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Themed maths

Big tournament statistics for IB Maths SL and HL

A ready-to-teach statistics lesson for the summer term, built on a made-up tournament: a 5-minute starter, a 35-minute main activity on shots and goals, goals per match and race times, an HL extension and full worked answers. All the data are made up for this lesson.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Descriptive statistics; Correlation and the regression line; The Poisson distribution (AI SL, AA HL); The normal distribution
Equipment
The starter is non-calculator. A GDC is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minShots and goals
Main: task B12 minGoals per match
Main: task C11 minThe 100 m final
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. Find the mean of 2, 4, 4, 5, 10.
  2. Find the median of 3, 8, 1, 9, 4, 7.
  3. Q1 = 11 and Q3 = 18. Find the interquartile range.
  4. Is a correlation coefficient of −0.9 strong or weak, positive or negative?

Main activity (35 minutes)

Task A: Shots and goals (12 min)

Made-up data for ten teams in a tournament. Shots on target x: 12, 15, 9, 20, 17, 11, 14, 22, 8, 18. Goals y: 4, 5, 2, 8, 6, 3, 5, 9, 2, 6.

  1. Find the mean number of shots on target and the mean number of goals.
  2. Find Pearson’s product-moment correlation coefficient, to 3 significant figures, and describe the correlation.
  3. Find the equation of the regression line of y on x, with coefficients to 3 significant figures. Use it to estimate the goals for a team with 16 shots on target.
  4. Why would it be unwise to use the line to predict the goals for a team with 40 shots on target?

Task B: Goals per match (12 min)

Made-up goals in 40 matches: 0 goals in 4 matches, 1 in 9, 2 in 11, 3 in 8, 4 in 5, 5 in 2 and 6 in 1.

  1. Find the mean and the standard deviation of the number of goals per match, each to 3 significant figures.
  2. Model the goals in a match by a Poisson distribution with mean 2.275. How many of the 40 matches would the model expect to have no goals? Give 3 significant figures.
  3. Using the model, find P(X ≥ 4), to 3 significant figures, and compare with the data.

Task C: The 100 m final (11 min)

In a made-up championship, sprinters’ 100 m times are modelled by T ~ N(10.2, 0.152) seconds.

  1. Find P(T < 10.0), to 3 significant figures.
  2. The fastest 5% of times win a medal place. Find the qualifying time, to 3 significant figures.
  3. Eight sprinters run independently. Find the probability that at least one runs under 10.0 s.

Extension (10 minutes)

HL: sums of Poisson variables.

  1. Goals in different matches are independent, each Po(2.275). Find the probability of exactly 6 goals in total in 3 matches, to 3 significant figures.

For teachers

Teacher notes and full worked answers
  • Every number in this pack is made up for the lesson; say so when you show it.
  • Task A (d) is the key discussion: correlation within the data does not license prediction outside it.
  • Ask the class what the Poisson model assumes about goals (independent, at a constant rate) and whether that is realistic.

Starter

  1. 5
    • 25 ÷ 5 = 5
  2. 5.5
    • In order: 1, 3, 4, 7, 8, 9. The median is (4 + 7) ÷ 2 = 5.5.
  3. 7
    • 18 − 11 = 7
  4. Strong and negative
    • |−0.9| is close to 1, and the sign is negative.

Task A: Shots and goals

  1. 14.6 and 5
    • 146 ÷ 10 = 14.6 and 50 ÷ 10 = 5
  2. r = 0.989: strong positive correlation
    • GDC: r = 0.98894…
  3. y = 0.499x − 2.29; about 5.70 goals
    • GDC: y = 0.49898…x − 2.2851…
    • 0.49898… × 16 − 2.2851… = 5.698…
  4. Extrapolation
    • 40 is far outside the data (8 to 22 shots), so the linear pattern may not hold there.

Task B: Goals per match

  1. Mean 2.275; SD 1.45
    • Total goals: 0 + 9 + 22 + 24 + 20 + 10 + 6 = 91, so the mean is 91/40 = 2.275.
    • GDC: σ = 1.4489…
  2. 4.11
    • P(X = 0) = e−2.275 = 0.1027…
    • 40 × 0.1027… = 4.11 (observed: 4)
  3. 0.196; the data give 8/40 = 0.2
    • P(X ≥ 4) = 1 − P(X ≤ 3) = 0.1955…
    • Observed: (5 + 2 + 1)/40 = 0.2. The model fits well.

Task C: The 100 m final

  1. 0.0912
    • GDC normal cdf: 0.09121…
  2. 9.95 s
    • Inverse normal: P(T < t) = 0.05 gives t = 9.953…
  3. 0.535
    • 1 − (1 − 0.09121…)8 = 0.5347…

Extension

  1. 0.152
    • The total is Po(3 × 2.275) = Po(6.825).
    • P(Y = 6) = e−6.825 × 6.8256/6! = 0.1524…

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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