfrac{\sum fx}{\sum f}\) uses every value, so one extreme value can drag it. The median and interquartile range use only positions, so they resist outliers and skew. Quote a centre AND a spread, and compare them in context.
Averaging the two means. \(\tfrac{1}{2}(64 + 59) = 61.5\) is wrong because the groups are different sizes; always go back to the totals \(\sum x\).
Using the upper class boundary (or the class width) instead of the midpoint in \(\sum fx\). For \(30 \le t < 50\) the midpoint is 40, not 50 and not 20.
Quoting a quartile POSITION instead of a value: for 11 values \(Q_3\) is the 9th value (26), not 9.
Saying 'Ana's data are higher' without context. In this question a lower time is better, so say 'faster', and refer to times, not just numbers.
\(\text{Variance} = \dfrac{\sum (x - \bar{x})^2}{n} = \dfrac{\sum x^2}{n} - \bar{x}^2\) and standard deviation \(= \sqrt{\text{variance}}\). Coding by subtracting \(a\) shifts the mean by \(a\) but leaves the standard deviation unchanged.
Writing \(\dfrac{\sum x^2}{n} - \bar{x}\) or \(\dfrac{(\sum x)^2}{n}\). It is the MEAN that is squared and subtracted: \(\dfrac{\sum x^2}{n} - \bar{x}^2\).
Using \((\sum fx)^2\) or \(\sum (fx)^2\) in place of \(\sum fx^2\). Square the value \(x\) first, then multiply by its frequency.
Thinking \(\sum x^2 = \sum (x - a)^2 + na^2\). The cross term \(-2a\sum x\) matters; expand the bracket properly or work via the variance.
The combined standard deviation (3.95) is bigger than both separate ones (2 and 3), because the two means are far apart. Averaging the standard deviations misses this completely.