Making Predictions
Learn to make predictions from bivariate data for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-06). Once a line of best fit is drawn, you can use it or its equation to estimate a value that was never measured.
You will substitute into the line’s equation, work backwards from a given value, and see the limits of interpolation (inside the data, more reliable) and extrapolation (outside the data, less reliable) — so you can judge whether a prediction, such as a household’s next electricity bill, really makes sense.
Theory
To make a prediction, substitute a value into the equation of the line of best fit. This Year 12 Standard 1 (NSW) guide shows how to predict \(y\) from \(x\), work backwards from \(y\), and decide whether a prediction is interpolation (inside the data, more reliable) or extrapolation (outside the data, less reliable).
A prediction uses the equation of the line of best fit. Given \(x\), substitute and work it out, e.g. \(y=4x+10\) at \(x=5\) gives \(y=4(5)+10=30\). Given \(y\), substitute, take away the constant, then divide by the gradient.
Interpolation is a prediction inside the data range (between the smallest and largest data values) — it is more reliable.
Extrapolation is a prediction outside the data range — it is less reliable, because the linear pattern may not continue. Give every answer in context with units.
Predict \(y\) by substituting \(x\) into the line of best fit:
Work backwards from a given \(y\): subtract \(c\), then divide by \(m\):
How to make and judge a prediction
- Substitute the known value into the equation of the line.
- Solve: given \(x\), evaluate; given \(y\), subtract the constant then divide by the gradient.
- Compare the value with the data range: inside is interpolation, outside is extrapolation.
- State the answer in context with units, and comment if it is not sensible.
Substitute \(u=450\), then check the data range.
| \(B\) | \(=\) | \(0.3u+20\) |
| \(B\) | \(=\) | \(0.3(450)+20\) |
| \(B\) | \(=\) | \(135+20\) |
| \(B\) | \(=\) | \(155\) |
\(100\le 450\le 800\) so this is interpolation (reliable). The predicted bill is \(\$155.00\).
Substitute \(C=131\), subtract the constant, then divide by the gradient.
| \(131\) | \(=\) | \(12d+35\) |
| \(131-35\) | \(=\) | \(12d\) |
| \(96\) | \(=\) | \(12d\) |
| \(d\) | \(=\) | \(\dfrac{96}{12} = 8\) |
Check: \(12(8)+35=131\) \(\checkmark\). The delivery was about 8 km.
Substitute each value, then compare it with the data range.
| \(\text{(i) } B\) | \(=\) | \(100-7.5(6) = 100-45 = 55\) |
| \(1\le 6\le 10\) | \(\Rightarrow\) | \(\text{interpolation}\) |
| \(\text{(ii) } B\) | \(=\) | \(100-7.5(16) = 100-120 = -20\) |
| \(16>10\) | \(\Rightarrow\) | \(\text{extrapolation}\) |
(i) 55%. (ii) −20%: a battery cannot be below 0%, so the model breaks down outside the data.
Solve for \(h\), then change the decimal part to minutes.
| \(615\) | \(=\) | \(75h+120\) |
| \(495\) | \(=\) | \(75h\) |
| \(h\) | \(=\) | \(\dfrac{495}{75} = 6.6\) |
| \(0.6 \text{ h}\) | \(=\) | \(0.6\times 60 = 36 \text{ min}\) |
\(1\le 6.6\le 8\) so this is interpolation. About 6 h 36 min.
Common pitfalls
Frequently asked questions
How do you make a prediction from a line of best fit?
Substitute the known value into the equation of the line and work out the other value. For example, y = 4x + 10 at x = 5 gives y = 4 times 5 plus 10 = 30.
What is the difference between interpolation and extrapolation?
Interpolation predicts a value inside the range of the collected data, so it is more reliable. Extrapolation predicts outside the data range, so it is less reliable because the pattern may not continue.
Why is extrapolation less reliable?
Outside the data range nobody has checked that the straight-line pattern continues. Extrapolating can give silly answers, such as a negative battery level or a negative price.
How do you find x when you are given y?
Substitute the y-value into the equation, subtract the constant first, then divide by the gradient. For C = 12d + 35 and C = 131: 131 - 35 = 96, and 96 divided by 12 = 8.
How do you change a decimal number of hours into hours and minutes?
Keep the whole hours and multiply the decimal part by 60. For example, 6.6 hours is 6 hours plus 0.6 times 60 = 36 minutes, so 6 hours 36 minutes.