Mathematical Statistics 214
Chapter 7 — Complete Study Guide
Sampling Distributions & The Central Limit Theorem
Contents
• Random sampling, statistics, and sampling distributions
• Sampling distribution of the mean — full proofs
• Normal, Chi-square (χ²), Student's t, and Fisher F distributions
• Law of Large Numbers
• Central Limit Theorem (CLT) — theorem and worked examples
• Normal approximation to the Binomial / Bernoulli
• Distribution identification map
• Exam strategy, tips per distribution, and common pitfalls
• Quick-reference formula card
Stellenbosch University | MS214 | Chapter 7
1. Key Definitions
Random Variable (Y)
A measurement or experiment that may produce a different numerical outcome for each
trial. Examples: height, coin toss, error count.
Random Sample (Y₁, Y₂, …, Yₙ)
A specific collection of independent and identically distributed (i.i.d.) random variables
drawn from the same population distribution.
Sample Observations (y₁, y₂, …, yₙ)
The actual realised (known) numerical values obtained from a random sample.
Statistic
A function of the observable random variables in a sample (may include known constants).
A statistic is itself a random variable and therefore has a probability distribution — called
its sampling distribution.
Common statistics and their formulas
Sample mean: Ȳ = (1/n) Σᵢ Yᵢ
Sample variance: S² = 1/(n−1) Σᵢ (Yᵢ − Ȳ)²
Minimum: Y(1) = min(Y₁, …, Yₙ)
Maximum: Y(n) = max(Y₁, …, Yₙ)
Range: R = Y(n) − Y(1)
Median: M = Y((n+1)/2)
2. Sampling Distribution of the Mean
Two fundamental results
E(Ȳ) = E(Y) = μ
Var(Ȳ) = σ²/n
The standard error of the mean is σ/√n (not σ/n — a very common error).
Proof 1 — E(Ȳ) = μ
E(Ȳ) = E[ (1/n) Σᵢ Yᵢ ] = (1/n) Σᵢ E(Yᵢ) [linearity of expectation]
= (1/n) · n · E(Y) [i.i.d. ⟹ all E(Yᵢ) equal E(Y)]
= E(Y) = μ ∎
Proof 2 — Var(Ȳ) = σ²/n
Var(Ȳ) = Var[ (1/n) Σᵢ Yᵢ ] = (1/n²) Var( Σᵢ Yᵢ )
= (1/n²) [ Σᵢ Var(Yᵢ) + Σᵢ≠ⱼ Cov(Yᵢ, Yⱼ) ]
= (1/n²) [ n · σ² + 0 ] [i.i.d. ⟹ Cov = 0]
= σ²/n ∎
If the parent population is Normal
If Y₁, …, Yₙ are i.i.d. Normal(μ, σ²), then for any sample size n:
Ȳ ~ Normal(μ, σ²/n)
Standardising — Z transformation (σ known)
Derivation:
E(Ȳ − μ) = μ − μ = 0 and Var(Ȳ − μ) = σ²/n
Dividing by σ/√n scales the variance to 1:
Z = (Ȳ − μ) / (σ/√n) ~ Normal(0, 1)
Worked Example — Battery Quality Assurance
Given: Y ~ Normal(μ = 100, σ² = 144). Sample size n = 5. Find P(Ȳ ≥ 110).
Step 1: Ȳ ~ Normal(100, 144/5) ⟹ SE = √(144/5) = 12/√5 ≈ 5.367
Step 2: P(Ȳ ≥ 110) = P( Z > (110 − 100) / 5.367 ) = P(Z > 1.863) ≈ 0.0312
By the Law of Large Numbers, larger n ⟹ Ȳ concentrates around μ = 100 < 110 ⟹
probability of passing the test decreases.
3. Law of Large Numbers
Theorem
lim_{n→∞} P( |Ȳ − μ| < ε ) = 1 for any ε > 0
As the sample size increases, the sample mean Ȳ converges in probability to the true
population mean μ. The variance Var(Ȳ) = σ²/n → 0 as n → ∞.
4. Chi-Square (χ²) Distribution
Definition
If X₁, …, Xₙ are i.i.d. Normal(μ, σ²), then:
(n − 1)S² / σ² ~ χ²(n−1)
where S² is the sample variance. Degrees of freedom = n − 1.
Note: The proof (Theorem 7.3) is not examinable.
Key properties
• Shape: right-skewed (not symmetric like the Normal distribution).
• Mean = ν (degrees of freedom).
• Variance = 2ν.
• Additive: if W₁ ~ χ²(ν₁) and W₂ ~ χ²(ν₂) are independent, then W₁ + W₂ ~ χ²(ν₁ + ν₂).
• Critical values from table: read P(χ²(ν) > c) in the right tail.
Worked Example — Species Diversity
X ~ Normal(μ, 0.2), n = 9 species. Find P(σ² − 0.1 < S² < σ² + 0.1).
With σ² = 0.2: find P(0.1 < S² < 0.3).
Multiply through by (n−1)/σ² = 8/0.2 = 40:
P( 40 × 0.1 < χ²(8) < 40 × 0.3 ) = P( 4 < χ²(8) < 12 )
= P(χ²(8) > 4) − P(χ²(8) > 12) ≈ (0.85–0.90) − (0.15–0.20)
Answer lies in the interval (0.65, 0.75). Interpolation not required in this subject.
5. Student's t Distribution
Definition
Let Z ~ Normal(0, 1) and W ~ χ²(ν) be independent. Then:
T = Z / √(W/ν) ~ t(ν)
Most important result — population σ unknown
When X ~ Normal(μ, σ²) and σ is unknown, replace σ with the sample standard deviation
S:
T = (X̄ − μ) / (S/√n) ~ t(n−1)
Degrees of freedom = n − 1.
Full proof
Let B = (X̄ − μ)/(σ/√n) ~ Normal(0,1) and A = (n−1)S²/σ² ~ χ²(n−1).
By the t-definition: B / √(A/(n−1)) ~ t(n−1) …(i)
Note: √(A/(n−1)) = √[ (n−1)S²/σ² × 1/(n−1) ] = S/σ …(ii)
Substitute (ii) into (i):
(B) / (S/σ) = [(X̄−μ)/(σ/√n)] × (σ/S) = (X̄ − μ) / (S/√n) ~ t(n−1) ∎
Shape notes
• Symmetric and bell-shaped, but with heavier tails than Normal(0,1).
• As ν → ∞, t(ν) → Normal(0,1).
• When table doesn't give the exact value, state the bound: e.g. 0.025 < P < 0.050.
Worked Example — Tomato Paste
n = 8, S = 4.9 g, advertised weight = 50 g, tolerance = 7.5%.
Threshold: 50 × (1 − 0.075) = 46.25 g.
T = (X̄ − 50) / (4.9/√8) ~ t(7)
P(X̄ < 46.25) = P( T(7) < (46.25 − 50)/(4.9/√8) )
= P( T(7) < −2.165 ) ≈ 0.0336
From t-table bounds: 0.025 < P < 0.050 (both answers accepted).
Exercise — Load Shedding (2024 A1)
X ~ Normal(μ = 9.25, σ² unknown). Capacity = 8.50 kWh. Each 0.25 kWh beyond
capacity = 1 extra load-shedding stage. Stage 4 begins at 8.50 + 4×0.25 = 9.50 kWh.
Sample of n = 10, S = 0.4 kWh.
P(Stage ≥ 4) = P(X̄ ≥ 9.50) = P( T(9) ≥ (9.50−9.25)/(0.4/√10) ) ≈ P(T(9) ≥ 1.976) ≈ 0.0398
6. Fisher (F) Distribution
Definition
Let W₁ ~ χ²(ν₁) and W₂ ~ χ²(ν₂) be independent. Then:
F = (W₁/ν₁) / (W₂/ν₂) ~ F(ν₁, ν₂)
Most important result — comparing two sample variances
If Sa² and Sb² are sample variances from two independent Normal(μ, σ²) populations
(same σ²):
Sa² / Sb² ~ F(nₐ−1, n_b−1)
Proof sketch
(nₐ−1)Sa²/σ² ~ χ²(nₐ−1) and (n_b−1)Sb²/σ² ~ χ²(n_b−1).
Forming the ratio of each divided by its degrees of freedom:
[ Sa²/σ² ] / [ Sb²/σ² ] = Sa² / Sb² ~ F(nₐ−1, n_b−1) ∎
Reciprocal property (very useful in exams)
If F ~ F(ν₁, ν₂), then 1/F ~ F(ν₂, ν₁).
Use this to flip the ratio so it is > 1, allowing you to look it up in the table.
Worked Example — Takealot vs Amazon (2024 A2)
Is Var(S.A.) > 4 × Var(U.S.)? Set σ²SA = 4σ²US under H₀.
SSA² / (4 · SUS²) ~ F(7, 5)
Computed: S²SA ≈ 11.72, S²US ≈ 4.007
Test statistic = 11.72 / (4 × 4.007) ≈ 0.731
Flip using reciprocal: P(bad choice) = P( F(5,7) > 4×4.007/11.72 ) = P( F(5,7) > 1.368 ) >
0.10
7. Central Limit Theorem (CLT)
Theorem
Let Y₁, …, Yₙ be i.i.d. with E(Y) = μ and Var(Y) = σ². For large n (rule of thumb: n ≥ 30):
Σᵢ Yᵢ → Normal(nμ, nσ²)
Ȳ → Normal(μ, σ²/n)
This holds regardless of the shape of the original distribution.
Standardised CLT form
Z = (Ȳ − μ) / (σ/√n) ~ Normal(0, 1)
Z = (Σᵢ Yᵢ − nμ) / (σ√n) ~ Normal(0, 1)
Why it matters
The CLT is the reason that Normal-distribution methods (Z-tests, confidence intervals) work
even when you do not know the parent distribution — provided n is large enough. It is also
the theoretical foundation for the Binomial normal approximation.
Worked Example — Poisson Accidents (2024 A2)
Y ~ Poisson(λ = 0.65) accidents per km. 32 independent road stretches each of 2.5 km.
Capacity = 60 accidents.
Accidents on one stretch: W_i = (sum of Y over 2.5 km)
E(W_i) = 2.5 × 0.65 = 1.625, Var(W_i) = 2.5 × 0.65 = 1.625
Total W = Σᵢ W_i: E(W) = 32 × 1.625 = 52, Var(W) = 32 × 1.625 = 52
By CLT: W ≈ Normal(52, 52)
P(W > 60) = P( Z > (60 − 52)/√52 ) = P(Z > 1.109) ≈ 0.1335
8. Normal Approximation to Bernoulli / Binomial
By the CLT applied to i.i.d. Bernoulli(π) random variables:
Approximation for the SUM Y = ΣXᵢ ~ Binomial(n, π):
Y* ~ Normal(nπ, nπ(1−π))
Approximation for the MEAN X̄ ~ Normal(π, π(1−π)/n):
X̄ ~ Normal(π, π(1−π)/n)
Validity criteria — ALWAYS check before approximating!
1. nπ ≥ 5
2. n(1−π) ≥ 5
3. π − 3 × √(π(1−π)/n) > 0
4. π + 3 × √(π(1−π)/n) < 1
If any criterion fails, use the exact Binomial formula.
Continuity corrections
For the SUM Y:
P(Y ≤ y) ≈ P(Y* < y + 0.5)
P(Y ≥ y) ≈ P(Y* > y − 0.5)
P(y₁ ≤ Y ≤ y₂) ≈ P(y₁ − 0.5 < Y* < y₂ + 0.5)
For the MEAN X̄:
P(Y ≤ y) ≈ P( X̄ < y/n + 1/(2n) )
P(Y ≥ y) ≈ P( X̄ > y/n − 1/(2n) )
P(y₁ ≤ Y ≤ y₂) ≈ P( y₁/n − 1/(2n) < X̄ < y₂/n + 1/(2n) )
Worked Example — Insurance (fully worked, all three cases)
Setup: P(male aged 50 dies before 60) = 0.23. Female risk is 13% lower relatively: π =
0.23 − 0.13×0.23 = 0.20. n = 37.
Validity check:
nπ = 37 × 0.20 = 7.4 ≥ 5 ✓
n(1−π) = 37 × 0.80 = 29.6 ≥ 5 ✓
π − 3√(π(1−π)/n) ≈ 0.003 > 0 ✓
π + 3√(π(1−π)/n) ≈ 0.397 < 1 ✓
Case 1 — Exact Binomial:
P(Y ≤ 5) = Σᵢ₌₀⁵ C(37,i) × 0.20ⁱ × 0.80^(37−i) ≈ 0.2220
Case 2 — Normal approximation of the SUM (continuity correction):
Y* ~ Normal(37×0.20, 37×0.20×0.80) = Normal(7.4, 5.92)
P(Y ≤ 5) ≈ P(Y* < 5.5) = P( Z < (5.5 − 7.4)/√5.92 ) ≈ P(Z < −0.781) ≈ 0.2174
Case 3 — Normal approximation of the MEAN (continuity correction):
X̄ ~ Normal(0.20, 0.20×0.80/37)
P(Y ≤ 5) ≈ P( X̄ < 5/37 + 1/74 ) = P( Z < (0.1486 − 0.20)/√(0.16/37) ) ≈ P(Z < −0.781) ≈
0.2174
9. Distribution Identification Map
Use this table to identify the correct distribution for every exam question.
Clue in the question
Distribution
Test statistic
df / params
σ (population std) is
KNOWN; asking about
sample mean Ȳ
Z (Standard
Normal)
Z = (Ȳ − μ) / (σ/√n)
—
σ is UNKNOWN; given S
(sample std); asking about
mean
Student's t
T = (X̄ − μ) / (S/√n)
ν = n−1
S² (sample variance) given;
asking probability about S²;
σ² known
χ² (Chi-square)
(n−1)S²/σ² ~ χ²(n−1)
ν = n−1
Ratio of TWO sample
variances Sa²/Sb²; testing
equality of variances
Fisher F
Sa²/Sb² ~ F(nₐ−1, n_b−1)
ν₁ = nₐ−1 ν₂
= n_b−1
Parent IS Normal; any
sample size n; σ² known
Normal (exact)
Ȳ ~ Normal(μ, σ²/n) exact for all n
—
n ≥ 30; non-normal parent
(Poisson, Uniform,
Exponential…)
Normal via CLT
Ȳ ~ N(μ, σ²/n) or Σ ~ N(nμ, nσ²)
n ≥ 30
Binomial / Bernoulli;
'approximate the
probability'; n large
Normal approx
to Binomial
(mean)
(sum)
(s um)
Y* ~ N(nπ, nπ(1−π)) or X̄ ~ N(π,
(mean)
π(1−π)/n)
Means and variances for CLT calculations
Distribution
E(Y)
Var(Y)
Bernoulli(π)
π
π(1−π)
Binomial(n, π)
nπ
nπ(1−π)
Poisson(λ)
λ
λ
Uniform(a, b)
(a+b)/2
(b−a)²/12
Exponential(λ)
1/λ
1/λ²
Normal(μ, σ²)
μ
σ²
Check 4
criteria
10. Exam Strategy & Question-Solving Framework
General 7-step approach for every question
Step 1 — Identify the random variable (X or Y) and state its distribution.
Step 2 — Identify what you're asked: probability about a mean? sum? variance? ratio?
Step 3 — Choose the correct distribution (use the identification map).
Step 4 — Write down the test statistic formula and substitute all values.
Step 5 — Simplify to a standard form (Z, t, χ², or F).
Step 6 — Read the probability from the appropriate table.
Step 7 — Write a one-sentence conclusion in the context of the question.
Tips for Z questions (σ known)
• Always standardise: Z = (X̄ − μ)/(σ/√n). Draw a sketch and shade the area.
• P(X̄ > value) = right tail; P(X̄ < value) = left tail.
• For a SUM question use W = Σ Yᵢ ~ N(nμ, nσ²).
Tips for t questions (σ unknown)
• Key clue: you are given S (sample std), not σ (population std).
• Always state T ~ t(n−1) and write the degrees of freedom explicitly.
• If the exact t-value is not in the table, state the interval bound: e.g. '0.025 < P <
0.050'.
• t has heavier tails than Normal — tail probabilities are larger for the same Z-score.
Tips for χ² questions (sample variance)
• Key clue: question mentions S² and asks for a probability involving S².
• Transform: multiply both sides of the inequality by (n−1)/σ².
• Result lies between two χ² critical values — read both from the table.
• χ² is right-skewed — P(χ²(ν) > a) is the right-tail probability.
Tips for F questions (comparing variances)
• Key clue: ratio of two sample variances, or 'is variability of A greater than B?'
• Always label which sample is in the numerator vs denominator.
• If your F-ratio < 1, flip it using the reciprocal property and swap df.
• If σ²a = k × σ²b under H₀, include the factor k in the denominator.
Tips for CLT questions (non-normal parent)
• Key clue: parent is Poisson / Uniform / Exponential AND n ≥ 30.
• Look up E(Y) and Var(Y) for that distribution in the reference table.
• For a sum: W = Σ Yᵢ ~ N(nμ, nσ²).
• Standardise: Z = (W − nμ) / (σ√n).
Tips for Binomial approximation questions
• The phrase 'approximate the probability' almost always signals this method.
• Always check all four validity criteria before approximating — show the working.
• Apply the continuity correction: add/subtract 0.5 to the sum boundary.
• You may use Case 2 (sum) or Case 3 (mean) — they give the same answer.
Common pitfalls — do NOT make these mistakes
✗ Using Z when σ is unknown — you must use t.
✗ Writing SE = σ/n instead of SE = σ/√n.
✗ Using df = n for t and χ² — the correct df is n−1.
✗ Skipping the validity check for the Binomial approximation.
✗ Forgetting the continuity correction when approximating Binomial.
✗ Confusing Var(Ȳ) = σ²/n with Var(Y) = σ².
✗ Getting the wrong tail direction — always sketch the distribution.
✗ Forgetting to flip the F ratio when it is < 1.
11. Quick-Reference Formula Card
Keep this page open during revision.
Distribution
Test statistic / formula
df
Condition
Z (Standard
Normal)
Z = (X̄ − μ) / (σ/√n)
—
σ known
Student's t
T = (X̄ − μ) / (S/√n)
n−1
σ unknown; X
Normal
χ² (Chi-square)
(n−1)S²/σ²
n−1
X Normal
Fisher F
Sa²/Sb²
(nₐ−1,
n_b−1)
Two indep. normal
pops
Normal (exact)
Ȳ ~ N(μ, σ²/n)
—
X Normal; any n
CLT — mean
Ȳ ~ N(μ, σ²/n)
—
n ≥ 30; any parent
CLT — sum
ΣYᵢ ~ N(nμ, nσ²)
—
n ≥ 30; any parent
Binomial approx
(sum)
Y* ~ N(nπ, nπ(1−π))
—
nπ ≥ 5; n(1−π) ≥ 5
Binomial approx
(mean)
X̄ ~ N(π, π(1−π)/n)
—
Same validity
criteria
Bernoulli approximation validity criteria — always show all four lines of working:
1. nπ ≥ 5
2. n(1−π) ≥ 5
3. π − 3 × √(π(1−π)/n) > 0
4. π + 3 × √(π(1−π)/n) < 1
MS214 Chapter 7 Study Guide • Stellenbosch University • All content derived from official course material.