Representing, summarising and comparing data

What you’ll learn

• Interpret location, spread, skewness and outliers.
• Use frequency density and coding correctly.
• Recognise estimates from grouped data.

Understand the idea

A mean uses every observation and is sensitive to extreme values. The median and interquartile range describe the centre and middle half with less sensitivity to extremes. Compare two distributions using both location and spread, with units and context; a larger mean alone does not show greater consistency.

In a histogram, bar area represents frequency. Unequal class widths therefore require frequency density on the vertical axis. A stem-and-leaf diagram retains individual observations, while a box plot foregrounds quartiles and unusual values. A longer upper tail suggests positive skew, but use the actual plot or numerical evidence rather than assuming a fixed relationship for every dataset.

Grouped means use class midpoints and are estimates. A grouped percentile may be estimated by linear interpolation within its class, which assumes observations are spread uniformly inside that interval. Follow the outlier rule specified in the question. Coding $X=aU+b$ changes the mean to $a\bar u+b$ and the standard deviation to $|a|s_u$.

Key formulae

$$\bar x=\frac{\sum fx}{\sum f}$$

$$\operatorname{Var}(X)=\frac{\sum fx^2}{\sum f}-\bar x^2$$

$$\mathrm{IQR}=Q_3-Q_1$$

$$\text{frequency density}=\frac{\text{frequency}}{\text{class width}}$$

Worked example

Find the mean and population standard deviation of the five recorded waiting times 2, 4, 4, 6 and 9 minutes.

1. The totals are $\sum x=25$ and $\sum x^2=153$, with $n=5$.
2. The mean is $25/5=5$ minutes.
3. The population variance is $153/5-5^2=5.6$ minutes squared.
4. Take the square root: $\sigma=\sqrt{5.6}=2.37$ minutes to three significant figures.

Answer: Mean $5$ minutes; population standard deviation $2.37$ minutes.