The mean and the weighted mean

Statistics and probability, grade 10 (4 units) Lesson 6 of 20Learning

The mean and the weighted mean

The mean of a list

The mean is the sum of the data divided by the number of data. It is the value each datum would have if the sum were shared out equally.

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

Rami’s English marks during the year:
60, 65, 70, 62, 87, 77.

\bar{x}=\frac{60+65+70+62+87+77}{6}=\frac{421}{6}\approx70.17

The mean is not a whole number, so it is rounded – here to two decimal places – and written with $\displaystyle \approx$ .

The mean from a frequency table

In a frequency table each value appears several times, so instead of adding it again and again it is multiplied by its frequency. The sum is divided by the sum of the frequencies – the number of all the data.

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

The number of fouls of 40 high jumpers in a competition:

Number of fouls	0	1	2	3	4
Frequency	4	16	9	6	5

\bar{x}=\frac{0\cdot4+1\cdot16+2\cdot9+3\cdot6+4\cdot5}{40}=\frac{72}{40}=1.8

A jumper had 1.8 fouls on average. No jumper fouled 1.8 times: the mean need not be one of the values.

The mean from a bar chart

In a bar chart of a quantitative variable, the values of the variable are written under the bars, and each bar’s height is the frequency. Read a frequency table off the chart, and work it out as from a table.

The number of goals in each of the season’s 20 league matches:

\bar{x}=\frac{0\cdot4+1\cdot6+2\cdot5+3\cdot3+4\cdot2}{20}=\frac{33}{20}=1.65

On average 1.65 goals were scored in a match.

The properties of the mean

The mean is worked out from all the data, and that is both its strength and its weakness.

The mean lies between the least datum and the greatest (or equals one of them, when all the data are equal).
It need not be one of the data: 1.8 fouls.
It has the same units as the data: marks, goals, minutes.
One extreme datum moves it a lot: had Rami also had a mark of 0, the mean would drop to $\displaystyle \frac{421}{7}\approx60.14$ .
A qualitative variable has no mean: “by bike” and “on foot” cannot be added.

A missing datum from a known mean

When the mean is known, so is the sum: the sum is the mean times the number of data. That gives a missing datum.

The mean hourly pay of six temporary workers is 50 ₪.
Five of them earn 50, 48, 53, 52 and 45 ₪.

Call the sixth worker’s pay $\displaystyle x$ :

\frac{50+48+53+52+45+x}{6}=50

248+x=300

x=52

The sixth worker earns 52 ₪ an hour.

The weighted mean

In a weighted mean each value has a weight – how much it counts. Multiply each value by its weight, and divide by the sum of the weights.

\displaystyle \bar{x}=\frac{w_1x_1+w_2x_2+\dots+w_kx_k}{w_1+w_2+\dots+w_k}

Dana’s maths mark is made of two parts:

the test – 60% of the mark; Dana got 85 in it.
the project – 40% of the mark; Dana got 70 in it.

\bar{x}=\frac{60\cdot85+40\cdot70}{60+40}=\frac{5100+2800}{100}=79

The plain mean of 85 and 70 is 77.5. Dana’s mark is higher, because the test, where her mark is higher, has the greater weight. The mean from a frequency table is a weighted mean too: the weights are the frequencies.

Exercise 1

Tamar’s English marks during the year:
10, 85, 92, 87, 100, 93.
She got the 10 in a test she sat while ill.

Work out the mean of Tamar’s marks. Round to two decimal places.

/ – 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

Work out the mean of the other five marks, without the 10.

/ – 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

Which of the two means describes better how well Tamar knows English? Explain.

Exercise 2

At a car park’s entrance the number of people in each of 40 cars was counted:

Number of people in the car	1	2	3	4
Frequency	12	15	8	5

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

25 students were asked how many books they read during the holiday:

Work out the mean number of books per student.

/ – 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 4

The mean of Ido’s five tests is 82.
His marks in four of them: 78, 90, 74, 85.

Solve the equation: what is the mark $\displaystyle x$ in the fifth test?

Exercise 5

A coffee roaster mixes three kinds of coffee:

20 kg of coffee at 40 ₪ a kg;
30 kg of coffee at 50 ₪ a kg;
50 kg of coffee at 60 ₪ a kg.

What is the mean price of one kg of the blend?

/ – 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

Write “true” or “false”. For “true” explain why; for “false” give a counterexample.

The mean must be one of the data.
The mean lies between the least datum (included) and the greatest (included).
Adding a datum 0 to the data does not change the mean.

Formulas: statistics and probability, grade 10 (4 units)

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Presenting data

Relative frequency – $\displaystyle f$ a value’s frequency, $\displaystyle n$ the number of all the data
$\displaystyle \frac{f}{n}$
Relative frequency in percent
$\displaystyle \frac{f}{n}\cdot100\%$
The angle of a slice in a pie chart
$\displaystyle \frac{f}{n}\cdot360^{\circ}$

Measures of centre and position

The mean of $\displaystyle n$ data
$\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$
The mean from a frequency table – $\displaystyle x_1, \dots, x_k$ the values, $\displaystyle f_1, \dots, f_k$ their frequencies
$\displaystyle \bar{x}=\frac{x_1f_1+x_2f_2+\dots+x_kf_k}{f_1+f_2+\dots+f_k}$
A weighted mean – $\displaystyle w_1, \dots, w_k$ the weights of the values $\displaystyle x_1, \dots, x_k$
$\displaystyle \bar{x}=\frac{w_1x_1+w_2x_2+\dots+w_kx_k}{w_1+w_2+\dots+w_k}$
The mean of data in classes – $\displaystyle m_1, \dots, m_k$ the classes’ middles, $\displaystyle f_1, \dots, f_k$ their frequencies
$\displaystyle \bar{x}\approx\frac{m_1f_1+m_2f_2+\dots+m_kf_k}{f_1+f_2+\dots+f_k}$
The median’s place in an ordered list of $\displaystyle n$ data
$\displaystyle \frac{n+1}{2}$
The lower decile’s place in an ordered list; the upper decile’s – $\displaystyle 0.9\cdot n$
$\displaystyle 0.1\cdot n$

Measures of spread

The range – the greatest value minus the least
$\displaystyle x_{max}-x_{min}$
The variance – the mean of the squared deviations from the mean
$\displaystyle \sigma^2=\frac{(x_1-\bar{x})^2+(x_2-\bar{x})^2+\dots+(x_n-\bar{x})^2}{n}$
The standard deviation – the square root of the variance
$\displaystyle \sigma=\sqrt{\sigma^2}$
The variance from a frequency table
$\displaystyle \sigma^2=\frac{(x_1-\bar{x})^2f_1+\dots+(x_k-\bar{x})^2f_k}{f_1+\dots+f_k}$

The rules of probability

Equally likely outcomes – $\displaystyle k$ the number of outcomes in the event $\displaystyle A$ , $\displaystyle n$ the number of outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
The probability of any event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
The union of disjoint events
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)$ $\displaystyle A\cap B=\emptyset$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A|B)$
The law of total probability – a weighted mean of the conditional probabilities
$\displaystyle P(A)=P(B)\cdot P(A|B)+P(\overline{B})\cdot P(A|\overline{B})$
Bayes’ formula – from $\displaystyle P(B|A)$ to $\displaystyle P(A|B)$
$\displaystyle P(A|B)\cdot P(B)$ $\displaystyle =$ $\displaystyle P(B|A)\cdot P(A)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A|B)$ $\displaystyle =$ $\displaystyle P(A)$