Average, total and number of values

Redistribute without changing the total

Four stacks contain 6, 8, 10 and 12 counters. The total is 36. Redistribute them equally: each stack gets 36÷4 = 9 counters. The arithmetic mean, or average, is 9, even though no original stack had 9.

Add all the values, then divide by how many values there are. Do not divide by the largest value or by the number of different values. Keep the same unit: if the values are distances in kilometres, their mean is in kilometres too.

A mean can be fractional even when the recorded values are whole numbers. Two pupils read 1 and 2 books. Their mean is 3÷2 = 11/2 books. This does not say that either pupil read exactly 11/2 books; it describes the pair using an equal share of their total.

Count every observation

For 0, 4, 5 and 7, the total is 16 and there are four values. The mean is 4. Zero contributes nothing to the sum, but it still counts as one observation. Ignoring it would change the set and give 16÷3, a different mean.

Repeated values count separately too. A score recorded twice belongs to two observations, not one.

Check your understanding 1 Keep every value

Four daily counts are 0, 4, 4 and 12. Find the total, number of values and mean. Explain how you treated the zero and the repeated 4.

Your total, count, mean and explanation

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The total is 0 + 4 + 4 + 12 = 20. There are four values, so the mean is 20÷4 = 5. Zero is one recorded daily count, and each 4 is a separate day’s count. Neither should be removed.

Decide which quantity is missing

The mean is an equal share, so putting all the shares back together gives:

If five quiz scores have mean 14, their total is 14 × 5 = 70. If the first four sum to 53, the fifth score is 70-53 = 17. Recover the whole total before subtracting the known part.

To find how many values there are, divide the total by the mean. If the total distance is 84 km and the mean daily distance is 12 km, there are 84÷12 = 7 days. Check 7 × 12 = 84.

Read whether the missing quantity is the mean, total or count before choosing an operation. A count of observations must be a positive whole number. A result such as 21/2 days cannot be the number of daily observations in this task: recheck the givens and calculation rather than rounding it.

Check your understanding 2 Find the right unknown

Six measurements have mean 9 cm. Find their total. In a separate set, the total is 45 cm and the mean is 5 cm. How many measurements are in that set? Check each answer using the relationship between mean, total and count.

Your total, number of measurements and checks

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The first total is 6 × 9 = 54 cm; 54÷6 = 9 cm checks the mean. The second count is 45÷5 = 9 measurements; 9 × 5 = 45 cm checks the total. The count has no centimetre unit.

Compare the totals before and after a change

A changed mean can come from a changed total, a changed number of values, or both. If you add one reading to four readings, there are now five readings. Using the old count with the new mean would give the wrong new total.

Check your understanding 3 How much must a new value contribute?

Four readings have mean 18. A fifth reading is added and the new mean is 20. Find the fifth reading. Show both totals and check the new mean.

Your old total, new total and fifth reading, or work on paper

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The old total is 4 × 18 = 72. The new total is 5 × 20 = 100. The new reading is 100-72 = 28. Check (72 + 28)÷5 = 20. The added reading must exceed 20 to raise the old mean of 18 to 20.

Combine totals, not unweighted means

A group of 3 pupils has mean score 8, so its total is 24. A group of 7 other pupils has mean score 6, so its total is 42. All 10 pupils have mean

The calculation (8 + 6)÷2 = 7 gives equal weight to the two group means. That does not give equal weight to the pupils: the second mean represents seven pupils, while the first represents only three. Recover each group’s total and combine the totals and counts. Averaging the two means directly works when the group sizes are equal.

Check your understanding 4 Use the group sizes

Two bags have mean mass 5 kg. Four other bags have mean mass 8 kg. Find the mean mass of all six bags. Show the two group totals before combining them.

Your group totals and combined mean mass

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The group totals are 2 × 5 = 10 kg and 4 × 8 = 32 kg. The combined mean is (10 + 32)÷6 = 7 kg. Averaging 5 and 8 directly would give 6.5 kg and ignore the unequal numbers of bags.

Is the mean a complete description?

The sets 0, 0, 12 and 4, 4, 4 both have mean 4. What does the mean hide? Also decide whether knowing only “total 0, mean 0” would tell you how many nonnegative values there are.

Your explanation of what each mean can and cannot tell you