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IB Diploma Mathematics: Applications & Interpretation SL · Statistics & Probability
Mini-Lesson

Statistics & Probability

This SL mini-lesson is the AI heartland: sampling methods and data, measures of centre & spread, correlation & regression, probability (Venn/tree diagrams, conditional), the binomial and normal distributions, and the χ² test for independence.

AI flavour: statistics is the largest strand — you design samples, run tests on your GDC and interpret results in context.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Every number on the calculation screens has been re-derived and checked. Press Start when you are ready.

Data & sampling

Sampling methods

  • Simple random — every member equally likely; needs a sampling frame.
  • Systematic — every k-th member from a list.
  • Stratified — sample each subgroup in proportion to its size.
  • Quota / convenience — quick but potentially biased.

Bias makes a sample unrepresentative — watch for self-selection and undercoverage.

Centre & spread

Averages & standard deviation

The mean is the balance point; the standard deviation σ measures spread. For data 4, 7, 7, 9, 10, 13:

Worked example

mean = (4+7+7+9+10+13) ÷ 6 = 50 ÷ 6 = 8.33

population σ = √(Σ(x−x̄)² ÷ n) = 2.81 (from GDC 1-Var Stats)

Quartiles give the IQR = Q3 − Q1; an outlier lies beyond Q1 − 1.5·IQR or Q3 + 1.5·IQR.

Calculate

Your turn — the mean

1Find the mean of the data 4, 7, 7, 9, 10, 13, to 2 decimal places.
Hint: add them (= 50) then divide by 6.
Probability

Probability rules

For events A and B:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)P(A|B) = P(A ∩ B) ÷ P(B)
  • Independent: P(A ∩ B) = P(A)·P(B).
  • Mutually exclusive: P(A ∩ B) = 0.
  • Use tree diagrams for sequences and Venn diagrams for overlaps.
Quick check

Independent events

?Two fair coins are tossed. What is the probability of two heads?
Binomial

The binomial distribution

X ~ B(n, p) counts successes in n independent trials each with probability p.

P(X = r) = nCr pr(1−p)n−rmean = np · variance = np(1−p)
Worked example

X ~ B(10, 0.3).

P(X = 4) = 10C4(0.3)⁴(0.7)⁶ = 0.2001

mean = 10 × 0.3 = 3; P(X ≤ 2) = 0.3828 (GDC binomcdf)

Calculate

Your turn — binomial

2For X ~ B(10, 0.3), find P(X = 4) to 4 decimal places using the binomial formula (or GDC binompdf).
Hint: 10C4 × 0.3⁴ × 0.7⁶.
Normal

The normal distribution

A continuous, symmetric, bell-shaped model X ~ N(μ, σ²). Areas under the curve are probabilities (use GDC normalcdf / inverse normal).

Worked example

X ~ N(50, 8²).

P(X < 60) = 0.8944

P(45 < X < 60) = 0.6284

90th percentile (inverse normal) = 60.25

Calculate

Your turn — normal probability

3For X ~ N(50, 8²), find P(X < 60) to 4 decimal places (GDC normalcdf, lower = −1E99).
Hint: standardise z = (60 − 50) ÷ 8 = 1.25, then read the area.
χ² test

Chi-square test for independence

The χ² test checks whether two categorical variables are independent. From a contingency table of observed O, compute expected E = (row total × column total) ÷ grand total, then:

χ²calc = Σ (O − E)² ÷ Edegrees of freedom = (rows − 1)(columns − 1)
Worked example — 2×3 table

Observed: 20,30,10 / 30,20,40 (grand total 150).

χ²calc = 16.67, df = (2−1)(3−1) = 2

Critical value at 5% = 5.991. Since 16.67 > 5.991, reject H₀ — the variables are associated.

Calculate

Your turn — degrees of freedom

4A χ² test uses a 3 × 3 contingency table. State the number of degrees of freedom, (r − 1)(c − 1).
Hint: (3 − 1) × (3 − 1).
Calculate

Your turn — χ² decision

5A χ² test gives χ²calc = 16.67 with critical value 5.991 at the 5% level. Enter 1 if you reject H₀ (association exists), or 0 if you do not.
Hint: reject H₀ when χ²calc is greater than the critical value.
Sort it

Sort the statistics idea

Tap a term, then the strand it belongs to.

🎲 Probability

🔔 Distribution

🧪 Hypothesis test

Quick check

χ² conclusion

?In a χ² test for independence, you reject H₀ when:
Quick check

Stratified sampling

?A school has 60% junior and 40% senior students. A stratified sample of 50 should contain how many juniors?
Match it

Match the tool to its job

Tap an item on the left, then its match on the right.

Item
Match
Correlation vs causation

Interpreting correlation with care

A strong r shows association, not cause. Confounding variables (a hidden third factor) can create correlation without any causal link.

In your exam, state associations cautiously and avoid claiming one variable causes the other.

Quick check

Correlation caution

?Ice-cream sales and drowning rates are strongly positively correlated. The best explanation is:
Recap

The big ideas to take away

Sampling: random, systematic, stratified; beware bias

Centre & spread: mean, median, σ, IQR; outliers beyond 1.5·IQR

Probability: P(A∪B)=P(A)+P(B)−P(A∩B); conditional P(A|B)

Binomial: B(n,p): mean np, variance np(1−p)

Normal: N(μ,σ²): GDC normalcdf / inverse normal

χ² test: Σ(O−E)²⁄E, df=(r−1)(c−1); reject H₀ if beyond critical value

You have worked through the whole topic. Press Finish to see your score.

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