Measures of Centre — Solutions

Fluency
Find the mean, median and mode of: 4, 7, 3, 7, 9, 2, 7, 5

Step 1 — Order the data: 2, 3, 4, 5, 7, 7, 7, 9

Mean: x̄ = (2 + 3 + 4 + 5 + 7 + 7 + 7 + 9) ÷ 8 = 44 ÷ 8 = 5.5

Median: n = 8 (even), so median = average of 4th and 5th values = (5 + 7) ÷ 2 = 6

Mode: 7 appears three times — more than any other value. Mode = 7
Fluency
Find the median of: 12, 5, 18, 3, 21, 9, 15, 7

Step 1 — Order the data: 3, 5, 7, 9, 12, 15, 18, 21

Step 2 — Find the middle: n = 8 (even), so median = average of the 4th and 5th values.

4th value = 9, 5th value = 12

Median = (9 + 12) ÷ 2 = 21 ÷ 2 = 10.5
Fluency
A dataset: 8, 12, 15, 8, 20, 8, 11, 14, 9, 10. Find the mean, median and mode.

Step 1 — Order the data: 8, 8, 8, 9, 10, 11, 12, 14, 15, 20

Mean: x̄ = (8 + 8 + 8 + 9 + 10 + 11 + 12 + 14 + 15 + 20) ÷ 10 = 115 ÷ 10 = 11.5

Median: n = 10 (even), median = average of 5th and 6th values = (10 + 11) ÷ 2 = 10.5

Mode: 8 appears three times. Mode = 8
Fluency
Test scores: 65, 72, 78, 80, 84, 85, 88, 92, 95, 100. The teacher adds a bonus 5 marks to every score. How does this affect the mean, median and mode?
Original measures:

Mean = (65 + 72 + 78 + 80 + 84 + 85 + 88 + 92 + 95 + 100) ÷ 10 = 839 ÷ 10 = 83.9

Median = (84 + 85) ÷ 2 = 84.5 | No mode (all values unique)

Effect of adding 5 to every value:

When a constant is added to every value, all measures of centre shift up by that constant.
- New mean = 83.9 + 5 = 88.9
- New median = 84.5 + 5 = 89.5
- Mode: still no single mode (all values unique), but each would increase by 5.
The spread (range, IQR, standard deviation) is unchanged because all values shift by the same amount.
Understanding
Salaries (thousands of dollars): 42, 45, 48, 50, 55, 110. Find the mean and median. Which better represents a ‘typical’ salary? Explain.

Mean: x̄ = (42 + 45 + 48 + 50 + 55 + 110) ÷ 6 = 350 ÷ 6 ≈ $58,300

Median: n = 6 (even), median = average of 3rd and 4th values = (48 + 50) ÷ 2 = $49,000

Which is more representative?

The median ($49,000) better represents a typical salary. The $110,000 salary is an outlier that pulls the mean up to $58,300 — well above most workers’ actual pay. Five of the six employees earn between $42,000 and $55,000, which aligns much more closely with the median.

Rule of thumb: When outliers are present, the median is the preferred measure of centre.
Understanding
Five values have a mean of 12. Four of them are 8, 10, 14, 16. Find the fifth value.

Step 1 — Find the required total:

If mean = 12 and n = 5, then Σx = 5 × 12 = 60

Step 2 — Find the known sum:

8 + 10 + 14 + 16 = 48

Step 3 — Find the fifth value:

Fifth value = 60 − 48 = 12

Check: (8 + 10 + 12 + 14 + 16) ÷ 5 = 60 ÷ 5 = 12 ✓
Understanding
A dataset has mean = 20, median = 18, mode = 15. Describe the likely shape (skew) of the distribution. Explain your reasoning.
Relationship: Mode (15) < Median (18) < Mean (20)

Shape: This distribution is positively skewed (right-skewed).

Reasoning:
- The mode (15) is the most common value — most data clusters at the lower end.
- The mean (20) is pulled upward by a small number of high values (the long right tail).
- The median (18) sits between them, less affected by the extreme high values than the mean.
The pattern mean > median > mode is a reliable indicator of positive (right) skew.
Understanding
Two classes sat the same test. Class A: mean = 72, n = 25. Class B: mean = 68, n = 15. Find the combined mean of all 40 students.

Step 1 — Find the total marks for each class:

Class A total = 72 × 25 = 1800

Class B total = 68 × 15 = 1020

Step 2 — Find the combined total and mean:

Combined total = 1800 + 1020 = 2820

Combined mean = 2820 ÷ 40 = 70.5

Note: You cannot simply average the two means (72 + 68) ÷ 2 = 70, because the classes have different sizes. Class A has more students, so it contributes more to the overall mean, pulling it slightly above 70.
Problem Solving
House prices in a suburb (in $000s): 480, 520, 490, 510, 500, 495, 1200. Calculate the mean and median. A real estate agent advertises the ‘average price’. Which measure should they use, and why might a buyer and seller prefer different measures?
Step 1 — Order the data: 480, 490, 495, 500, 510, 520, 1200

Mean: x̄ = (480 + 490 + 495 + 500 + 510 + 520 + 1200) ÷ 7 = 4195 ÷ 7 ≈ $599,300

Median: n = 7 (odd), median = 4th value = $500,000

Analysis of perspectives:
- Seller might prefer the mean ($599,300) because it sounds more impressive — the outlier property ($1.2M) inflates it significantly.
- Buyer would prefer the median ($500,000) because it truly reflects what most homes in the suburb cost. The $1.2M property is not representative of the typical purchase.
The median is the more honest representation for advertising purposes — it is not distorted by the single high-value outlier. Six of seven houses sold for $480,000–$520,000.
Problem Solving
A student’s assessment comprises 3 tasks weighted 20%, 30% and 50%. Scores: Task 1 = 75, Task 2 = 82, Task 3 = 68. Calculate the weighted mean. Compare with the simple (unweighted) mean. Which is the student’s actual percentage?
Weighted mean formula: x̄ₗ = (w₁x₁ + w₂x₂ + w₃x₃) ÷ (w₁ + w₂ + w₃)

Since the weights already sum to 100%, we multiply directly:

x̄ₗ = (0.20 × 75) + (0.30 × 82) + (0.50 × 68)

= 15.0 + 24.6 + 34.0

= 73.6%

Simple (unweighted) mean:

x̄ = (75 + 82 + 68) ÷ 3 = 225 ÷ 3 = 75.0%

Comparison:
- The student’s actual percentage is 73.6% (weighted mean).
- The simple mean (75%) is higher because it ignores weighting — it treats all tasks equally.
- Task 3 (scored 68%) carries the most weight (50%), so it pulls the true result below the simple average.
- The weighted mean is always the correct measure when tasks have different levels of importance.