Average

A crick­eter scores 20 in one match and 40 in the next. How well is he bat­ting? We say his aver­age is 30. An aver­age gives one num­ber that stands for a whole group of num­bers. In this les­son you will learn how to find an aver­age, a short way to write it, and a use­ful rule for what hap­pens to the aver­age when every num­ber is changed in the same way.

What an aver­age is#

What is the aver­age of 2 and 4?

Add them and divide by how many there are:

$$\displaystyle \frac{2 + 4}{2} = \frac{6}{2} = 3$$

The aver­age is 3. If both num­bers were 3, they would add up to the same total, 6.

Aver­age.

$$\displaystyle \text{average} = \frac{\text{sum of the observations}}{\text{number of observations}}$$

Each num­ber in the list is called an obser­va­tion. The aver­age is also called the mean.

A short way to write it#

Sup­pose there are $n$ obser­va­tions, called $x_1, x_2, x_3, \ldots, x_n$. Their aver­age is writ­ten $\bar{x}$ and read "x bar".

$$\displaystyle \bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n}$$

The Greek let­ter $\Sigma$, called sigma, means "add them all up". So we can also write

$$\displaystyle \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$$

Mul­ti­ply both sides by $n$ and you get a fact that is used in almost every prob­lem:

$$x_1 + x_2 + x_3 + \cdots + x_n = n\bar{x}$$

In words: sum = num­ber of obser­va­tions × aver­age.

Exam­ple 1#

Find the aver­age of the first nine prime num­bers.

The first nine primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

$$\displaystyle \text{average} = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23}{9} = \frac{100}{9} = 11\frac{1}{9}$$

The aver­age does not have to be one of the num­bers. It does not even have to be a whole num­ber.

Exam­ple 2#

The aver­age of five con­sec­u­tive odd num­bers is 61. Find the largest of them.

Con­sec­u­tive odd num­bers go up by 2. Let the small­est be $x$. Then the five num­bers are $x$, $x + 2$, $x + 4$, $x + 6$ and $x + 8$.

$$\displaystyle \begin{aligned} \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} &= 61 \\ \frac{5x + 20}{5} &= 61 \\ x + 4 &= 61 \\ x &= 57 \end{aligned}$$

The largest num­ber is $57 + 8 = 65$. The num­bers are 57, 59, 61, 63 and 65, and 61 is right in the mid­dle.

Using sum = num­ber × aver­age#

Many prob­lems give you aver­ages of groups. Turn each aver­age into a sum first. Sums can be added; aver­ages can­not.

Exam­ple 3#

A crick­eter's aver­age score in two matches is 27. His aver­age in three other matches is 32. Find his aver­age in all five matches.

Let the scores be $m_1$ to $m_5$.

$$m_1 + m_2 = 2 \times 27 = 54$$

$$m_3 + m_4 + m_5 = 3 \times 32 = 96$$

$$\displaystyle \text{average of all five} = \frac{54 + 96}{5} = \frac{150}{5} = 30$$

Notice that the answer is not the aver­age of 27 and 32, which would be 29.5. The group of three matches counts for more.

Exam­ple 4#

The aver­age weight of 29 stu­dents in a class is 48 kg. When the teacher's weight is included, the aver­age goes up by 500 g. Find the teacher's weight.

500 g is $\displaystyle \frac{1}{2}$ kg. So the new aver­age is $\displaystyle 48\frac{1}{2}$ kg, and there are now 30 peo­ple.

Sum of the 29 stu­dents' weights $= 48 \times 29$ kg. Call this (1).

Sum of the 30 weights $\displaystyle = 48\frac{1}{2} \times 30 = 48 \times 30 + \frac{1}{2} \times 30$ kg. Call this (2).

The teacher's weight is (2) minus (1):

$$\displaystyle \begin{aligned} T &= 48 \times 30 + \frac{1}{2} \times 30 - 48 \times 29 \\ &= 48 \times (30 - 29) + 15 \\ &= 48 + 15 = 63 \end{aligned}$$

The teacher weighs 63 kg.

A quick way to think about it: the teacher brings his own 48 kg, plus an extra half kilo­gram for each of the 30 peo­ple, which is 15 kg more.

Chang­ing every obser­va­tion#

Rule. If the aver­age of $x_1, x_2, \ldots, x_n$ is $\bar{x}$, then the aver­age of $ax_1 + b, ax_2 + b, \ldots, ax_n + b$ is $a\bar{x} + b$.

So if you mul­ti­ply every num­ber by $a$ and then add $b$, the aver­age is mul­ti­plied by $a$ and then has $b$ added.

Proof. We are given $\displaystyle \frac{x_1 + x_2 + \cdots + x_n}{n} = \bar{x}$.

$$\displaystyle \begin{aligned} \text{new average} &= \frac{(ax_1 + b) + (ax_2 + b) + \cdots + (ax_n + b)}{n} \\ &= \frac{a(x_1 + x_2 + \cdots + x_n) + nb}{n} \\ &= a \times \frac{x_1 + x_2 + \cdots + x_n}{n} + \frac{nb}{n} \\ &= a\bar{x} + b \end{aligned}$$

There are $n$ copies of $b$, which is why we get $nb$.

Exam­ple 5#

The aver­age of 10 num­bers is 23. Each num­ber is increased by 4. What is the new aver­age?

Here $a = 1$ and $b = 4$. The new aver­age is $23 + 4 = 27$.

Exam­ple 6#

The aver­age of 10 num­bers is 23. Each num­ber is mul­ti­plied by 3. What is the new aver­age?

Here $a = 3$ and $b = 0$. The new aver­age is $3 \times 23 = 69$.

Prac­tice#

1. Find the aver­age of 12, 15, 18, 21 and 24.
2. Find the aver­age of the first five even num­bers.
3. The aver­age of 8 num­bers is 14. Find their sum.
4. The aver­age of 6 num­bers is 20. Each num­ber is mul­ti­plied by 2 and then 5 is added. Find the new aver­age.
5. The aver­age of four con­sec­u­tive even num­bers is 27. Find the num­bers.
6. The aver­age mark of 20 boys is 62 and the aver­age mark of 30 girls is 72. Find the aver­age mark of all 50 stu­dents.
7. The aver­age weight of 24 stu­dents is 40 kg. When the teacher is included, the aver­age goes up by 1 kg. Find the teacher's weight.
8. Find the aver­age of the first 10 nat­ural num­bers.
9. The aver­age age of 15 play­ers is 22 years. When the coach's age is included, the aver­age becomes 23 years. Find the coach's age.

Answers#