Practice Maths

Statistical Problem Solving — Solutions

  1. Full Statistical Analysis — Race Times

    1. Display choice: Stem-and-leaf plot — shows individual values and distribution shape  /  11 | 9  /  12 | 4 6 8  /  13 | 1 5 8  /  14 | 3 7  /  15 | 2 ▶ View Solution
    2. Measures: Mean = 13.43 s, Median = 13.3 s, no mode, Range = 3.3 s ▶ View Solution
    3. Three conclusions: 1. Average race time ~13.4 s. 2. Range 3.3 s shows notable variability between fastest and slowest. 3. Most runners clustered 12–14 s — typical performance range ▶ View Solution

  2. Comparing Two Groups — Quiz Scores

    1. Measures: Class A: Mean 30.6, Median 30.5, Range 16  |  Class B: Mean 28.0, Median 28.5, Range 25 ▶ View Solution
    2. Better performance: Class A — higher mean (30.6 vs 28.0) and higher median (30.5 vs 28.5) ▶ View Solution
    3. More consistent: Class A — range 16 vs Class B’s 25 ▶ View Solution
    4. Comparison paragraph: Class A outperformed (mean 30.6 vs 28.0) and was more consistent (range 16 vs 25); Class B had some very high scorers (up to 40) but also very low (down to 15), suggesting uneven preparation ▶ View Solution

  3. Misleading Statistics — Bar Chart

    1. Actual percentage increase: 8.5% increase from Q1 to Q4 ▶ View Solution
    2. Misleading feature: Y-axis starts at 80 not 0 — makes the small difference look much larger than it is ▶ View Solution
    3. Honest representation: Start y-axis at 0; bars would show only a small proportional difference ▶ View Solution
    4. Why companies mislead: To exaggerate growth and impress investors; a small real increase looks dramatic with a manipulated scale ▶ View Solution

  4. Dice Experiment

    1. Frequency table: Count each face value from the raw data to complete the table (total must equal 40) ▶ View Solution
    2. Expected frequency: 40 ÷ 6 ≈ 6.67 (approximately 7 times each) ▶ View Solution
    3. Mean of rolled values: Theoretical mean = 3.5; actual mean from the data should be close to 3.5 ▶ View Solution
    4. Is the die fair?: 40 rolls is too few for certainty; if all frequencies close to 6–7, consistent with a fair die — many more rolls needed for a definitive conclusion ▶ View Solution

  5. Design a Statistical Inquiry — Bedtime

    1. Statistical question: e.g. “What time do Year 7 students at this school go to bed on school nights?” ▶ View Solution
    2. Data collection method: Random sample of 60–80 students from the roll; anonymous paper questionnaire; collect on a Wednesday to represent a typical school night ▶ View Solution
    3. Display type: Stem-and-leaf plot or grouped histogram — shows distribution of bedtimes and reveals spread ▶ View Solution
    4. Measures and meaning: Mean: average bedtime; Median: middle bedtime less affected by outliers; Mode: most common bedtime; Range: earliest to latest bedtime ▶ View Solution

  6. Interpret a Data Table

    1. Mean study hours: Aisha 3.0, Ben 2.0, Chloe 3.0, Dan 2.0, Ela 3.0 ▶ View Solution
    2. Range per student: Aisha 2, Ben 4, Chloe 0, Dan 4, Ela 5 ▶ View Solution
    3. Most consistent: Chloe — range of 0 (exactly 3 hours every day, no variation) ▶ View Solution
    4. Ben’s claim: Accurate — his mean is exactly 2.0 hours per day ▶ View Solution

  7. True or False? Statistics in the Media

    1. “Average $120 000” vs median $65 000: Possibly misleading — mean is pulled up by very high earners; median better represents a typical Australian ▶ View Solution
    2. 90% satisfaction from 10 customers: Possibly misleading — sample of 10 is too small to be representative; 9 out of 10 is not statistically reliable ▶ View Solution
    3. 50% crime drop (2 → 1): Possibly misleading — technically true but the absolute numbers are so small that a single event causes a 50% change; not meaningful ▶ View Solution
    4. “Average 3 days” recovery: Possibly misleading — if recovery ranges from 1 day to 2 weeks the mean is heavily distorted; median would be a fairer summary ▶ View Solution

  8. Spot the Error — Plant Heights

    1. Mean = (12+15+11+14+13+16+12+60) ÷ 8 = 153 ÷ 8 = 19.125 ≈ 19.1 — yes, the calculation is correct ▶ View Solution
    2. Error in conclusion: Plant 8 (60 cm) is a clear outlier — 7 out of 8 plants are 11–16 cm; the mean of 19.1 cm is higher than every plant except the outlier, so it does not represent the typical plant ▶ View Solution
    3. Better measure: Median — sorted: 11, 12, 12, 13, 14, 15, 16, 60 → Median = (13+14)÷2 = 13.5 cm. Better conclusion: “A typical plant in this experiment is about 13.5 cm tall (median); one plant grew unusually tall at 60 cm.” ▶ View Solution

  9. Which Display is Best?

    Monthly rainfall over 12 months: Line graph — shows change over time  |  Test scores for 25 students: Stem-and-leaf plot — shows distribution of numerical data  |  Favourite sports (categories): Column/bar chart — compares category frequencies  |  Books read (small whole numbers): Dot plot — shows each value, suits small whole-number range

  10. Full Investigation — Screen Time

    1. Sorted: {2, 3, 3, 4, 4, 5, 5, 5, 6, 7, 8, 9} — Mean = 61÷12 ≈ 5.08, Median = 5, Mode = 5, Range = 7 ▶ View Solution
    2. Mode (5) and Median (5) both support the claim; mean (5.08) also close; claim is reasonably accurate ▶ View Solution
    3. 11 out of 12 students exceed 2 hours — fraction: 11⁄12, percentage ≈ 91.7% ▶ View Solution
    4. Para 1: Mean ≈ 5.1 hrs, median = mode = 5 hrs, range = 7 hrs — typical teenager in this group uses screens for about 5 hours daily. Para 2: 91.7% exceed the 2-hour guideline; however the sample is only 12 teenagers so broader conclusions should be made cautiously ▶ View Solution