Numeracy fundamentals

Example: Data relating to the favourite sports of students in a school can be summarised using the following frequency distribution table and column graph:

Use this information to determine the percentage of students who prefer tennis, as well as the most and least popular sports.

Solution: Using the data in the table we can determine that \(32 \div 200 \times 100 = 16\%\) of students prefer tennis. Using the graph or the table we can determine that football is the most and tennis is the least popular sport.

Example: Consider that the ages of ten education students are \(19\), \(20\), \(29\), \(28\), \(23\), \(18\), \(27\), \(22\), \(24\) and \(20\). What is the mean age of the students?

Solution: The mean age of the students can be calculated as follows:

Example: The following frequency distribution table shows the number of students who received different marks out of ten on a test (i.e. \(4\) students received a mark of \(6\), \(5\) students received a mark of \(7\), etc.):

Based on the data in this table, what is the mean mark?

Solution: To calculate the mean in this case you again need to divide the total by the number of values, however in order to find the total you need to take into account the frequency of each value. You can do this by multiplying each value by its frequency before adding together, and hence the calculation should be as follows:

Example: Consider that the ages of ten students are \(19\), \(20\), \(29\), \(28\), \(23\), \(18\), \(27\), \(22\), \(24\) and \(20\). What is the median age of the students?

Solution: The ages of the ten students in ascending order are \(18\), \(19\), \(20\), \(20\), \(22\), \(23\), \(24\), \(27\), \(28\) and \(29\). Since there are ten ages there are two middle values (the fifth and sixth ages), which are \(22\) and \(23\) respectively. Therefore the median age is \((22 + 23) \div 2 = 22.5\).

Example: The following frequency distribution table shows the number of students who received different marks out of ten on a test (i.e. \(4\) students received a mark of \(6\), \(5\) students received a mark of \(7\), etc.):

Based on the data in this table, what is the median mark?

Solution: Since there are \(30\) marks there are two middle values; the \(15\textrm{th}\) and \(16\textrm{th}\) marks. In order to find out what these are the cumulative frequencies (i.e. the sum of the frequencies to that point) should be calculated. It can be helpful to add another column to the table with these values, as follows:

From these cumulative frequencies we can see that the \(10\textrm{th}\) through to the \(16\textrm{th}\) students all have a mark of \(8\), and therefore that the \(15\textrm{th}\) and \(16\textrm{th}\) students both have a mark of \(8\). Hence \(8\) is the median.

Example: Consider that the ages of ten students are \(19\), \(20\), \(29\), \(28\), \(23\), \(18\), \(27\), \(22\), \(24\) and \(20\). What is the mode of the students’ ages?

Solution: The only age that occurs more than once in the list is \(20\), which occurs twice. Hence \(20\) is the mode.

Example: The following frequency distribution table shows the number of students who received different marks out of ten on a test (i.e. \(4\) students received a mark of \(6\), \(5\) students received a mark of \(7\), etc.):

Based on the data in this table, what is the mode mark?

Solution: The mode is \(9\), as that is the mark that occurs most frequently.

Example: Consider that the ages of ten students are \(19\), \(20\), \(29\), \(28\), \(23\), \(18\), \(27\), \(22\), \(24\) and \(20\). What is the range of the students’ ages?

Solution: The oldest person is \(29\) and the youngest is \(18\), so the range is \(11\) (i.e \(29 - 18\)).

Example 1: The following clustered column graphs compare the percentage of students who earned marks in four different ranges in a standardised test, for two different classes over two years:

Are the following statements true or false?

1. In \(2019\) more than three-quarters of students in class \(\textrm{A}\) had a mark over \(50\%\).
2. The distribution of marks was more similar for the classes in \(2019\) than in \(2020\).
3. The percentage of students who scored between \(26\%\) and \(50\%\) increased by \(11\%\) between \(2019\) and \(2020\) for Class \(B\).

Solution: Working and answers are as follows:

1. True. Since the question asks about \(2019\), we can focus on the first graph and ignore the second for this question. Furthermore, since it refers to Class \(\textrm{A}\) we can just look at the blue columns. The question asks whether more than three-quarters of the relevant students had a mark over \(50\%\), which equate to students in the third and fourth ranges on the graph (i.e. \(51 - 75\%\) and \(76 - 100\%\)). Adding the percentage of students in each of these columns together gives \(47\% + 29\% = 76\%\), which is just over three-quarters (since three-quarters is equivalent to \(75\%\)).
2. True. The column pattern for Class \(A\) and Class \(B\) is the same for \(2019\), but not for \(2020\).
3. False. This is true of Class \(A\), but the percentage of students for Class \(B\) actually decreased by \(4\%\).

Example 2: The following graph shows NAPLAN student achievement for years \(3\), \(5\), \(7\) and \(9\):

1. For how many year groups is Band \(6\) above the national minimum standard?
2. A student scores in Band \(5\), and is at the national minimum standard. What year must they be in?
3. How many bands are reported in common for years \(3\) and \(9\)?

Solution: Working and answers are as follows:

1. Looking at Band \(6\) in the graph we can see that it is green for three of the year groups, and the key indicates that green means that students are above the minimum standard. So Band \(6\) is above the national minimum standard for three year groups (years \(3\), \(5\) and \(7\)).
2. From the key we can see that a blue cell indicates that students are at the national minimum standard. Looking at Band \(5\) in the graph we can see that the only year group with a blue cell is Year \(7\), indicating that the student must be in Year \(7\).
3. Looking across at the different bands, we can see that there are only two reported in common for both years \(3\) and \(9\) (bands \(5\) and \(6\)).

Example 3: The following graph shows NAPLAN \(2018\) Reading results for year \(3\) students in eight states and territories and in Australia overall:

1. For which state or territory were the scores of more than \(20\%\) of students below Band \(3\)?
2. For how many states or territories were the scores of at least \(80\%\) of students at Band \(4\) or above?

Solution: Working and answers are as follows:

1. In order for the scores of more than \(20\%\) of students to be below Band \(3\), the entire dark orange section at the bottom of the column (which indicates the bottom \(20\%\) percentile) needs to be below Band \(3\). This is only true of the Northern Territory.
2. In order for the scores of at least \(80\%\) of students to be at Band \(4\) or above, the line indicating the \(20\textrm{th}\) percentile (the line above the the dark orange section at the bottom) must be in Band \(4\) or above. This is only true for Victoria and the ACT, so the answer is \(2\).

Example : The following spreadsheet shows the marks obtained by six students in a class:

In a spreadsheet such as this:

• Each cell is referred to by the column letter and row number. For example, ‘Matt’ is in cell \(\textrm{A}6\).
• ‘*’ stands for multiplication and ‘/’ stands for division
• Each formula starts with an equals sign

You can therefore use different cell references to perform different calculations. For example \(=\textrm{B}2 + \textrm{C}2\) is the formula for Michael’s total test score, while \(=(\textrm{B}4 + \textrm{D}4)/2\) gives Kate’s term \(1\) average score.

Example: What is the probability of rolling an even number when a die is rolled?

Solution: To calculate this consider that there are \(3\) outcomes in which an even number occurs (i.e. when a \(2\), \(4\) or \(6\) is rolled), and that there are \(6\) possible outcomes in total. So this gives a probability of \(\frac{3}{6}\), which should then be simplified to \(\frac{1}{2}\).

Example: What is the probability that when two dice are thrown the faces sum to \(7\)?

Solution: To calculate this you could create a sample space in the form of a table, as follows:

From this we can see that there are \(6\) possible outcomes where the faces sum to \(7\), out of a total of \(36\) equally likely outcomes. So the probability of the faces summing to \(7\) is \(\frac{6}{36}\), which simplifies to \(\frac{1}{6}\).