← Unit 4 Measures of Centre and Spread
MA5-DAT-C-01

Measures of Centre and Spread

⏱ 25 min📚 Year 9📈
Think First

For the data set 3, 7, 8, 8, 10, 12, 15, which measure of centre best represents the data? Why?

💡 Revision: Ensure you can order data and find the middle value.

Learning Intentions

Know

  • Mean
  • Median
  • Mode
  • Range
  • Interquartile range

Understand

  • When median is preferable to mean
  • How outliers affect each measure

Can Do

  • Calculate mean, median, mode
  • Find range and IQR
  • Choose appropriate measures for given data
MeanMedianModeRangeIQROutlier
Learn Phase
1

Mean

The average

The mean is the sum of all values divided by the number of values.

Mean

$ar{x} = dfrac{sum x}{n}$

The mean uses all data points but is affected by outliers.

Example: 2, 4, 6, 8, 100. Mean = 24, but most values are much smaller.

2

Median and Mode

Middle and most common

The median is the middle value when data is ordered.

For $n$ values: position = $(n+1)/2$

If $n$ is even, average the two middle values.

The mode is the most frequently occurring value.

Example: 3, 5, 7, 8, 8, 10. Median = $(7+8)/2 = 7.5$. Mode = 8.

3

Range and IQR

Measuring spread

Range = maximum − minimum

Interquartile Range (IQR) = Q3 − Q1

Where Q1 is the median of the lower half and Q3 is the median of the upper half.

The IQR measures the spread of the middle 50% of data and is not affected by outliers.

Outliers are often defined as values below Q1 − 1.5×IQR or above Q3 + 1.5×IQR.

Check Understanding

Try it yourself

For data 2, 5, 7, 8, 10, 12, 20: find mean, median, mode, range, and IQR. Identify any outliers.

Worked Example

Measures of Centre and Spread

1

Find mean, median, mode for: 4, 6, 6, 8, 10, 12.

Mean = $(4+6+6+8+10+12)/6 = 46/6 approx 7.67$

Median = $(6+8)/2 = 7$

Mode = 6

2

Find range and IQR for: 3, 5, 7, 9, 11, 13, 15.

Range = 15 − 3 = 12

Q1 = 5, Q3 = 13, IQR = 13 − 5 = 8

3

Data: 10, 12, 15, 18, 20, 22, 50. Is 50 an outlier?

Q1 = 12, Q3 = 22, IQR = 10

Upper fence = $22 + 1.5(10) = 37$

50 > 37, so yes, 50 is an outlier.

Common Misconceptions

Mean is always the best measure of centre. No — the median is better for skewed data or data with outliers.

Range and IQR are the same. No — range uses all data (max − min), while IQR uses only the middle 50%.

For even $n$, pick either middle value as median. No — average the two middle values.

Your Turn

Practice — Centre and Spread

Work through each question in your book or digitally. Answers are in the Questions phase.

1Find mean, median, mode for: 5, 8, 8, 10, 12, 15, 20.
2Find range and IQR for: 2, 4, 6, 8, 10, 12, 14, 16.
3For data 5, 10, 15, 20, 25, 100, which measure of centre is most appropriate? Explain.
Real-World Anchor

Economics and Policy

The Australian Bureau of Statistics reports both mean and median household income because the mean is skewed by high earners. The median gives a better picture of the typical Australian family. Similarly, median house prices are reported rather than mean prices.

📓 Copy Into Your Books

Mean

  • $ar{x} = sum x / n$
  • Affected by outliers

Median

  • Middle value
  • Better for skewed data

Spread

  • Range = max − min
  • IQR = Q3 − Q1
  • IQR ignores outliers
Questions Phase
Check Your Understanding
Answer all questions correctly to unlock the Game phase.
Mean of 4, 6, 8, 10:
Median of 3, 5, 7, 9, 11:
Mode of 2, 3, 3, 5, 7:
Range of 5, 10, 15, 20, 25:
IQR uses:
Best measure for skewed data:
Q1 for 2, 4, 6, 8, 10, 12, 14:
Outlier if above Q3 + 1.5×IQR:
1Find mean, median, mode, range, and IQR for: 3, 7, 8, 10, 12, 15, 20.
2A data set has mean 50 and median 45. What does this suggest about the distribution?
3Explain why IQR is preferred over range when data contains outliers.

Comprehensive Answers

1Mean, median, mode, range, IQR for 3,7,8,10,12,15,20.
Mean=75/7≈10.7, Median=10, Mode=none, Range=17, Q1=7, Q3=15, IQR=8.
2Mean 50, median 45.
Positive skew - tail extends to higher values, pulling mean above median.
3Why IQR preferred over range.
Range affected by outliers; IQR measures middle 50% spread, robust to extremes.
MC 1Mean of 4,6,8,10.
$28/4 = 7$. Answer: B
MC 2Median of 3,5,7,9,11.
7. Answer: C
MC 3Mode of 2,3,3,5,7.
3. Answer: B
MC 4Range of 5,10,15,20,25.
25-5=20. Answer: D
MC 5IQR uses.
Middle 50%. Answer: B
MC 6Best for skewed data.
Median. Answer: B
MC 7Q1 for 2,4,6,8,10,12,14.
Median of lower half (2,4,6) = 4. Answer: A
MC 8Outlier above fence.
Always (by definition). Answer: A
SA 1Stats for 3,7,8,10,12,15,20.
Mean≈10.7, Median=10, Mode=none, Range=17, IQR=8.
SA 2Mean 50, median 45.
Positive skew.
SA 3IQR vs range.
IQR robust to outliers.
Game Phase
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