Numeracy-fundamentals-printable-guide-UniSkills.pdf
Numeracy fundamentals
Table of Contents
Introduction
Welcome to the Numeracy fundamentals module. This module introduces important numeracy concepts that you may need for your studies and in everyday life. Some parts may review things you’ve already learned, while other parts might be new to you. The aim is to help you understand how to do calculations and how to use these ideas in real-world situations.
You can choose to work through the module in the order it’s presented, or you can skip to the sections that interest you most. If you’d like to explore certain topics further, there are separate modules on Algebra, Geometry and trigonometry, and Statistics.
We welcome your feedback at any time. You can share your thoughts on the feedback page. If you have any questions about the module, please contact Library-UniSkills@curtin.edu.au
What you will learn
- How to use BIMDAS to solve problems with more than one operation.
- How to round decimal numbers.
- How to work with percentages, fractions, ratios, proportions and rates.
- How to calculate averages.
- How to write and apply simple algebraic formulas.
- How to convert between units of measurement.
- How to solve problems using diagrams.
- How to perform calculations with time and speed.
- How to interpret directions.
- How to calculate the perimeter, area and volume of common shapes.
- How to find and understand key statistics for categorical and continuous data.
- How to interpret graphs.
- How to write and use Excel formulas.
- How to calculate probabilities and create sample spaces to solve probability problems.
Number & algebra
This page explains some key ideas about number and algebra. As you go through the content, you’ll find activities to help you practise and strengthen your understanding.
If you would like more example questions on these topics, especially if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you might like to visit the LANTITE page. If you’re interested in learning more advanced algebra skills, you can also check out the Algebra module.
As you work through the examples, remember that there is often more than one way to solve a problem. In some cases, we’ve shown different methods, but for many examples, only one way is explained. If you already know another correct method, that’s great! You don’t need to change how you do it. What matters most is that you use a method that makes sense to you and gives the right answer.
In brief, this page covers the following:
- How to use BIMDAS to solve problems with more than one operation.
- How to round decimal numbers.
- How to perform calculations involving percentages.
- How to carry out calculations involving fractions.
- How to perform calculations involving ratios, proportions and rates.
- How to calculate averages.
- How to write and apply simple algebraic formulas
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Order of operations (BIMDAS)
Understanding the correct order to complete calculations in a maths problem is very important. If you don’t follow the right order, you might get the wrong answer. It is also important to use the correct order when writing expressions, so that other people - and even your calculator - can understand and solve them the way you meant.
The rule for this is called the order of operations, and one way to remember it is with the acronym BIMDAS. This stands for:
- Brackets - Always do the calculations inside brackets first. When writing expressions, use brackets to make sure certain parts are done first. If you’re not sure, it’s okay to add brackets anyway - it won’t cause any problems.
- Indices - After brackets, solve any indices (also called exponents). These are not explained in this section, but if you want to learn more, check out the Algebra module.
- Multiply, Divide - Next, do any multiplication or division. These two are treated equally, so just go from left to right and do whichever comes first.
- Add, Subtract - Lastly, do any addition or subtraction. Like multiplication and division, these are also equal in priority, so work from left to right and solve whichever comes first.
Example: Michelle is a maths teacher. She wants to buy a licence for some maths software that her students will use during a \(10\) week school term. The licence costs \(\$5\) per student as a one-time registration fee. After that, it costs \(\$2\) per student each week. Michelle has \(27\) students in her class. How much will the total cost be?
Solution: To work this out, we first need to calculate the cost for one student. Each student pays \(\$5\) for registration, plus \(\$2\) per week for \(10\) weeks. So the cost for one student is \(5 + 2 \times 10\) (we don’t need brackets here because BIMDAS tells us that multiplication happens before addition, so the total weekly cost is calculated first).
Next we need to multiply the cost per student by \(27\) as there are \(27\) students in the class. To make sure the cost per student is calculated first, we use brackets, so the formula should be \(27 \times (5 + 2 \times 10)\)
You can enter this into a calculator, or work it out by hand. Either way, the total cost is \(\$675\)
Another way to write the formula is \(27 \times 5 + 27 \times 2 \times 10\) which gives the same total of \($675\)
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Once you have worked through the example above, have a go at using BIMDAS to solve the problems in the following activity:
Decimal numbers
A decimal number has two parts. The part on the left side of the decimal point is the whole number part. The part on the right side is the fractional part.
If the whole number part is \(0\) we usually write the \(0\) before the decimal point to make the number clearer. For example, we write \(0.35\) not just \(.35\)
If the fractional part is \(0\) we usually leave out the decimal point and write the number as a whole number. For example, instead of writing \(2.0\) we would usually just write \(2\)
Being able to round decimal numbers is important as often it is not practical or necessary to write every digit. In some cases, like with the number \(\pi\) (Pi), it is actually impossible to write the full number because the decimal goes on forever without repeating. These are called irrational numbers.
The number of decimal places you round to can depend on the situation. For example:
- when working with money, you usually round to two decimal places;
- if a question gives values to three decimal places, you should round your answer to three decimal places;
- sometimes, it makes more sense to round to the nearest whole number, such as when counting people, or;
- you might also be told exactly how many decimal places to round to.
When solving a problem with more than one step, it is important to keep all the exact values while you are woirking. You should only round your final answer. This helps make sure your answer is accurate.
When rounding a decimal number to a certain number of decimal places, or to the nearest whole number, follow these steps:
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First, find the ‘round-off place’. If you are rounding to \(n\) decimal places then your round-off place is the number in the \(n^{th}\) decimal place. If you are rounding to the nearest whole number, then your round-off place is the number to the left of the decimal point.
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Next, look at the digit immediately to the right of your round-off place.
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If this number is \(5\) or more, round up. This means you increase the digit in the round-off place by \(1\) and remove all digits after it. If this number is less than \(5\), round down. This means you keep the digit in the round-off place the same and remove all digits after it.
Example: Suppose you want to round \(4.576\) to two decimal places.
Solution: Following the steps above, we have:
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Since we are rounding to two decimal places the round-off place is the number in the second decimal place, which in \(4.576\) is \(7\)
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The number immediately to the right of this is the number in the third decimal place, which is \(6\)
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Since \(6\) is greater than \(5\) we increase the number in the round-off place from \(7\) to \(8\) and remove all digits after it. So we round up to \(4.58\)
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Once you have worked through the example above, have a go at the rounding problems in the following activity:
Percentages
A percentage is a way of showing a number out of \(100\). It is written using the % symbol. For example, if the tax rate is \(30\%\), it means you pay \(\$30\) in tax for every \(\$100\) you earn.
Here are some important facts about percentages:
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You can find a percentage of a quantity. This includes working out percentage increases and decreases. You can also show one quantity as a percentage of another.
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Sometimes, the percentage can be more than \(100\%\). This can happen when calculating a percentage increase or when comparing one quantity to another.
Percentages are often used to understand and compare data. They are also used to show things like discounts in sales, commissions, and interest rates.
Calculating percentages
A percentage shows an amount out of \(100\). To calculate a percentage of a quantity you need to divide the percentage by \(100\) and multiply by the quantity.
Example: If \(15\%\) of the students in a particular school are likely to need extra help with their numeracy each year, and there are \(360\) students in the school this year, how many students are likely to need extra help?
Solution: To calculate \(15\%\) of \(360\) you need to divide \(15\) by \(100\) and multiply by \(360\):
\(15 \div 100 \times 360 = 54\)
So \(54\) students are likely to need extra help with their numeracy. You can always do the operations in a different order if you prefer, as long as it gives you the correct answer and makes sense to you.
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Building on this idea, you will often need to work out the result of a percentage increase or decrease. For example, you may want to find:
- the new price of an item after a \(20\%\) discount, or
- the amount of money in your bank account after \(3\%\) interest has been added.
There are two main ways to find the new value after a percentage is applied:
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Find the percentage of the original value, using the method explained earlier.
- If it’s an increase, add this amount to the original value.
- If it’s a decrease, subtract this amount from the original value.
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Add or subtract the percentage from 100%, depending on whether it’s an increase or decrease.
- The calculate this new percentage of the original value, using the same method as before.
Both methods one and two will give you the same final result. You can choose the one that feels easier for you or the situation.
Example 1: Suppose you want to find the total cost of a \(\$350\) computer, once \(10\%\) GST has been added.
Solution: One way of calculating this would be:
\(10 \div 100 \times \$350 = \$35\) and then \(\$350 + \$35 = \$385\)
So the total cost of the computer is \(\$385\)
Example 2: Suppose you’d like to buy a TV, and you find one that has a \(25\%\) off deal. If the TV was originally priced at \(\$799\) how much will the sale price be?
Solution: One way of calculating this would be:
\(100 - 25 = 75\) and then \(75 \div 100 \times \$799 = \$599.25\)
So the new price of the TV will be \(\$599.25\)
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Alternatively, you may sometimes need to work out the original value before a percentage was applied, using the new value you already know. For example, you might need to find:
- the original price of an item before a \(20\%\) discount given the discounted price, or
- the original bank balance before \(3\%\) interest is added, given the current balance.
To do this, add the percentage to \(100\%\) if it was an increase, or subtract the percentage from \(100\%\) if it was a decrease. Then divide the new value by this percentage and multiply the result by \(100\).
Example: Suppose you know that the cost including GST of a tablet is \(\$451\) and you want to calculate the cost before \(10\%\) GST was added.
Solution: To calculate this you would do:
\(100 + 10 = 110\) and \(\$451 \div 110 \times 100 = \$410\)
So the original price of the tablet was \(\$410\)
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Once you have worked through the examples above, have a go at the percentage problems in the following activity:
Calculating percentages without a calculator
While you can usually use a calculator to work out percentages with the formulas above, it is also important to understand how to estimate (approximate) percentages without a calculator. You can do this by combining the methods below. You can also round numbers to make the calculations easier when needed:
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To calculate \(1\%\) of a number: divide by \(100\) by moving the decimal place two places to the left (For whole numbers, this is the same as removing two zeroes, if possible.)
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To calculate \(10\%\) of a number: divide by \(10\) by moving the decimal place one place to the left (For whole numbers, this is the same as removing one zero, if possible.)
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To calculate \(25\%\) of a number: divide by \(4\) (or divide by \(2\) and \(2\) again if that is easier)
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To calculate \(33\%\) of a number: divide by \(3\)
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To calculate \(50\%\) of a number: divide by \(2\)
Example: Approximate \(21\%\) of \(619300\)
Solution: You could do this as follows:
\(10\%\) of \(619300\) is \(61930 \approx 62000\) so \(20\%\) of \(619300 \approx 62000 + 62000 \approx 124000\)
\(1\%\) of \(619300\) is \(6193 \approx 6200\)
So \(21\%\) of \(619300 \approx 124000 + 6200 \approx 130200\)
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Once you have worked through the example above, have a go at completing the percentage problems in the activity below without using a calculator:
Expressing one quantity as a percentage of another
Another way you may need to use percentages is to show one quantity as a percentage of another. To do this, divide the first quantity by the second quantity and then multiply by \(100\)
Example: Suppose that a student scored \(34\) out of \(40\) on a test, and you want to give them their result as a percentage.
Solution: To calculate what percentage \(34\) is of \(40\) you would do:
\(34 \div 40 \times 100 = 85\%\)
So the student scored \(85\%\) on the test.
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Finally, you may need to calculate a percentage change. For example, you might calculate the percentage change in a power bill from the previous amount to the current amount.
To do this:
- First, subtract one value from the other to find the difference
- Next, divide this difference by the original amount
- Finally, multiply the result by \(100\) to convert it to a percentage.
Example: Suppose a student obtains a mark of \(72\) in a standardised test one year and a mark of \(81\) the next. What is the percentage change in the student’s mark?
Solution: To calculate the percentage change from \(72\) to \(81\) you would do:
\(81\) (new amount) \(- 72\) (original amount) \(= 9\)
\(9 \div 72\) (original amount) \(\times 100 = 12.5\%\)
So the student’s mark has increased by \(12.5\%\)
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Once you have worked through the examples above, have a go at the percentage problems in the activity below:
Fractions
A fraction is part of a whole number. The bottom number (called the denominator) shows how many equal parts the whole is divided into. The top number (called the numerator) shows how many of those parts you have.
Understanding fractions is important for many everyday tasks, such as cooking, telling the time, reading maps, and working out discounts. It is also an important skill for other areas of maths, including ratio and proportion, algebra, geometry, and statistics.
You may often need to:
- simplify fractions
- convert them into equivalent fractions
- add or subtract them
- multiply or divede them.
You can use a calculator to do many of these calculations. However, it is also important to understand how to work them out by hand. This section explains how to do this.
First, there are some important things to know about fractions:
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If the numerator is smaller than the denominator, the fraction is less than \(1\). This is called a proper fraction. For example, \(\frac{1}{4}\) is a proper fraction.
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If the numerator is larger than the denominator, the fraction is greater than \(1\). This is called an improper fraction. For example, \(\frac{5}{4}\) is an improper fraction. These fractions can also be written as mixed numbers (a whole number and a fraction together) or as whole numbers, depending on the situation. For example, \(\frac{5}{4}\) can also be written as \(1\frac{1}{4}\) (because \(5\) divided by \(4\) equals \(1\) with \(1\) left over). Another example is \(\frac{8}{4}\), which can be written as (2\) (because \(8 \div 4 = 2\) with no remainder). Unless you are told otherwise, you should usually give answers as mixed numbers or whole numbers instead of improper fractions.
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If the denominator of a fraction is \(1\) the fraction is equal to the whole number in the numerator. For example \(\frac{4}{1} = 4\)
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If the numerator and denominator are the same, the fraction is equal to \(1\) and we usually write it as \(1\). For example \(\frac{4}{4} = 1\)
Simplifying fractions
Unless you are told otherwise, fractions should always be written in their simplest form. This means using the smallest possible numbers. You can change a fraction to its simplest form by following these steps:
- Find the highest common factor (HCF) of the numerator and denominator. This is the largest number that divides exactly into both numbers. You may need to list all the factors of each number to find this, as there is no single method that always works.
- Another option is to use any common factor (any number that divides both numbers exactly). If you do this, you will need to repeat the steps until no common factors remain.
- When doing this, it is usually easiest to start with simple factors such as \(2, 3, 5\) or \(10\).
- Divide both the numerator and denominator by the highest common factor or the chosen common factor. This will give you a new numerator and denominator.
- If you used a factor that is not the highest, repeat the process until the fraction cannot be simplified any further.
Example: If \(8\) out of \(12\) participants who responded to a survey are female, what is this as a fraction in simplest form?
Solution: Following the steps above, we have:
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The factors of \(8\) are \(1, 2, 4, 8\) and the factors of \(12\) are \(1, 2, 3, 4, 6, 12\)
Since the largest number on both these lists is \(4\) it means the highest common factor is \(4\) -
Dividing the numerator of \(8\) by this highest common factor gives \(8 \div 4 = 2\) and dividing the denominator of \(12\) by this highest common factor gives \(12 \div 4 = 3\)
So \(\frac{8}{12}\) simplifies to \(\frac{2}{3}\)
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Finding equivalent fractions
The fractions \(\frac{8}{12}\) and \(\frac{2}{3}\) from the previous example are equivalent fractions, which means they have the same value.
Often, you will find equivalent fractions by simplifying, as shown earlier. However, sometimes you may need to find an equivalent fraction with a given denominator (or less often, a given numerator). This is useful when you need to compare, add, or subtract fractions that have different denominators. You can find an equivalent fraction with a chosen denominator by following these steps (you can reverse the steps if you are working with a chosen numerator):
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Divide the new denominator (from the incomplete fraction) by the denominator of the original fraction.
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Multiply the numerator of the original fraction by the value you found in the first step.
If you prefer, you can do the division the other way around in the first step. If you do that, you would then divide in the second step instead of multiplying.
If you are unsure whether to multiply or divide, think about whether the new denominator (or numerator) should be larger or smaller than the original. This will help you decide.
Another method is to:
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Multiply the know value in the incomplete fraction (for example, the denominator) by the “opposite” value in the original fraction (for example, the numerator),
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The divide by the remaining value (for example, the denominator of the orginal fraction).
Example: If a student scores \(\frac{32}{40}\) on a test, but the test only counts for \(15\) marks of her final grade, what is her mark out of \(15\) as an equivalent fraction?
Solution: Following the steps above, we have:
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The denominator of the incomplete fraction is \(15\) and the denominator of the complete fraction is \(40\)
So \(15 \div 40\) gives \(0.375\) -
The numerator of the complete fraction is \(32\)
So \(32 \times 0.375\) gives \(12\) and the equivalent fraction is \(\frac{12}{15}\)
Alternatively, you could multiply \(15\) and \(32\) (the ‘opposite’ values in the two fractions) and divide by the remaining value of \(40\)
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Once you have worked through the examples above, have a go at the fraction problems in the activity below:
Adding and subtracting fractions
When adding or subtracting fractions, there are two main types of problems to consider.
The first type is when the fractions have like denominators (the same denominator). In this case, the process is simple. You add or subtract the numerators, and keep the denominator the same as in the original fractions.
Example: Nick eats \(\frac{1}{8}\) of a pizza while his brothers Jo and Ryan each eat \(\frac{3}{8}\) of the pizza. What fraction has been eaten altogether?
Solution: Adding the three fractions simply requires adding the numerators, as follows:
\[\frac{1}{8} + \frac{3}{8} + \frac{3}{8} = \frac{1 + 3 + 3}{8} = \frac{7}{8}\]So \(\frac{7}{8}\) of the pizza has been eaten altogether.
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Adding or subtracting fractions with unlike denominators (different denominators) requires more steps. You must convert the fractions into equivalent fractions with the same denominator before you can add or subtract them. You can do this by following these steps:
- Find the lowest common denominator of all the fractions This is the smallest number that each denominator can divide into exactly.
- If you cannot find the lowest common denominator, you can use the product of the denominators. This will always work, but it may not be the simplest option.
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Convert each fraction into an equivalent fraction with the chosen denominator.
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Add or subtract the fractions as usual once they all have the same denominator.
- Simplify your final answer if needed.
Example: David has a big pile of assignments to mark, which he spreads over a number of days. He marks \(\frac{1}{4}\) of the assignments on Monday, \(\frac{1}{3}\) of the assignments on Tuesday and \(\frac{2}{5}\) of the assignments on Wednesday. What fraction of the assignments has he marked over the three days?
Solution: Following the steps above, we have:
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The three denominators in the problem are \(4, 3\) and \(5\) so the lowest common denominator is \(60\) (which in this case is the same as the product of \(4, 3\) and \(5\) since \(4 \times 3 \times 5 = 60\))
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The fractions can be converted to equivalent fractions with denominators of \(60\) as follows:
\(\frac{1}{4} = \frac{15}{60}\) since \(60 \div 4 = 15\) and \(1 \times 15 = 15\)
\(\frac{1}{3} = \frac{20}{60}\) since \(60 \div 3 = 20\) and \(1 \times 20 = 20\)
\(\frac{2}{5} = \frac{24}{60}\) since \(60 \div 5 = 12\) and \(2 \times 12 = 24\) -
The sum now becomes \(\frac{15}{60} + \frac{20}{60} + \frac{24}{60} = \frac{15 + 20 + 24}{60} = \frac{59}{60}\)
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This fraction cannot be simplified, so the fraction of assignments marked is \(\frac{59}{60}\)
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Multiplying and dividing fractions
Sometimes you may need to find a fraction of a number, such as \(\frac{2}{3}\) of another fraction, or \(\frac{2}{3}\) of an integer. In both cases, you do this by multiplying the fraction by the number.
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If the number is a fraction, multiply the two fractions by multiplying the numerators together and the denominators together to form a new fraction. Simplify the result if needed.
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If the number is an integer, multiply the integer by the numerator of the fraction, and then divide by the denominator. You can do this in the order that is easiest for you to calculate.
Example: Suppose you are considering taking a job that is part-time, and you would be working \(\frac{4}{5}\) of a normal load. If the full time salary is \(\$60000\) what would your part-time salary be?
Solution: This requires you to calculate \(\frac{4}{5}\) of \(\$60000\) which is the same as multiplying \(\frac{4}{5}\) and \(\$60000\)
You can do this in a number of ways (note it doesn’t matter which way around you write the numbers being multiplied, so do it in a way that makes sense to you). For example:
Method 1: \(\$60000 \times \frac{4}{5} = \$60000 \times 4 \div 5 = \$240000 \div 5 = \$48000\)
Method 2: \(\$60000 \times \frac{4}{5} = \$60000 \div 5 \times 4 = \$12000 \times 4 = \$48000\)
Method 3: \(\frac{4}{5} \times \$60000 = 4 \div 5 \times \$60000 = 0.8 \times \$60000 = \$48000\)
So your part-time salary would be \(\$48000\)
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Furthermore, if you can multiply fractions, you are already halfway there when it comes to dividing fractions. Dividing by a fraction is the same as multiplying by its reciprocal. This means you swap the numerator and denominator of the fraction, and then multiply.
This works because multiplication and division are opposite operations.
Example: Suppose you have a piece of string that is \(30\)m long, and you want to cut it into lengths that are \(\frac{3}{4}\)m each. How many pieces will you end up with?
Solution: This requires you to divide \(30\) by \(\frac{3}{4}\) which is the same as multiplying \(30\) by \(\frac{4}{3}\)
You can do this in any of the ways detailed above, for example:
\(30 \times \frac{4}{3} = 30 \times 4 \div 3 = 120 \div 3 = 40\)
So you would end up with \(40\) pieces of string.
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Once you have worked through the examples above, have a go at the fraction problems in the activity below:
Ratios, rates and proportions
A ratio is another way of comparing two or more numbers, similar to fractions and percentages. For example, if a class has \(17\) male and \(13\) female students, the ratio of males to females can be expressed as \(17:13\) or as \(17\) to \(13\) or as \(\frac{17}{13}\). The most important thing is to make sure the order of the items matches the order of the numbers.
Ratios compare values that use the same units, so the units are not included. We are only comparing the numbers. A rate, however, is a special type of ratio that compares two quantities with different units. When describing a rate, we use the word “per”. For example, the price of a food item might be \(\$6\) per \(2\) kg.
Like fractions, ratios are usually written in their simplest form, or lowest terms. For example, if there are \(10\) male and \(20\) female students then the ratio usually be written as \(1:2\) not \(10:20\)
Rates are often written as unit rates, where the second number is one. For example, the cost of the same food item as a unit rate is \(\$3\) per kg. You can simplify ratios and rates in the same way as fractions.
When two ratios or rates are equal, we say they are in proportion. For example, the ratios \(1:2\) and \(10:20\) are in proportion as they show the same relationship. The rates \(\$6\) per \(2\) kg and \(\$3\) per kg are also in proportion.
Finding missing values in ratios or rates so they stay in proportion is a common task. You can do this in a similar way to finding missing values in equivalent fractions. The key idea is that all parts of the ratio or rate must be multiplied or divided by the same number to keep the relationship the same.
Example: Consider that an excursion is planned for a year two class of \(30\) students. Given that there is a requirement for a ratio of one adult to every \(6\) students, how many adults are required on the excursion?
Solution: To calculate this, first note that there is one complete ratio (\(1:6\)) and one incomplete ratio (\(?:30\))
To find the missing value so that the ratios are in proportion, one way is to write the ratios on separate lines:
\(1:6\)
\(?:30\)
This way it is easy to see that you need to divide the two numbers that are above one another to find out the relationship between the ratios, and then apply this relationship to the other side of the ratios.
So in this case \(30 \div 6 = 5\) meaning the second ratio is five times the first, and therefore that \(1 \times 5 = 5\) adults are required:
Alternatively, you could multiply \(1\) and \(30\) (the ‘opposite’ values in the two ratios) and divide by the remaining value of \(6\) which also gives \(5\)
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Once you have worked through the example above, have a go at the ratio, rate and proportion problems in the activity below:
Averages
Sometimes you may need to calculate a simple average using a total and a number of people or items. To do this, divide the total by the number of people or items.
Example: Suppose that in a school reading challenge a class of \(22\) students reads \(99\) books. What was the average number of books read by each individual student?
Solution: To calculate the average books per student when \(99\) books are read by \(22\) students you would do:
\[99 ÷ 22 = 4.5\]So the average number of books read per student is \(4.5\) books.
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Once you have worked through the example above, have a go at the average problems in the activity below:
Algebra
Having a basic understanding of algebra can help you solve problems that involve unknown values. This section explains how to write simple equations and how to substitute values to solve them. If you want to learn more about algebra, including how to rearrange equations, refer to the Algebra module.
Writing a simple algebraic equation involves using letters or symbols to represent unknown values. These are called variables. You then combine these variables with numbers, an equals sign, and mathematical operations to show the relationship between the values in the equation.
Example: Suppose you are a teacher planning a school incursion. You investigate prices and find an incursion you like which costs \(\$5\) per student, plus an additional one-off booking fee of \(\$50\) for the event. What is an equation that represents the cost of the incursion for different numbers of students?
Solution: In this case our unknown values are the number of students and the total cost. We could use the letter \(s\) (for example) to represent the number of students, and the letter \(t\) (for example) to represent the total cost. Two examples of equations to represent the total cost (\(t\)) of the incursion for \(s\) students are then as follows:
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\(t = 5 \times s + 50\) \((\)or you can omit the \(\times\) sign and just write \(t = 5s + 50)\)
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\(t = 50 + 5 \times s\) \((\)or you can omit the \(\times\) sign and just write \(t = 50 + 5s)\)
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In general, the purpose of writing algebraic equations is to solve them and find unknown values. You can solve equations with more than one variable, but this is more complex and requires the same number of equations as unknowns. This is not covered here.
Instead, we will focus on solving one equation with one unknown value. In particular, we will look at how to do this by substituting values for the other variables, so that only one variable remains. The goal is to have this variable by itself on one side of the equals sign.
Example: Consider the equation from the previous example. If there are \(20\) students in the class, how much would the incursion cost?
Solution: You can solve to find the total cost by substituting (i.e. replacing) the \(s\) with \(20\) which gives:
\[t = 5 \times 20 + 50 = 100 + 50 = 150\]So the equation becomes \(t = 150\) and this tells us that the total cost of the incursion for \(20\) students would be \(\$150\)
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Alternatively, you may want to find how many students can attend an incursion with a set budget. To do this, you would need to rearrange the equation so that the \(s\) is by itself on one side of the equals sign. For more information on rearranging equations, refer to the Algebra module.
Once you have worked through the examples above, complete the algebra problems in the following activity:
Are you a Curtin student who would like to work through a module and gain a certificate for learning more about key numerical skills? Access the Maths 4 University module.
</div> Measurement & geometry
This page explains important ideas about measurement and geometry. It also includes activities to help you practise and build your understanding as you go. If you would like more practice questions, especially if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you should visit the LANTITE page. If you want to develop more advanced skills in geometry, you can visit the Geometry and trigonometry module.
As you work through the content, remember that there are often different ways to solve the same problem. In some cases, alternative methods are shown, but many examples only include one method. If you already know a different way to solve a problem, that is fine, as long as your method is correct. You do not need to change your approach. The most important thing is to use a method that makes sense to you.
In summary, this page explains how to:
- Convert units of measurement
- Draw diagrams to help solve measurement problems
- Solve problems involving time
- Use distance and time to calculate speed
- Understand and follow directions
- Calculate the perimeter, area and volume of simple shapes
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Converting between units
The metric system is an international system of measurement that is based on the decimal (base 10) number system. This means measurements are organised in powers of ten, making them easier to convert.
The system uses two types of units: base units and derived units. The base units for three common types of measurement are as follows:
| Measurement | Unit | Abbreviation |
|---|---|---|
| Length | metre | m |
| Volume | litre | L |
| Weight | gram | g |
From these base units of measurement, other units are created to show smaller or larger amounts of the base unit. This is done by adding a prefix to the base unit to show how it has changed. The most commonly used prefixes are:
| Metric prefix | Relation to base unit | Abbreviation | Derived units |
|---|---|---|---|
| kilo | 1000 times bigger | k | kilometre (km), kilolitre (kL), kilogram (kg) |
| centi | 100 times smaller | c | centimetre (cm) |
| milli | 1000 times smaller | m | millimetre (mm), millilitre (mL), milligram (mg) |
When solving a problem that involves measurements, you need to make sure that all measurements of the same type are written in the same unit. If they are not, you will need to convert them into the same unit.
You can convert from one unit to another by using the relationships shown in the table above.
Example: Suppose you wish to tile the floor of a room that is \(3\textrm{m}\) wide with small tiles of width \(50\textrm{mm}\). How many tiles will you need to cover the width of the room?
Solution: To solve this you first need both measurements to be in the same unit. If you wish to have them both in metres, for example, you could convert the measurement that is in millimetres to metres by dividing by \(1000\):
\(50\textrm{mm} \div 1000 = 0.05\textrm{m}\)
Alternatively, if you wish to have them both in millimetres you could convert the measurement that is in metres by multiplying it by \(1000\):
\(3\textrm{m} \times 1000 = 3000\textrm{mm}\)
Once the measurements are both in the same unit, you can simply divide one by the other:
Method 1: \(3\textrm{m} \div 0.05\textrm{m} = 60\)
Method 2: \(3000\textrm{mm} \div 50\textrm{mm} = 60\)
Either way, \(60\) tiles are required.
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Some other common equivalent measurements that you may need to convert between are:
- \(1\textrm{cm} = 10\textrm{mm}\)
- \(1\textrm{ metric tonne} = 1000\textrm{kg}\)
- \(1\textrm{ hectare} = 10000\textrm{m}^2\)
- \(1\textrm{ hectare} \approx 2.471\textrm{ acres}\) (an imperial measurement)
f you are unsure whether to divide or multiply when converting units, think about the size of each unit and whether your answer makes sense. For example, millimetres are much smaller than metres, so you need many millimetres to make a small number of metres.
You can also use your understanding of rates and proportions to check that your conversion is correct. If needed, refer to the Ratios, rates and proportions section.
Example: Suppose you are considering purchasing a property that is listed as being \(5\) acres. How big is the property in hectares?
Solution: If you wish to convert this to the metric unit of hectares, you have the following complete and incomplete rates:
\[\begin{aligned} 1 \textrm{ hectare} &= 2.471 \textrm{ acres} \\\ ? \textrm{ hectares} &= 5 \textrm{ acres} \end{aligned}\]Therefore, since \(5 \div 2.471 \approx 2.023\), it follows that \(5\textrm{ acres}\) is equivalent to \(2.023\textrm{ hectares}\).
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Once you have worked through the examples above, have a go at the conversion problems in the activity below:
Drawing diagrams
Never underestimate how helpful a simple diagram can be when solving a measurement problem. Diagrams can support your thinking and make it easier to understand and solve the problem.
Example: Suppose you have a piece of cardboard measuring \(1.4\textrm{m}\) by \(1.2\textrm{m}\) and you want to cut it into squares of size \(25\textrm{cm} \times 25\textrm{cm}\) for a craft activity. You need \(23\) squares - will you be able to cut enough, or will you require some extra cardboard?
Solution: The first thing to do here is to convert the measurements to have them all in the same unit. Having all lengths in metres only requires one conversion, so doing this gives:
\(25\textrm{cm} \div 100 = 0.25\textrm{m}\)
Next, you might be tempted to find the area of the card and divide it by the area of the pieces you want to cut - but this doesn’t take into account the fact that you might have some unused strips of paper at the sides. A diagram illustrates this, and shows that instead you need to treat the length and width separately:
Doing this gives:
\(1.4 \div 0.25 = 5.6\), meaning \(5\) pieces can be cut from the length
\(1.2 \div 0.25 = 4.8\), meaning \(4\) pieces can be cut from the width
(Note you need to truncate - cut off - rather than round in both instances.)
Therefore \(5 \times 4 = 20\) pieces can be cut altogether, so you will need some extra cardboard.
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Once you have worked through the example above, have a go at using diagrams to solve the problems in the activity below:
Time
Another idea you may need to use when working with measurements is time. Some common calculations with time include finding how long an event lasts and working out the difference between two times. There are a few different methods you can use to do this.
Example: Suppose you are timing a race, and the times for the first three competitors in minutes and seconds are \(23\):\(56\), \(24\):\(01\) and \(25\):\(12\). How much slower was the third placed competitor compared to the first?
Solution: There are a few different methods to find out the difference in minutes and seconds between \(23\):\(56\) (first) and \(25\):\(12\) (third).
For example, one approach is to convert both times to seconds by multiplying the minutes by 60 and adding the existing seconds, as follows:
\(23 \times 60 + 56 = 1436\textrm{ seconds}\), and \(25 \times 60 + 12 = 1512\textrm{ seconds}\)
You can then calculate the difference in seconds, which is:
\(1512 – 1436 = 76 \textrm{ seconds}\)
Finally, you can convert this difference back to minutes and seconds by dividing through by \(60\). When you do this the whole number part gives the number of minutes, while the decimal part can be multiplied by \(60\) again to give the number of seconds, as follows (note that the dot over the \(6\) indicates that the decimal is recurring - meaning it repeats forever):
\(76 \div 60 = 1.2\dot{6}\), and \(0.2\dot{6} \times 60 = 16\)
So the answer is \(1 \textrm{ minute}\) and \(16 \textrm{ seconds}\).
Another approach is to work it out in stages. For example, first of all you could calculate how many seconds are between \(23\):\(56\) and \(24\):\(00\), by subtracting \(56\) from \(60\) (i.e. \(60 – 56 = 4 \textrm{ seconds}\)). Then you could work out how many minutes are between \(24\):\(00\) and \(25\):\(00\) (i.e. \(25 – 24 = 1 \textrm{ minute}\)), and finally you could add on the extra \(12 \textrm{ seconds}\) from \(25\):\(12\) to give \(4 + 12 = 16 \textrm{ seconds}\).
So again the answer is \(1 \textrm{ minute}\) and \(16 \textrm{ seconds}\).
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Once you have worked through the example above, have a go at the time problems in the following activity:
Speed
OOnce you understand how distance and time work, you can use them to calculate how fast something is travelling. This is called speed. You can calculate speed using the following formula:
\[\textrm{speed} = \frac{\textrm{distance}}{\textrm{time}}\]Example: Suppose you have just started a new job that is \(18\textrm{km}\) from your house, and you plan to ride your bike there. If you need to be at work at \(8\textrm{am}\) and you plan to leave your house at \(7.15\textrm{am}\), what will be the average speed you need to ride in kilometres per hour?
Solution: To calculate this, first note that you need to work out how much time you have in hours. Since you leave at \(7.15\textrm{am}\) and need to arrive by \(8\textrm{am}\) this gives a total time of \(45 \textrm{ minutes}\), and \(45 \div 60 = 0.75 \textrm{ hours}\).
Therefore your average speed will need to be: \(\frac{18}{0.75} = 24 \textrm{ km/hr}\)
Alternatively, \(\frac{18}{45} = 0.4 \textrm{ km/min}\) and \(0.4 \times 60 = 24 \textrm{ km/hr}\)
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Once you have worked through the example above, have a go at the speed problem in the following activity:
Directions
Being able to read maps and understand directions is an important skill. There are two main types of directions you may need to use: the four cardinal directions (north, east, south, and west), and left and right directions.
When reading cardinal directions on a map, there is often an arrow that shows which way is north. If there is no arrow, you can usually assume that north is at the top of the map. Once you know where north is, you can work out the other directions. There are simple memory aids to help you remember the order of directions. For example, the phrase “Never Eat Soggy Weetbix” shows the clockwise order: North, East, South, West.
There are also directions between the main cardinal directions. These are called intermediate directions and sit halfway between them, as shown below.
For left and right directions, it can help to imagine yourself facing the same direction as you are ‘travelling’ on the map. You can also turn the map or your body to match this direction. This will make it easier to work out which way is left and right.
If you are unsure which side is left or right, you can hold both hands in front of you and make an ‘L’ shape with your thumb and fingers. The hand that forms a correct ‘L’ shape is your left hand.
Example: Suppose you are travelling south down Conway St and turn left on Allen St on the map shown. Will you make it to the shopping centre?
Solution: To figure this out, first note that north is at the top of the map, which means that south is at the bottom and that if you are travelling south down Conway St you are travelling ‘down’ the map. So try to imagine that when you get to Allen St you are coming at it from the ‘top’ side, facing downwards, and therefore that if you go left, you would go off the map and not towards the shopping centre.
Alternatively, you might like to try and remember that if you are travelling ‘down’ the map, the directions will be the opposite of what they would be if you were travelling ‘up’. So while it looks like the shopping centre is to the left of Conway St, actually you would need to turn right.
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Once you have worked through the example above, have a go at the direction problems in the following activity:
Perimeter
The perimeter of a shape is the total distance around its outer edge. For example, the perimeter of a rectangle is found by adding together the lengths of all four sides. This can also be calculated by multiplying the length by two and the width by two, then adding these values together. As a formula this looks like:
\(\textrm{Perimeter of rectangle} = 2 \times \textrm{length} + 2 \times \textrm{width}\)
You may also need to find the perimeter of irregular shapes. In some cases, you will need to add up the lengths of all the individual sides. In other cases, there may be a quicker method based on the perimeter of one or more shapes that the irregular shape is made from or based on.
Example: What is the perimeter of the following shape?
Solution: One way of calculating the perimeter of this shape is to start by calculating the total perimeter of both the squares, which is \(12\textrm{cm} \times 4 + 12\textrm{cm} \times 4 = 96\textrm{cm}\)
Next, since both squares have been pushed together, there is a small part of the perimeter of each square that needs to be subtracted from this total. From the \(9\textrm{cm}\) distance on the diagram, we can work out that the distance that needs to be subtracted is two lots of \(3\textrm{cm}\) (since \(12\textrm{cm} - 9\textrm{cm} = 3\textrm{cm}\))
Therefore the perimeter of the irregular shape is \(96\textrm{cm} - 3\textrm{cm} - 3\textrm{cm} = 90\textrm{cm}\)
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Once you have worked through the example above, have a go at the perimeter problems in the activity below:
Area
The area of a shape is the amount of space inside its boundary. The most common area you will need to calculate is the area of a rectangle. This is found by multiplying the length by the width:
\(\textrm{Area of rectangle} = \textrm{length} \times \textrm{width}\)
To calculate the area of a triangle, you multiply the base by the height and then divide the result by two. Dividing the result by two is the same as multiplying it by a half:
\(\textrm{Area of triangle} = \frac{1}{2} \times \textrm{base} \times \textrm{height}\)
Example: What is the area of the following triangle?
Solution: The area of this triangle is \(\frac{1}{2} \times 7\textrm{cm} \times 3\textrm{cm} = 10.5\textrm{cm}^2\)
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Once you have worked through the example above, have a go at the area problems in the activity below:
Volume
The volume of a shape is the amount of three-dimensional (3D) space inside it. Volume can be measured in cubic units, such as cubic metres, cubic centimetres, or cubic millimetres. It can also be measured in litres, such as millilitres or kilolitres.
There is a relationship between these two types of measurements, which is shown below:
\(1 \textrm{litre} = 1000\textrm{cm}^3\)
Example: Suppose you have a rectangular fish tank that measures \(50\textrm{cm}\) long by \(30\textrm{cm}\) wide and \(40\textrm{cm}\) high. If you want to fill it three quarters of the way up, what would be the volume of water required in litres?
Solution: To do this you would first calculate the volume by multiplying the three measurements:
\(50\textrm{cm} \times 30\textrm{cm} \times 40\textrm{cm} = 60000\textrm{cm}^3\)
Next, three quarters of this volume is:
\(60000\textrm{cm}^3 \times \frac{3}{4} = 45000\textrm{cm}^3\)
Finally, you can convert to litres by dividing through by \(1000\):
\(45000\textrm{cm}^3 \div 1000 = 45 \textrm{ litres}\) required
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Once you have worked through the example above, have a go at the volume problem in the activity below:
H5P interactive activity
Statistics & probability
This page explains important concepts in statistics and probability. It also includes activities to help you practise and build your understanding as you work through the content. If you would like more sample questions, especially if you are preparing for the Literacy and Numeracy Test for Initial Teacher Education (LANTITE), you should visit the LANTITE page. If you want to develop more advanced skills in statistics, you can visit the Introduction to statistics module. If you would like to improve your skills in Microsoft Excel, you can visit the Microsoft Excel essentials module.
As you work through the content, remember that there are often different ways to solve the same problem. In some cases, alternative methods are explained, but many examples only show one method. If you already know a different way to solve a problem, that is fine, as long as your method is correct. You do not need to change your approach. The most important thing is to use a method that makes sense to you.
In summary, this page explains how to:
- Calculate and interpret statistics for categorical data
- Calculate and interpret statistics for continuous data
- Interpret graphs
- Write and interpret Excel formulas
- Calculate probabilities and create sample spaces to solve probability problems
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Statistics for categorical data
Statistics are used to summarise, display, and describe data (observations and measurements). There are many different types of statistics, and the type you use depends on the type of data you have.
If the data is grouped into categories, it is called categorical data. This type of data is usually summarised using frequencies (counts) and/or percentages. It can also be displayed using graphs such as column graphs or pie charts.
Example: Data relating to the favourite sports of students in a school can be summarised using the following frequency distribution table and column graph:
| Favourite Sports | Frequency |
|---|---|
| Netball | 55 |
| Football | 67 |
| Tennis | 32 |
| Basketball | 46 |
| TOTAL | 200 |
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.
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Once you have worked through the example above, have a go at analysing the categorical data in the following activity:
Statistics for continuous data
If data is measured on a continuous numerical scale, it is called continuous data. In fact, most numerical data is treated as continuous, even if it is not perfectly measured on a continuous scale.
This type of data is summarised using measures of central tendency and measures of dispersion. It is often displayed using graphs such as histograms and box plots.
Common measures of central tendency include the mean, median, and mode. Common measures of dispersion include the range, interquartile range, and standard deviation, although the last two are not covered here.
Continuous data can also be grouped into categories. When this happens, it can be displayed and analysed in the same way as categorical data. For example, age data can be grouped into age ranges.
The mean
The mean (arithmetic average) of a set of values is found by adding all the values together and then dividing by the total number of values.
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:
\[\frac{19 + 20 + 29 + 28 + 23 + 18 + 27 + 22 + 24 + 20}{10} = 23\]</div>
Sometimes data is shown in a frequency table. In this case, you need to use the frequencies when calculating the mean, as they show how often each value occurs.
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.):
| Marks (out of 10) | Frequency |
|---|---|
| 6 | 4 |
| 7 | 5 |
| 8 | 7 |
| 9 | 8 |
| 10 | 6 |
| TOTAL | 30 |
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:
\[\frac{6 \times 4 + 7 \times 5 + 8 \times 7 + 9 \times 8 + 10 \times 6}{30} = \frac{247}{30} = 8.23\]</div>
Once you have worked through the examples above, have a go at calculating means in the following activity:
The median
The median is the middle value in an ordered set of data. If there is an even number of values, the median is found by calculating the mean of the two middle values. To find the median, you must first arrange the data in ascending order (from smallest to largest).
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 \((22 + 23) \div 2 = 22.5\).
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Sometimes data is shown in a frequency table. In this case, you need to calculate cumulative frequencies to help you find the median.
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.):
| Marks (out of 10) | Frequency |
|---|---|
| 6 | 4 |
| 7 | 5 |
| 8 | 7 |
| 9 | 8 |
| 10 | 6 |
| TOTAL | 30 |
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:
| Marks (out of 10) | Frequency | Cumulative Frequency | |
|---|---|---|---|
| 6 | 4 | 4 | |
| 7 | 5 | 9 | |
| 8 | 7 | 16 | |
| 9 | 8 | 24 | |
| 10 | 6 | 30 | |
| TOTAL | 30 |
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.
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Once you have worked through the examples above, have a go at calculating medians in the following activity:
The mode
The mode in a set of scores is the value that appears most often. This value may not be near the middle of the data set, but it is useful when you want to find the most common value. If two or more values appear the same highest number of times, the data set has more than one mode. These values should not be averaged.
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.
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If the data is shown in a frequency table, it is even easier to identify the mode. The mode is simply the value with the highest frequency (the largest count).
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.):
| Marks (out of 10) | Frequency |
|---|---|
| 6 | 4 |
| 7 | 5 |
| 8 | 7 |
| 9 | 8 |
| 10 | 6 |
| TOTAL | 30 |
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.
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Once you have worked through the examples above, have a go at the following activity:
The range
The range is the difference between the highest and lowest values in a data set.
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 \(29 - 18\)).
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Once you have worked through the example above, have a go at the following activity:
Interpreting graphs
Being able to read and compare different types of graphs is an important skill. These graphs can include pie charts, column graphs, histograms, box plots, and line graphs. When working with graphs, make sure you read the question carefully and focus only on the part of the graph that is relevant to the question.
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?
- In \(2019\) more than three-quarters of students in class \(\textrm{A}\) had a mark over \(50\%\).
- The distribution of marks was more similar for the classes in \(2019\) than in \(2020\).
- 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:
- 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 \(75\%\)).
- True. The column pattern for Class \(A\) and Class \(B\) is the same for \(2019\), but not for \(2020\).
- False. This is true of Class \(A\), but the percentage of students for Class \(B\) actually decreased by \(4\%\).
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Example 2: The following graph shows NAPLAN student achievement for years \(3\), \(5\), \(7\) and \(9\):
- For how many year groups is Band \(6\) above the national minimum standard?
- A student scores in Band \(5\), and is at the national minimum standard. What year must they be in?
- How many bands are reported in common for years \(3\) and \(9\)?
Solution: Working and answers are as follows:
- 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 \(7\)).
- 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 \(7\).
- 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 \(6\)).
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Example 3: The following graph shows NAPLAN \(2018\) Reading results for year \(3\) students in eight states and territories and in Australia overall:
- For which state or territory were the scores of more than \(20\%\) of students below Band \(3\)?
- 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:
- 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.
- 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 \(2\).
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Once you have worked through the examples above, have a go at the following activity:
Calculations in Excel
Using a spreadsheet is often the easiest way to complete calculations. You can also use it to create graphs and organise your data clearly.
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.
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Once you have read through the example above, have a go at the following activity:
Probability
The probability of an event is the chance that it will happen. It is measured as a value between \(0\) and \(1\). A probability of \(0\) means the event can never happen, and a probability of \(1\) means the event is certain to happen.
One of the simplest examples of probability is tossing a coin. When you toss a single coin, there are two possible outcomes: heads or tails. These outcomes are equally likely. This means the probability of getting a head is \(\frac{1}{2}\) and the probability of geting a tail is also \(\frac{1}{2}\).
In general, to calculate the probability of an event, you divide the number of outcomes where the event occurs by the total number of possible outcomes.
\[\frac{\textrm{Number of outcomes in which event occurs}}{\textrm{Total number of possible outcomes}}\]Finally, simplify the fraction if needed, or write the probability as a decimal if that is what the question asks for.
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 \(\frac{1}{2}\).
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If a probability calculation is more complex, you may need to create a sample space. A sample space shows all possible outcomes of an event. It can be written as a list, shown in a table, or drawn as a tree diagram. A tree diagram is especially helpful when the outcomes are not equally likely.
Creating a sample space makes it easier to see both the total number of possible outcomes and the number of outcomes where the event occurs.
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:
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 1, 1 | 1, 2 | 1, 3 | 1, 4 | 1, 5 | 1, 6 |
| 2 | 2, 1 | 2, 2 | 2, 3 | 2, 4 | 2, 5 | 2, 6 |
| 3 | 3, 1 | 3, 2 | 3, 3 | 3, 4 | 3, 5 | 3, 6 |
| 4 | 4, 1 | 4, 2 | 4, 3 | 4, 4 | 4, 5 | 4, 6 |
| 5 | 5, 1 | 5, 2 | 5, 3 | 5, 4 | 5, 5 | 5, 6 |
| 6 | 6, 1 | 6, 2 | 6, 3 | 6, 4 | 6, 5 | 6, 6 |
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 \(\frac{1}{6}\).
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Once you have worked through the examples above, have a go at the probability problems in the following activity:
H5P interactive activity
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