Calculate the mean, median and mode, find a weighted mean and the mean from a frequency or grouped frequency table, identify the modal class, and choose the most appropriate average for the data.

What this dot point is asking

An average summarises a data set with one typical value. CCEA expects you to find the mean, median and mode, calculate a weighted mean, find the mean from a frequency table and an estimated mean from a grouped table, identify the modal class, and choose the most suitable average for a given context. These are bread-and-butter marks, but the grouped-mean and weighted-mean methods and the "which average" reasoning are where care pays off.

The three averages

The mean uses every value but is pulled by outliers. The median ignores extreme values, so it suits skewed data. The mode is the only average for non-numerical (qualitative) data, such as the most popular colour.

The weighted mean

A weighted mean is used when some values count more than others.

Mean from frequency and grouped tables

For a frequency table of discrete values, multiply each value by its frequency, total these, and divide by the total frequency. For a grouped table, the exact values are unknown, so you estimate the mean using the midpoint of each class.

Choosing the right average

The exam often asks which average is most appropriate and why. The mean is best when you want to use all the data and there are no extreme outliers. The median is best for skewed data or data with outliers, because it is not distorted by them, for example house prices or salaries where a few very large values would inflate the mean. The mode is best for the most common item (such as the shoe size a shop should stock most) and is the only average for categorical data such as favourite sport. A good answer names the average and gives the reason tied to the data, not just a definition.

The weighted mean also lets you combine the means of two groups of different sizes correctly. If a class of 30 has a mean of 60 and a class of 20 has a mean of 70, the overall mean is not 65 but $\dfrac{(30 \times 60) + (20 \times 70)}{30 + 20} = \dfrac{1800 + 1400}{50} = \dfrac{3200}{50} = 64$, because the larger class carries more weight. Treating each group's mean equally is a common and costly slip.

Why this matters

Averages are the most-used statistics in everyday life and the foundation of the analysis stage of the enquiry cycle. The grouped-mean method links straight to histograms and frequency tables, and the choice of average connects to spread and skewness: an outlier that distorts the mean is exactly why the median is sometimes preferred. The weighted mean reappears in index numbers, and the mean and standard deviation together define standardised scores at Higher tier.