Measures of centre and spread

Mean, median and mode, choosing the best measure of centre, modality, range and population standard deviation, and comparing two data sets.

Two classes both average $70\%$ on a test. Are they the same? Not necessarily. In one class everybody might score close to $70$. In the other, half the class might be in the $40$s and half in the $90$s. A measure of centre on its own hides that. You also need a measure of spread.

Measures of centre

Mean $\bar{x}$: add the values and divide by how many there are. It uses every value, so it's pulled towards outliers and long tails.
Median: the middle value when the data is in order. For $n$ values, it's at position $\dfrac{n + 1}{2}$. Outliers barely affect it.
Mode: the most common value. It's the only measure of centre for categorical data.

Which to use? If the data is roughly symmetric with no outliers, the mean is best. If it's skewed or has outliers, the median gives a fairer "typical" value. That's why house prices and incomes are almost always reported as medians.

The modality of a distribution describes its peaks. Uniform means roughly flat, with no peak. Unimodal has one peak, bimodal has two and multimodal has more. Two peaks often mean two different groups have been mixed together, like the heights of a combined Year 7 and Year 12 class.

Range and standard deviation

The range $=$ largest $-$ smallest. It's quick, but it depends on only two values, so one outlier changes it completely.

The standard deviation measures how far, on average, the values sit from the mean. Each value's distance from the mean is its deviation. The standard deviation squares the deviations (so negatives don't cancel positives), averages them, then takes the square root to get back to the original units.

Population standard deviation

$$\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$$

In Standard Maths you find $\sigma$ with your calculator's statistics mode. You only need the formula to understand what $\sigma$ means: small $\sigma$ means the data is bunched near the mean, and large $\sigma$ means it's spread out.

Example 1●

Find the standard deviation of $4$, $7$, $8$, $9$, $12$, first by hand to see how it works, then with a calculator.

$\bar{x} = \dfrac{4 + 7 + 8 + 9 + 12}{5} = \dfrac{40}{5} = 8$

Find the mean.

Deviations: $-4, -1, 0, 1, 4$. Squares: $16, 1, 0, 1, 16$

Deviations from the mean, then square them.

$\sigma = \sqrt{\dfrac{34}{5}} = \sqrt{6.8} \approx 2.61$

Average the squares, then take the square root.

Calculator: $\sigma_x \approx 2.6077$, which rounds to $2.61$ ✓

On a calculator: STAT mode, enter the data, then read σx (sometimes written σn). Don't use sx, which divides by n − 1.

InteractiveDrag the data, watch the deviations

Try this

Tick off all four challenges in the checklist. For "mean 10, median 8", what must the data look like?

Press Add 5 to every value. Which of the mean, median, range and $\sigma$ change? Explain why.

Comparing data sets

To compare two groups, comment on centre (which is higher, by how much) and spread (which is more consistent), and always use the context.

Example 2●●

Two classes sat the same test. Class A: $62, 65, 68, 70, 71, 73, 75, 76$. Class B: $45, 52, 60, 70, 78, 82, 85, 88$. Compare the classes.

Class A: $\bar{x} = 70$, $\sigma \approx 4.53$, range $= 14$

Calculator, statistics mode, for each class.

Class B: $\bar{x} = 70$, $\sigma \approx 15.02$, range $= 43$

Same for Class B.

Both classes have a mean of $70$. Class B's standard deviation is more than three times Class A's, so Class B's results are far more spread out. Class A performed consistently, while Class B had some very high and some very low results.

Same centre, very different spread. Write it in context.

Your turn

Without a calculator, find the mean and $\sigma$ of $10$, $12$, $14$, $16$, $18$. Then predict $\sigma$ for $20$, $22$, $24$, $26$, $28$.

Show the working

Mean $= 70 \div 5 = 14$. Deviations $-4, -2, 0, 2, 4$, squares $16, 4, 0, 4, 16$, total $40$. $\sigma = \sqrt{40 \div 5} = \sqrt8 \approx 2.83$. The second set is the first plus $10$: the mean moves to $24$, but the spread is unchanged, so $\sigma \approx 2.83$ again.

Flashcards

Mean or median for skewed data?

Median: it isn't dragged by the tail or outliers

Calculator standard deviation for NESA

Population: $\sigma_x$ (or $\sigma_n$), not $s_x$

Adding a constant to every value

Centre moves by that constant; spread ($\sigma$, range) doesn't change

Practice

Find the population standard deviation of $2, 4, 6, 8$, to $2$ decimal places.
What is the standard deviation of $7, 7, 7, 7$?
Find the mean of $12, 15, 15, 16, 18, 19, 21, 24$.
Find the median of the same data.
Use your calculator to find $\sigma$ for the same data, to $2$ decimal places.
Every value in a data set is increased by $5$. The standard deviation:
The best measure of centre for Sydney house prices is the:
Exam style: Two cafés have mean waiting times of $6.2$ and $5.8$ minutes, with standard deviations of $1.1$ and $3.4$ minutes. Which statement is best?
Exam style: A histogram of the heights of everyone at a primary school assembly (students and teachers) has two clear peaks. It is best described as: