Subjects
Grades
The mean (Avg) is a measure of central tendency that represents the centre of equilibrium of a distribution.
There are 2 types of means.
The arithmetic mean is the value that each data value in the distribution would have if the total was equally divided among them. It is calculated by adding all the data and dividing the sum by the number of data values in the distribution. It is calculated differently depending on how the data are presented.
|\text{Mean} = \dfrac{\text{Sum of all the data}}{\text{Number of data values}}|
To simplify notation, different symbols can be used.
When referring to the mean of a sample, the symbol |\overline x| is used.
When referring to the mean of a population, the Greek letter |\mu| is used.
The arithmetic mean is calculated in the same way in both cases.
The following is the number of goals scored by the Montreal Canadiens in their last |15| games:
|0,| |1,| |3,| |2,| |3,| |1,| |3,| |4,| |5,| |2,| |5,| |1,| |3,| |4,| |2|
What is the average number of goals scored by the Canadiens in their last 15 games?
||\begin{align} \text{Mean} &= \dfrac{\small0+1+3+2+3+1+3+4+5+2+5+1+3+4+2}{15} \\\\ &= \dfrac{39}{15} \\\\ &= 2.6\ \text{goals per game} \end{align}||
Answer: During the last |15| games, the Canadiens averaged |2.6| goals per game.
It is possible to find a missing data value if the mean of a distribution and all other data values are known. The definition of the mean can be used to solve this kind of problem.
For her Term 3 report card, Marie-Claude has set a goal of getting an |85\ \%| mean (average) in math. So far, she has earned the following results: |90\ \%,| |82\ \%,| and |81\ \%.|
If all the evaluations are equally weighted, what grade does Marie-Claude need to have on her last evaluation to reach her goal?
Since the mean is the value that each data would have if the total was equally divided among them, we can rephrase this and conclude that Marie-Claude needed to get |85\ \%| on all |4| evaluations. So if we add |85\ \%| together |4| times, we get |85\ \% \times 4 = 340\ \%.|
She must therefore accumulate a total of |340\ \%| on her |4| evaluations to reach her goal. However, she has already received |3| grades: |90\ \%,| |82\ \%,| and |81\ \%.| So, after |3| evaluations, she has accumulated a total of |90\ \% + 82\ \% + 81\ \% = 253\ \%.| This means that the missing value can be found by calculating |340\ \% - 253\ \% = 87\ \%.|
Answer: Marie-Claude needs a grade of |87\ %| to have an |85\ \%| average in Term 3.
It is also possible to solve this kind of problem using algebra. To do so, replace the unknown data value with |x| and calculate its value.
The mean of |5| data values is |35,| but only |4| of the |5| data values are known: |20,| |40,| |45| and |29.|
What is the missing data value?
Replace the missing data value with |x| and use the arithmetic mean formula.
||\begin{align} \text{Mean} &= \dfrac{\text{Sum of all data}}{\text{Number of data values}} \\ 35 &= \dfrac{20 + 40 + 45 + 29 + x}{5} \\ 35 &= \dfrac{134+x}{5} \end{align}||
Next, isolate |x.|
||\begin{align}35\boldsymbol{\color{#ec0000}{\times5}}&=\dfrac{134+x}{5}\boldsymbol{\color{#ec0000}{\times5}}\\175&=134+x\\175\boldsymbol{\color{#ec0000}{- 134}}&=134+x\boldsymbol{\color{#ec0000}{- 134}}\\41&=x\end{align}||
Answer: The missing data is |41.|
Sometimes a distribution contains values that are repeated several times. In this case, it is useful to group them together in a condensed data table. In this situation, the mean (average) is calculated as follows:
|\text{Mean} = \dfrac{\text{Sum of all data values multiplied by their frequencies}}{\text{Total number of data}}|
The age of 30 players on a sports team is represented in the following table:
| Age | |7| | |8| | |9| | |10| |
|---|---|---|---|---|
| Frequency | |13| | |9| | |6| | |2| |
What is the mean (average) age of the players on this team?
Note that the age of |7| repeats |13| times |(7 \times 13),| age |8| repeats |9| times |(8 \times 9),| age |9| is present |6| times |(9 \times 6)| and age |10| is present |2| times |(10 \times 2).|
||\begin{align}\text{Mean}&= \dfrac{7 \times 13 + 8 \times 9 + 9 \times 6 + 10 \times 2}{13+9+6+2}\\ &= \dfrac{91+72+54+20}{30}\\&=\dfrac{237}{30}\\&= 7.9\ \text{years old}\end{align}||
Answer: The average age of the students on the team is |7.9| years old.