4.2 Presenting dataIB Maths: Applications and Interpretation SL: Revision notes

A frequency table lists each value (discrete data) or each class interval (continuous or grouped data) with its frequency. Class intervals are written as inequalities with no gaps, for example $0\le x<10$, $10\le x<20$, $20\le x<30$. Each value belongs to exactly one class: a mass of 10 kg belongs to $10\le x<20$. The modal class is the class with the greatest frequency (when the class widths are equal). Example: classes $0<m\le2$ to $8<m\le10$ with frequencies $4,11,17,12,6$ have modal class $4<m\le6$.

A frequency histogram (equal class widths) has the variable on the horizontal axis, with touching bars whose heights are the frequencies. At this level the class widths are equal, so you do not need frequency density. Bars touch because the data are continuous; the shape shows the distribution: symmetrical, or skewed to the right (long tail of high values) or left.

Cumulative frequency is a running total of the frequencies. For a cumulative frequency graph, plot each cumulative frequency against the upper boundary of its class and join the points with a smooth curve (or straight lines if told to assume uniform spread). Reading the graph: the median is at $\frac12n$, the lower quartile $Q_1$ at $\frac14n$, the upper quartile $Q_3$ at $\frac34n$. The $p$th percentile is at $\frac{p}{100}n$. Then $\text{IQR}=Q_3-Q_1$ and $\text{range}=\text{max}-\text{min}$.

Ages of 40 visitors: $0\le a<20$: 5; $20\le a<40$: 13; $40\le a<60$: 15; $60\le a<80$: 7. Cumulative frequencies at $20,40,60,80$: $5,18,33,40$. Median (20th value, in $40\le a<60$): $40+\frac{20-18}{15}\times20\approx42.7$. $Q_1$ (10th value): $20+\frac{10-5}{13}\times20\approx27.7$. $Q_3$ (30th value): $40+\frac{30-18}{15}\times20=56$. $\text{IQR}\approx56-27.7=28.3$. 90th percentile (36th value): $60+\frac{36-33}{7}\times20\approx68.6$. These are estimates because individual values within a class are unknown.

A box and whisker diagram uses the five-number summary: minimum, $Q_1$, median, $Q_3$, maximum. The box runs from $Q_1$ to $Q_3$ with a line at the median; whiskers extend to the least and greatest values that are not outliers. Any outlier (more than $1.5\times\text{IQR}$ beyond a quartile) is drawn as a cross, and the whisker stops at the most extreme non-outlier value. Example: $Q_1=50$, $Q_3=66$, $\text{IQR}=16$, upper boundary $66+24=90$; a maximum of $110$ is a cross.

To compare two data sets, comment on a measure of centre (median) and a measure of spread (IQR or range), in context: for example, 'The median for A is higher, so A typically has more customers; A has the greater IQR, so its numbers are more variable.' Quote the numbers. Data may be normally distributed if the diagram is symmetrical: the median is in the middle of the box and the whiskers are of similar length. If the median is nearer one end of the box, or one whisker is much longer, the data are skewed.