The mean

Data and probability, grade 8 Lesson 6 of 14Learning

The mean

The mean of data is their sum divided by the number of data.

When the data are $\displaystyle x_1, x_2, \dots, x_n$ – $\displaystyle n$ numbers – their mean $\displaystyle \bar{x}$ (read “x bar”) is:

\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}

The points Noa scored in five basketball games:
$\displaystyle 12, 18, 9, 15, 21$

The sum: $\displaystyle {12+18+9+15+21}={75}$ .
There are 5 data, so the mean: $\displaystyle {\frac{75}{5}}={15}$ points a game.

The mean as sharing equally

The mean is what each one would get if the total were shared out equally.

Four friends picked lemons: 3, 9, 5 and 7 kg.

Together they picked $\displaystyle {3+9+5+7}={24}$ kg.
Had each picked the same amount: $\displaystyle {\frac{24}{4}}={6}$ kg each – and that is the mean.
In the chart: what the bars taller than 6 have above it fills exactly what the shorter bars lack.

A mean of 2.4 children

The mean can come out a number that is not among the data, even a number that is not whole.

The number of children in five families on a street: $\displaystyle 3, 2, 1, 4, 2$ .
The mean: $\displaystyle {\frac{12}{5}}={2.4}$ children a family.

No family has 2.4 children, but the number describes all five families together: 10 such families have about 24 children.

The mean from a frequency table

In a frequency table each value comes up as many times as its frequency. So each value is multiplied by its frequency, the products are added, and the sum is divided by the number of data – the sum of the frequencies.

The marks of 20 pupils in a quiz (out of 10):

Mark	6	7	8	9	10
Frequency	2	5	6	5	2

The sum of the marks: $\displaystyle {6\cdot2+7\cdot5+8\cdot6+9\cdot5+10\cdot2}={12+35+48+45+20}={160}$
The number of pupils: $\displaystyle {2+5+6+5+2}={20}$ .
The mean: $\displaystyle {\frac{160}{20}}={8}$ .

When the values $\displaystyle x_1, \dots, x_k$ have the frequencies $\displaystyle f_1, \dots, f_k$ :

\displaystyle \bar{x}=\frac{x_1f_1+x_2f_2+\dots+x_kf_k}{f_1+f_2+\dots+f_k}

A missing value from the mean

When the mean is known and only one value is missing, the missing value is the solution of an equation.

Noam has two marks: 70 and 85.
He wants the mean of his three marks to be 80.
What does he need in the third test?

We call the third mark $\displaystyle x$ , and write that the mean is 80:

\frac{70+85+x}{3}=80

We multiply both sides by 3: $\displaystyle {70+85+x}={240}$ .
$\displaystyle {155+x}={240}$ , so $\displaystyle {x}={85}$ .

A check: $\displaystyle {\frac{70+85+85}{3}}={\frac{240}{3}}={80}$ . Noam needs 85.

Adding a value to the data

A new value larger than the mean raises it, a value smaller than the mean lowers it, and a value equal to the mean leaves it as it is.

The mean of four numbers is 15, so their sum is $\displaystyle {4\cdot15}={60}$ . The number 20 is added:

The new sum: $\displaystyle {60+20}={80}$ , and there are now 5 numbers.
The new mean: $\displaystyle {\frac{80}{5}}={16}$ – larger, since 20 is larger than 15.

Exercise 1

Work out the mean.

How many minutes Maya read each day of a week: $\displaystyle 25, 40, 30, 15, 35, 45, 20$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The price of a notebook in six shops, in shekels: $\displaystyle 12, 15, 9, 14, 10, 18$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

At a junction, the people in 30 passing cars were counted:

People in the car	1	2	3	4
Frequency	14	8	5	3

Work out the mean number of people in a car.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

Yoav’s marks in four quizzes: 8, 6, 9 and 7.
He wants a mean of 8 in the five quizzes.

Solve the equation: what mark $\displaystyle x$ does Yoav need in the fifth quiz?

Exercise 4

Maya ran on three days: 4 km, 6 km and 5 km.
She wants the mean of the four days to be 6 km.

How many km must she run on the fourth day? Write an equation and solve it.

Exercise 5

The mean height of six pupils is 150 cm.
A seventh pupil, 157 cm tall, joins them.

Work out the mean height of the seven pupils.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 6

The mean mark of the 25 pupils of a class in a test is 75.
A new pupil took the same test.

Will the class’s mean go up, go down or stay as it is, if the new pupil’s mark is:

Explain without working it out.

Exercise 7

The mean of the numbers $\displaystyle 3, 5, 10$ is 6.
2 is added to each of them.

What is the mean now? Explain why this happens.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: data and probability, grade 8

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Relative frequency

The relative frequency of an answer – $\displaystyle f$ its frequency, $\displaystyle n$ the total of all the frequencies
$\displaystyle \frac{f}{n}$

Measures of centre and the range

The mean of $\displaystyle n$ data
$\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$
The mean from a frequency table – $\displaystyle f_1, \dots, f_k$ the frequencies of the values $\displaystyle x_1, \dots, x_k$
$\displaystyle \bar{x}=\frac{x_1f_1+x_2f_2+\dots+x_kf_k}{f_1+f_2+\dots+f_k}$
The place of the median in an ordered list of $\displaystyle n$ data
$\displaystyle \frac{n+1}{2}$
The range – the largest value minus the smallest
$\displaystyle R=x_{\max}-x_{\min}$

Probability

Equally likely outcomes – $\displaystyle k$ the outcomes in the event $\displaystyle A$ , $\displaystyle n$ all the outcomes
$\displaystyle P(A)=\frac{k}{n}$
An impossible event – 0, a certain event – 1
$\displaystyle 0\le P(A)\le1$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
How many times an event $\displaystyle A$ is expected to happen in $\displaystyle n$ experiments
$\displaystyle n\cdot P(A)$