Practice Maths

Topic Review — Measures of Centre — Solutions

  1. Calculate all four measures

    1. {6,9,4,8,3,9,7}: Ordered: 3,4,6,7,8,9,9. Mean=46÷7≈6.57. Median=7. Mode=9. Range=6. ▶ View Solution
    2. {20,25,22,28,25,30,18,22}: Ordered: 18,20,22,22,25,25,28,30. Mean=190÷8=23.75. Median=(22+25)÷2=23.5. Mode=22 and 25 (bimodal). Range=12. ▶ View Solution
    3. {11,11,13,15,17,19,11}: Ordered: 11,11,11,13,15,17,19. Mean=97÷7=13.86. Median=13. Mode=11. Range=8. ▶ View Solution
    4. {50,55,60,65,70,75,80}: Already ordered. Mean=455÷7=65. Median=65. No mode. Range=30. ▶ View Solution
    5. {4,4,4,4,4}: Mean=4. Median=4. Mode=4. Range=0. ▶ View Solution

  2. Find missing values

    1. Mean of {8,12,x,10} = 11: Total = 11 × 4 = 44. 8+12+x+10 = 44. 30+x = 44. x = 14. ▶ View Solution
    2. 5th number (mean=14, four known): Total = 14 × 5 = 70. Known sum = 10+16+13+18 = 57. Fifth = 70−57 = 13. ▶ View Solution
    3. n in {n,5,9,7,4} with mean 7: Total = 7 × 5 = 35. n+5+9+7+4 = 35. n+25 = 35. n = 10. ▶ View Solution
    4. Missing value with median 10: 5 values, median is the 3rd. Ordered: 6, 8, ?, 12, 15. For median = 10, the 3rd value = 10. Missing value = 10. ▶ View Solution

  3. Read from a stem-and-leaf plot

    1. All values: 15, 18, 20, 23, 23, 27, 31, 34, 34, 39, 42, 45 (12 values) ▶ View Solution
    2. All measures: Sum = 351. Mean = 351÷12 = 29.25 km. Median = (27+31)÷2 = 29 km. Mode = 23 and 34 (bimodal). Range = 45−15 = 30 km. ▶ View Solution
    3. What median tells us: The median (29 km) tells us that the typical runner covered about 29 km in a month — half ran more and half ran less. It is a good summary of a "typical" month because it is not affected by extreme values like the 15 km minimum or 45 km maximum. ▶ View Solution

  4. Choose the right measure

    1. House prices with $2.1M: Median. The $2.1M property is an extreme outlier. Mean = (350+380+390+420+440+2100)÷6 = 4080÷6 = $680k, which is much higher than five of the six properties. Median = (390+420)÷2 = $405k, far more representative of a typical house in this suburb. ▶ View Solution
    2. Baker's croissants: Mode (20). The baker wants to know the most common daily sales figure to stock the right amount. Mode = 20 (appears 3 times). This is the most practically useful measure for ordering purposes. ▶ View Solution
    3. Range of 5 vs range of 42: Range of 5: all students scored within 5 marks of each other — very consistent, uniform performance. Range of 42: scores vary enormously — some students scored very high, others very low. This class has very mixed performance levels. ▶ View Solution
    4. Mean equals median: Yes. Example: {2, 4, 6, 8, 10}. Mean = 30÷5 = 6. Median = 6 (middle value). Mean = Median when data is symmetric. ▶ View Solution

  5. Outliers and context

    1. {14,16,15,17,13,16,62}: Outlier = 62. With outlier: Mean = (14+16+15+17+13+16+62)÷7 = 153÷7 ≈ 21.9; Ordered: 13,14,15,16,16,17,62 → Median = 16. Without outlier: Mean = (14+16+15+17+13+16)÷6 = 91÷6 ≈ 15.2; Median = (15+16)÷2 = 15.5. ▶ View Solution
    2. Mean vs median change: The mean changed much more (from ≈21.9 to ≈15.2, a change of 6.7) compared to the median (from 16 to 15.5, a change of only 0.5). The mean uses the actual value of all numbers in its calculation, so an extreme outlier has a large effect. The median only depends on position, so one extreme value changes it very little. ▶ View Solution
    3. Removing outlier always improves mean: False. If the outlier is a genuine data point (e.g. one very tall student in a class), removing it actually makes the mean less accurate as a description of the full dataset. Removing an outlier is only appropriate if it was a recording error. Otherwise it should be kept, and median should be used instead. ▶ View Solution

  6. Interpret and compare

    A: 33,34,34,35,35,36,36,37 (ordered)  |  B: 28,29,30,31,40,41,42,43 (ordered)

    1. Measures: Swimmer A: Sum=280, Mean=35s, Median=(35+35)÷2=35s, Mode=34 and 35 and 36, Range=4s.  |  Swimmer B: Sum=284, Mean=35.5s, Median=(31+40)÷2=35.5s, Mode=none, Range=15s. ▶ View Solution
    2. Similar means: Their similar means (35s vs 35.5s) show that both swimmers are roughly equivalent in overall performance — neither is significantly faster on average. However, mean alone doesn't reveal how consistent their times are. ▶ View Solution
    3. More consistent: Swimmer A is far more consistent — range of only 4 seconds (33s to 37s). Swimmer B has a range of 15 seconds (28s to 43s), meaning their performance varies enormously between races. ▶ View Solution
    4. Coach selection: Swimmer A. In a race where consistency is critical, you need a swimmer who reliably performs well. Swimmer A's range of 4 seconds guarantees a time between 33–37 seconds. Swimmer B might swim 28 seconds (excellent) or 43 seconds (poor) — too unpredictable for a crucial race. ▶ View Solution
    5. Comparison paragraph: Both swimmers have almost identical mean times (A: 35s, B: 35.5s) and medians, suggesting equivalent overall speed. However, Swimmer A is dramatically more consistent, with a range of just 4 seconds compared to Swimmer B's range of 15 seconds. Swimmer A's times cluster tightly between 33 and 37 seconds, while Swimmer B alternates between outstanding performances (28s) and very slow races (43s). For any race where reliable performance matters, Swimmer A is the clear choice. ▶ View Solution