Representation of data

5.1 Representation of data

Choosing a diagram

Stem-and-leaf diagrams

Values are split into a stem and a leaf, with the leaves in order and a key.

$\begin{array}{c|l} 1 & 2\ 5\ 8 \\ 2 & 0\ 3\ 3\ 7 \\ 3 & 1\ 6 \end{array} \qquad \text{Key: } 2 \mid 3 \text{ means } 23$

A back-to-back diagram shows two sets on a shared stem, with each one's leaves increasing outward. Always give a key for each side.

Median and quartiles

For $n$ ordered values, the median is at position $\dfrac{n + 1}{2}$. The lower and upper quartiles are the medians of the lower and upper halves.

There are 11 values. The median is the 6th, so 10. The lower half is 3, 5, 7, 8, 9, so $Q_1 = 7$. The upper half is 12, 13, 15, 18, 20, so $Q_3 = 15$. The interquartile range is $15 - 7 = 8$.

Box-and-whisker plots

A box plot shows the five-number summary: least value, $Q_1$, median, $Q_3$ and greatest value.

When comparing data sets, make one statement on the centre and one on the spread, in context: "The median mark of group $A$ is higher, so it did better on average. Its interquartile range is smaller, so its marks are more consistent."

Histograms

In a histogram, the area of each bar represents the frequency. The bars can have different widths, and the height is:

Continuous data has no gaps, so use the true class boundaries. Ages given as "10–19" years run from 10 to 20, and a length to the nearest cm given as "10–19" runs from 9.5 to 19.5.

$\dfrac{30}{10} = 3$ and $\dfrac{12}{4} = 3$. The bars are the same height, although the frequencies differ.

Cumulative frequency graphs

Plot the cumulative frequency against the upper class boundary, and join the points with a smooth curve.

From the curve you can estimate the median, the quartiles and any percentile, as well as how many values lie above or below any point.

Measures of centre and spread

Mean and standard deviation

The variance is the square of the standard deviation.

$\bar{x} = 12$, and $\sigma = \sqrt{164 - 144} = \sqrt{20} = 4.47$.

Grouped data

Use the midpoint of each class for $x$, with $f$ for frequency:

$\bar{x} = \frac{\sum fx}{\sum f} \qquad \sigma = \sqrt{\frac{\sum fx^2}{\sum f} - \bar{x}^2}$

These are estimates, because the midpoints stand in for the actual values.

Coded totals

Subtracting a constant $a$ moves the mean by $a$ but does not change the spread.

$\bar{x} = 50 + \dfrac{-40}{20} = 48$, and $\sigma = \sqrt{\dfrac{480}{20} - (-2)^2} = \sqrt{20} = 4.47$.

Combining two data sets

Totals add, but means and standard deviations do not. Go back to $\sum x$ and $\sum x^2$ for each set, then add them.

For $A$: $\sum x = 120$ and $\sum x^2 = 10(2^2 + 12^2) = 1480$.

For $B$: $\sum x = 300$ and $\sum x^2 = 15(3^2 + 20^2) = 6135$.

Combined: $n = 25$, $\sum x = 420$ and $\sum x^2 = 7615$.

$\bar{x} = \frac{420}{25} = 16.8 \qquad \sigma = \sqrt{\frac{7615}{25} - 16.8^2} = \sqrt{22.36} = 4.73$