Finding and Using the Mean (Average)

1. The mean as a fair share

Imagine five friends who have 4, 6, 8, 10 and 12 oranges. If they put all their oranges together and shared them out equally, each friend would get the same number. That equal share is the mean. Together there are 4 + 6 + 8 + 10 + 12 = 40 oranges. Shared among 5 friends, each gets 40 ÷ 5 = 8.

This gives us the rule for the mean of an ungrouped data set:

$$\text{mean}=\frac{\text{sum of all the values}}{\text{number of values}}$$

▶ The red dashed line is the mean. Bars taller than the line (9, 10) 'give' to the shorter bars (7, 6) until all are level at 8.

2. Calculating the mean step by step

1. 1Add all the values to get the sum.
2. 2Count how many values there are.
3. 3Divide the sum by the number of values.
4. 4State the answer with its unit (cedis, kg, marks) and check that it lies between the smallest and largest values.

Example. A shopkeeper sold 25, 48, 25, 33, 57 and 50 loaves of bread over six days. Sum = 25 + 48 + 25 + 33 + 57 + 50 = 238 loaves. Number of days = 6. Mean = 238 ÷ 6 = 39.67 loaves per day (correct to 2 decimal places). Notice the mean does not have to be a whole number or equal to any one of the values, because it is a fair-share value.

The mean is useful in everyday life. A farmer finds the mean mass of cocoa bags, a coach finds the mean number of goals per match, and a teacher finds the class mean mark. Because it uses every value, an unusually big or small value pulls the mean towards it. For instance, the mean of 2, 3, 3, 4 and 48 is 12, which is not typical of most of the data.

3. The mean from a frequency table

When the same value occurs many times, the data are given in a frequency table. Instead of adding the same number again and again, multiply each value $x$ by its frequency $f$. Then:

$$\text{mean}=\frac{\sum fx}{\sum f}=\frac{\text{total of all }f\times x}{\text{total frequency}}$$

Here are the marks out of 5 scored by 20 learners in a class test.

Total frequency = 2 + 6 + 4 + 5 + 3 = 20 learners. Total of f × x = 2 + 12 + 12 + 20 + 15 = 61. Mean = 61 ÷ 20 = 3.05 marks.

▶ The bars show the frequencies. The mean of 3.05 is a little above the middle score 3, because more learners scored 4 than scored 2.

4. Solving problems with the mean

Because mean × number of values = sum, we can work backwards. If you know the mean and the number of values, you can find the total. This is the key to many BECE questions.

Missing value. Sena has scores 75, 87, 90, 88 and 79 in five subjects. She wants an average of 85 over six subjects. Total needed = 85 × 6 = 510. Total so far = 75 + 87 + 90 + 88 + 79 = 419. Score needed = 510 − 419 = 91. In equation form: (419 + x) ÷ 6 = 85.

Adding a value to a data set. The mean of 8 learners' mass is 40 kg. Their total mass is 8 × 40 = 320 kg. If a learner of mass 49 kg joins, the new total is 369 kg for 9 learners, so the new mean is 369 ÷ 9 = 41 kg.

Worked examples

Try each one yourself first, then open the solution and compare every step.