Practice Maths

Topic Review — Statistics — Solutions

  1. Calculating All Four Measures

    1. Cricket squad ages: Sorted: 19,20,21,22,23,24,25,26,27,28,29,30,31,33,35 | Sum=393, Mean=393÷15=26.2 | Median=26 (8th value) | Mode=none | Range=35−19=16 ▶ View Solution
    2. Homework minutes: Sorted: 30,40,45,45,45,45,50,55,55,60,65,70,75,80,90 | Sum=850, Mean=850÷15=56.7 min | Median=55 (8th value) | Mode=45 (appears 4 times) | Range=90−30=60 ▶ View Solution
  2. Using Technology Features

    1. Range formula: =MAX(B2:B20)−MIN(B2:B20) ▶ View Solution
    2. Mean of 0.48 for data 40–60: Most likely a decimal error — values like 45 were entered as 0.45. The formula calculated the mean of the wrongly entered values. ▶ View Solution
    3. =MODE() error: False. =MODE() returns an error when all values are unique (no value repeats). This is not a data entry error — it simply means the dataset has no mode. ▶ View Solution
    4. Estimate then calculate: Estimate: values are 12–16, middle is around 13–14. Exact: Sum=(12+14+13+15+12+14+16+13)=109, Mean=109÷8=13.625 ▶ View Solution
  3. Full Inquiry Process — Travel Distances

    1. Measures: Sum=1.2+3.5+0.8+5.1+2.4+1.8+4.2+0.5+3.1+2.7+1.5+6.3+2.0+3.8+1.1+4.5+2.9+0.7+3.3+2.2=53.6 | Mean=53.6÷20=2.7 km | Sorted: 0.5,0.7,0.8,1.1,1.2,1.5,1.8,2.0,2.2,2.4,2.7,2.9,3.1,3.3,3.5,3.8,4.2,4.5,5.1,6.3 | Median=(2.4+2.7)÷2=2.55 km | Mode=none | Range=6.3−0.5=5.8 km ▶ View Solution
    2. Display type: A dot plot or stem-and-leaf plot — shows the distribution of continuous values for a moderately sized dataset and allows individual values to be read ▶ View Solution
    3. Two conclusions: 1. "On average, students travel about 2.7 km to school, suggesting most live reasonably close." 2. "The range of 5.8 km indicates considerable variation; some students travel more than five times as far as others." ▶ View Solution
    4. Limitation: Only 20 students surveyed — this may not represent the full Year 7 cohort or all transport methods. Students who were absent that day are excluded. ▶ View Solution
  4. Interpreting and Communicating Results

    1. Swimming squads:
      1. Mean and range: Squad A: Sum=248.5, Mean=248.5÷8=31.1 s, Range=33.2−28.9=4.3 s | Squad B: Sum=256.0, Mean=256.0÷8=32.0 s, Range=36.1−27.9=8.2 s ▶ View Solution
      2. Faster / more consistent: Squad A is faster on average (31.1 s vs 32.0 s) and more consistent (range 4.3 s vs 8.2 s). Squad B has more variability — some fast swimmers but also some who are considerably slower. ▶ View Solution
    2. 90% pass rate claim: This conclusion is not valid from this data alone. We need to know: what was the pass mark? (a low pass mark could explain high rates). What is the historical pass rate? (if it was usually 85%, 90% isn't dramatic). Was the exam harder or easier in content? Without context, we cannot conclude the exam was too easy. ▶ View Solution
  5. Extended Problem Solving

    1. Temperature analysis:
      1. Measures: Sorted: 18,19,21,22,23,24,25,25,26,27,28,29,30,31 | Sum=348, Mean=348÷14=24.9°C | Median=(25+25)÷2=25°C | Mode=25 | Range=31−18=13°C ▶ View Solution
      2. Stem-and-leaf: 1 | 8 9 | 2 | 1 2 3 4 5 5 6 7 8 9 | 3 | 0 1 | Key: 1|8 = 18°C ▶ View Solution
      3. Three conclusions: 1. "The average maximum temperature was about 25°C, suggesting warm conditions throughout the period." 2. "Most days had temperatures between 21 and 29°C, with two cooler days (18°C and 19°C) and two hotter days (30°C and 31°C)." 3. "The range of 13°C indicates noticeable day-to-day variation in temperature." ▶ View Solution
    2. Inquiry design — sport and academics:
      1. Statistical question: Sample: "Is there a difference in average academic scores between Year 7 students who participate in school sport and those who do not?" ▶ View Solution
      2. Data collection plan: Compare semester report card averages for all Year 7 students; split into two groups (sport participants vs non-participants) using school sport registration records. This avoids self-reporting bias. ▶ View Solution
      3. Confounding factor: Parental involvement: parents who support sport participation may also provide more academic support at home. If so, better academic results could be due to parent engagement, not sport itself. ▶ View Solution
      4. Correlation not causation: Even if sport participants score higher, we cannot prove sport causes better grades. Both sport participation and academic performance might be explained by a third factor (e.g. general motivation, family support). To prove causation, you would need a controlled experiment where students are randomly assigned to sport or no-sport conditions. ▶ View Solution