Comparing Means and Standard Deviations
Comparing means and standard deviations shows both the typical value and the consistency of data sets. A score can also be described by how many standard deviations it lies above or below the mean, which lets marks from different tests be compared fairly. This is NSW Year 10 Advanced maths (NESA outcome MA5‑DAT‑C‑01).
You will find a mean or standard deviation from a score's position, such as a mark 1.5 standard deviations below the mean, and decide which of two test results is better. You will compare cricket bowlers, netball shooters, train times, temperatures in Perth and Adelaide, and egg suppliers for a bakery. Harder questions find the mean and standard deviation from two students' positions, check a coffee machine is in control, and transform a data set. It is part of Year 10 Advanced Single Variable and Bivariant Statistics.
Every question with a fully worked solution.
- Comparing Means and Standard Deviations - Video - Interpret the Mean and Standard Deviation of Two Data Sets Watch
Theory
The mean tells you which group is higher on average; the standard deviation tells you which is more consistent. Measuring a score in standard deviations from the mean lets you compare results from different tests. This Year 10 Advanced maths page covers both comparisons, intervals such as \(\bar{x} \pm 2\sigma\), and problems where the mean or standard deviation is unknown.
To compare two data sets, compare the means (centre: which is higher on average) and the standard deviations (spread: the smaller one is more consistent, predictable or reliable). The best choice depends on the question.
A score that is \(2\) standard deviations above the mean is \(\bar{x} + 2\sigma\). The number of standard deviations a score is from the mean is \(\dfrac{x - \bar{x}}{\sigma}\): positive above the mean, negative below.
This lets you compare scores from different tests: the score more standard deviations above its mean is the better result relative to its group.
If \(\sigma = 0\), every score equals the mean (all the scores are the same).
Comparing two data sets
- Calculate the mean and population standard deviation of each set (statistics mode, \(\sigma_x\)).
- Centre: the larger mean is higher on average. Say by how much.
- Spread: the smaller standard deviation is more consistent.
- Decide in context: a higher average or a more reliable result? Intervals like \(\bar{x} \pm 2\sigma\) help with targets and limits.
Scores in standard deviations
- How many SDs? Divide the distance from the mean by \(\sigma\).
- Find a score: add (above) or subtract (below) the right multiple of \(\sigma\).
- Find \(\sigma\) or the mean: if one is unknown, rearrange \(x = \bar{x} + z\sigma\). If both are unknown, two scores give two simultaneous equations.
- Effect of a change: correcting a score towards the mean reduces \(\sigma\); away from it increases \(\sigma\).
| Subject | Noah's mark | Class mean | Standard deviation |
|---|---|---|---|
| History | \(66\) | \(58\) | \(4\) |
| Geography | \(75\) | \(69\) | \(5\) |
(i) In which subject did Noah do better compared with his class?
(ii) In Science the mean was \(70\) and the standard deviation \(6\). Noah was \(1.5\) standard deviations below the mean. What did he score?
(i) Count how many standard deviations each mark is above its mean.
| \(\text{History}\) | \(=\) | \(\dfrac{66 - 58}{4} = \dfrac{8}{4} = 2\) |
| \(\text{Geography}\) | \(=\) | \(\dfrac{75 - 69}{5} = \dfrac{6}{5} = 1.2\) |
\(2 > 1.2\): Noah did better in History, compared with his class, even though his Geography mark is higher.
(ii) Below the mean, so subtract.
| \(\text{mark}\) | \(=\) | \(70 - 1.5 \times 6 = \mathbf{61}\) |
Let the mean be \(m\) and the standard deviation \(s\). Each score gives an equation.
| \(m + 2s\) | \(=\) | \(88 \quad (1)\) |
| \(m - s\) | \(=\) | \(61 \quad (2)\) |
| \((1) - (2):\quad 3s\) | \(=\) | \(27\) |
| \(s\) | \(=\) | \(9\) |
| \(m\) | \(=\) | \(61 + 9 = 70\) |
Mean \(\mathbf{70}\), standard deviation \(\mathbf{9}\). Check: \(70 + 2(9) = 88\ \checkmark\)
Ruby: \(7,\ 8,\ 8,\ 9,\ 9,\ 9,\ 10,\ 10\)
Sam: \(5,\ 7,\ 9,\ 10,\ 10,\ 10,\ 10,\ 10\)
(i) Find each mean and population standard deviation, correct to \(2\) decimal places.
(ii) Compare them.
(iii) The team needs an archer who almost always scores at least \(6\). Using mean \(- 2 \times\) standard deviation, whom should it pick?
(i) Statistics mode:
| \(\text{Ruby}:\ \bar{x}\) | \(=\) | \(\dfrac{70}{8} = \mathbf{8.75},\quad \sigma = 0.968\ldots \approx \mathbf{0.97}\) |
| \(\text{Sam}:\ \bar{x}\) | \(=\) | \(\dfrac{71}{8} = 8.875 \approx \mathbf{8.88},\quad \sigma = 1.763\ldots \approx \mathbf{1.76}\) |
(ii) Sam's mean is slightly higher, so on average he scores a little more. Ruby's standard deviation is much smaller, so she is more consistent.
(iii)
| \(\text{Ruby}\) | \(=\) | \(8.75 - 2(0.968\ldots) = 6.81\ldots\) |
| \(\text{Sam}\) | \(=\) | \(8.875 - 2(1.763\ldots) = 5.35\ldots\) |
Ruby almost always scores above \(6\); Sam often may not. Pick Ruby.
Find \(\bar{x} \pm 2\sigma\) for each machine.
| \(\text{X}:\ 50.0 \pm 2(0.3)\) | \(=\) | \(49.4 \text{ to } 50.6~\text{mm}\) |
| \(\text{Y}:\ 50.1 \pm 2(0.15)\) | \(=\) | \(49.8 \text{ to } 50.4~\text{mm}\) |
X's range of usual lengths goes outside the limits at both ends, so some of its bolts will be rejected. Y's usual lengths all lie inside \(49.5\) to \(50.5\). Use machine Y: its mean is slightly off-centre, but its smaller standard deviation matters more here.
Common pitfalls
Frequently asked questions
What does a smaller standard deviation tell you?
The scores are closer to the mean, so the data are more consistent and predictable.
How do you compare marks from two different tests?
Work out how many standard deviations each mark is above or below its class mean: subtract the mean and divide by the standard deviation. The larger value is the better result compared with the class.
How do you find the mean and standard deviation from two students' scores?
Write each score as the mean plus or minus a number of standard deviations, for example m plus 2s equals 88 and m minus s equals 61. Solve the two equations simultaneously.
What does within two standard deviations of the mean mean?
It is the interval from the mean minus two standard deviations to the mean plus two standard deviations. For roughly symmetrical, hump-shaped data almost all scores lie in this interval.
What does a standard deviation of zero mean?
Every score is the same, and equal to the mean, because no score differs from the mean.