• Mean: $\dfrac{\text{sum of values}}{\text{number of values}}$.
• Median: the middle value once the data is ordered; position $\tfrac{n+1}{2}$. For an even count, average the two middle values.
• Mode: the most frequently occurring value.
• Range: largest value minus smallest value — a measure of spread, not an average.

Example: for $5,6,2,4,7,8,3,5,6,6$: mean $= 52\div10 = 5.2$; ordered data gives median $=5.5$ (average of the 5th and 6th values); mode $=6$ (appears three times); range $=8-2=6$.

For the mean: multiply each value by its frequency, sum these products, and divide by the total frequency — $\text{Mean} = \dfrac{\sum(x\times f)}{\sum f}$. For the mode: find the highest frequency, then state the data value it belongs to — not the frequency itself. For the median: use a running total to find which value sits at position $\tfrac{n+1}{2}$.

Grouping data loses the exact position of each value within its interval, so the mean can only be estimated by assuming every value sits at its interval's midpoint.

1. Find each interval's midpoint, $m$.
2. Multiply by frequency to get $f \times m$ for each row.
3. Estimated mean $= \dfrac{\sum(f \times m)}{\sum f}$.

The modal class is simply the interval with the highest frequency; the median class is found via a running total, just as with an ungrouped table.

Use $\text{total} = \text{mean} \times \text{count}$ to work backwards to a missing value.

Example: 5 players have a mean score of 12. One leaves, and the remaining 4 have a mean of 10. Find the score of the player who left.

1. Original total: $5 \times 12 = 60$.
2. New total: $4 \times 10 = 40$.
3. Missing score: $60 - 40 = 20$.

State which average you're comparing (usually the mean or median — match whichever the question gives you), say which group scores higher in context, then compare the range or spread and say which group was more consistent. A smaller range means less variation between the highest and lowest values, not that the data is "better".

Example: Class A has mean $65$ and range $20$; Class B has mean $58$ and range $8$. Class A scored higher on average, but Class B was more consistent — its results were closer together.

Find the mean, median and mode.

1. $\sum f = 2+7+6+4+1 = 20$. $\sum(x\times f) = 0+7+12+12+4 = 35$. Mean $= 35\div20 = 1.75$.
2. Median position: $\tfrac{20+1}{2}=10.5$, so average the 10th and 11th values. Running totals ($2,9,15,19,20$) show both the 10th and 11th values fall at $x=2$. Median $=2$.
3. Highest frequency is 7, at $x=1$. Mode $=1$.