Measures of Centre and Spread Revision Notes for VCE SSCE General Mathematics

Measures of Centre and Spread

Introduction

When describing a data distribution numerically, we use two types of measures:

Measures of centre: tell us about the typical or central value
Measures of spread: tell us how spread out the data values are

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The most versatile statistical tools are:

The median as a measure of centre
The range and interquartile range (IQR) as measures of spread

These measures can be determined exactly from dot plots and stem plots, but only estimated approximately from histograms.

The median

Definition

The median is the middle value in an ordered data set.

For $n$ data values, the median is located at the $\left(\frac{n+1}{2}\right)$ th position.

When:

$n$ is odd, the median will be the middle data value
$n$ is even, the median will be the average of the two middle data values

Finding the median from a data set

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Worked Example: Finding the Median

Given data set: $2, 9, 1, 8, 3, 5, 3, 8, 1$

Step 1: Write the data in order from smallest to largest

Ordered: $1, 1, 2, 3, 3, 5, 8, 8, 9$

Step 2: Locate the middle value(s)

There are $9$ values, so the median is at position $\frac{9+1}{2} = 5$

Step 3: Identify the median

Median $= 3$

Check: Count the values on each side - there should be $4$ values below and $4$ values above the median. ✓

General steps:

Write the data in order from smallest to largest
Locate the middle value(s)
If odd number of values: median is the middle value
If even number of values: median is the average of the two middle values

Finding the median from a dot plot

A dot plot already displays data in order, making it easy to locate the median by counting.

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Worked Example: Median from a Dot Plot

For $13$ cricket team members, the median age is the $7$ th value from either end.

Median age $= 22$ years

Finding the median from a stem plot

Stem plots also display ordered data. Count from either end to find the middle position.

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Worked Example: Median from a Stem Plot (Even Number of Values)

For the temperature data with $12$ values:

Stem	Leaf
1	8 9 9
2	0 2 5 7 8 9 9
3	1 3

Key: $0|8 = 8°C$

The median is the average of the $6$ th and $7$ th values:

$M = \frac{25 + 27}{2} = 26°C$

The range

Definition

The range, $R$ , is the simplest measure of spread. It shows the total spread of the data.

$R = \text{largest data value} - \text{smallest data value}$

Finding the range from a stem plot

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Worked Example: Finding the Range

Using the same temperature data:

Lowest value $= 18°C$ , Highest value $= 33°C$

Range $= 33 - 18 = 15°C$

Limitations of the range

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Important Limitations of the Range:

The range has important limitations that make it less useful than other measures:

It depends only on the two extreme values
It can be heavily affected by outliers
Data sets with the same highest and lowest values have the same range, regardless of how the middle values are distributed

For these reasons, the interquartile range is often a better measure of spread.

The interquartile range (IQR)

Understanding quartiles

Quartiles divide a distribution into quarters (four equal parts).

$Q_1$ (first quartile): the value with 25% of data below it and $75\%$ above it
$Q_2$ (second quartile): the median, with 50% of data on each side
$Q_3$ (third quartile): the value with 75% of data below it and $25\%$ above it

Determining the interquartile range

Steps:

Arrange all observations in order
Divide the observations into two equal-sized groups
If $n$ is odd, omit the median from both groups
Find $Q_1 =$ median of the lower half
Find $Q_3 =$ median of the upper half
Calculate: $IQR = Q_3 - Q_1$

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Interpretation: The IQR gives the spread of the middle 50% of data values.

Finding IQR when $n$ is even

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Worked Example: IQR When $n$ is Even

Given: Weights of $18$ cats (in kg)

Stem	Leaf
1	9
2	1 3 5 8
3	0 0 4 9 9
4	0 4 5 8
5	0 3
6	3 4

Key: $1|2 = 1.2$ kg

Step 1: With $18$ values, there are $9$ values in each half.

Step 2: Find $Q_1$ (median of lower half)

Lower half: $1.9, 2.1, 2.3, 2.5, 2.8, 3.0, 3.0, 3.4, 3.9$

$Q_1 = 2.8$ kg (the $5$ th value from bottom)

Step 3: Find $Q_3$ (median of upper half)

Upper half: $3.9, 4.0, 4.4, 4.5, 4.8, 5.0, 5.3, 6.3, 6.4$

$Q_3 = 4.8$ kg (the $5$ th value from top)

Step 4: Calculate IQR

$IQR = Q_3 - Q_1 = 4.8 - 2.8 = 2.0 \text{ kg}$

Finding IQR when $n$ is odd

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Worked Example: IQR When $n$ is Odd

Given: Life expectancy for $23$ countries

Step 1: Find the median

With $23$ values, the median is the $12$ th value $= 73$ years.

Step 2: Exclude the median (since $n$ is odd)

This leaves $11$ values below and $11$ values above.

Step 3: Find quartiles

$Q_1 = 66$ (middle of bottom $11$ values)

$Q_3 = 75$ (middle of top $11$ values)

Step 4: Calculate IQR

$IQR = 75 - 66 = 9 \text{ years}$

Check: The quartiles should divide the data into four equal groups of $5$ values each (with the median excluded). ✓

Finding median and quartiles from a histogram

For grouped data shown in a histogram, we can only find the interval containing each quartile, not exact values.

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Worked Example: Quartiles from a Histogram

For $23$ people's internet usage:

Median ( $12$ th value): in interval 60-70 hours
$Q_1$ ( $6$ th value): in interval 50-60 hours
$Q_3$ ( $6$ th value from top): in interval 80-90 hours

Why is IQR more useful than the range?

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Advantages of IQR over Range:

The IQR is superior to the range because:

It measures the spread of the middle 50% of data
The upper $25\%$ and lower $25\%$ are excluded
It is not affected by outliers
It provides a more stable measure of spread

The mean and standard deviation

The mean and standard deviation are alternative measures of centre and spread. They are most useful for symmetric distributions without outliers.

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While more restrictive than median and IQR, these measures can fully describe a symmetric distribution with just two numbers.

The mean

The mean is what most people call the 'average'. It is the balance point of a distribution.

$\text{mean} = \frac{\text{sum of data values}}{\text{total number of data values}}$

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Worked Example: Calculating the Mean

For the data set $2, 3, 3, 4$ :

$\text{mean} = \frac{2 + 3 + 3 + 4}{4} = \frac{12}{4} = 3$

The mean represents the balance point. In symmetric distributions, the mean and median are equal.

Effect of outliers on the mean

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Worked Example: How Outliers Affect the Mean

For the data set $2, 3, 3, 8$ :

Median remains $M = 3$ , but:

$\text{mean} = \frac{2 + 3 + 3 + 8}{4} = \frac{16}{4} = 4$

The mean is pulled toward the outlier ( $8$ ), while the median stays at $3$ . This shows that the mean is affected by extreme values but the median is not.

Statistical notation

To write formulas more compactly, we use:

$\sum$ (Greek capital sigma): means "sum of"
$x$ : represents a data value
$\bar{x}$ (pronounced "x bar"): represents the mean
$n$ : represents the total number of data values

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The formula for the mean becomes:

$\bar{x} = \frac{\sum x}{n}$

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Worked Example: Using Statistical Notation

Given: Reaction times (in milliseconds): $38, 36, 35, 43, 46, 64, 48, 25$

Step 1: Count the values

$n = 8$ (number of values)

Step 2: Sum the values

$\sum x = 38 + 36 + 35 + 43 + 46 + 64 + 48 + 25 = 335$

Step 3: Calculate the mean

$\bar{x} = \frac{335}{8} = 41.9 \text{ milliseconds}$

Relationship between mean and median

For symmetric distributions:

The median lies at the midpoint
The mean is the balance point
Mean $\approx$ Median (approximately equal)

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Worked Example: Symmetric Distribution

Mortality rates for $60$ US cities

Mean $= 940$ per $100{,}000$

Median $= 944$ per $100{,}000$

Very similar values, as expected for a symmetric distribution.

For skewed distributions:

Mean and median are different
The mean is pulled toward the tail
The mean may not represent a typical value

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Worked Example: Skewed Distribution

Population distribution (positively skewed)

Mean population $= 1.4$ million

Median population $= 0.9$ million

The mean has been affected by extreme values in the tail.

When to use median rather than mean

The median is a resistant statistic - it is relatively unaffected by extreme values.

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Use the median when:

The distribution is clearly skewed
There are outliers present
You want a measure that represents a typical value

Example: Median house prices are used in Australia because house price distributions are positively skewed (a few very expensive houses).

Use the mean when:

The distribution is symmetric
There are no outliers
You need to do further statistical calculations

Choosing between mean and median

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Decision Rule:

If the distribution is:

Symmetric with no outliers: Either mean or median can be used
Clearly skewed and/or has outliers: Use the median

The standard deviation

To measure spread around the median, we use the IQR.

To measure spread around the mean, we use the standard deviation ( $s$ ).

Definition

The formula for the standard deviation is:

$s = \sqrt{\frac{\sum(x - \bar{x})^2}{n-1}}$

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The standard deviation is an average of the squared deviations from the mean. We square the deviations because the sum of deviations around the mean (the balance point) always equals zero.

Calculating standard deviation using a calculator

You will normally use a CAS calculator to find the standard deviation.

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Worked Example: Standard Deviation Using TI-Nspire CAS

Given: Heights (in cm) of women: $176, 160, 163, 157, 168, 172, 173, 169$

Using TI-Nspire CAS:

Table

Step 1: Enter data into a list (e.g., named "height")

Step 2: Go to Statistics > Stat Calculations > One-Variable Statistics

Step 3: Select the list name as X1 List

Step 4: Press enter to generate results

Table

Results show:

$\sum x = 1338$
$n = 8$
$s_x = 6.67083$ (sample standard deviation)
Minimum $= 157$
$Q_1 = 161.5$

Answer: Mean height $\bar{x} = 167.25$ cm, Standard deviation $s = 6.67$ cm

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The sample standard deviation is denoted $s_x$ on the calculator.

Summary: choosing the right measures

Distribution type	Measure of centre	Measure of spread
Symmetric, no outliers	Mean ( $\bar{x}$ ) or Median	Standard deviation ( $s$ ) or IQR
Skewed or has outliers	Median	IQR

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Key principle: Use median and IQR together, or use mean and standard deviation together. Don't mix them inappropriately.

Remember!

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Key Points to Remember:

The median is the middle value that divides data in half - it is resistant to outliers
The range ( $R$ ) is the simplest measure of spread but is affected by extreme values
The IQR measures the spread of the middle $50\%$ of data and is not affected by outliers
The mean ( $\bar{x}$ ) is the balance point of a distribution and can be affected by outliers
The standard deviation ( $s$ ) measures spread around the mean and is best used with symmetric distributions
For symmetric distributions with no outliers, use mean and standard deviation
For skewed distributions or data with outliers, use median and IQR
Always check your quartiles by ensuring they divide the data into four equal groups

Measures of Centre and Spread (VCE SSCE General Mathematics): Revision Notes