Lesson 9 of 12
75% complete
1. The Normal Distribution S1
Normal Distribution: A continuous, symmetric, bell-shaped probability distribution characterised by two parameters — mean μ (location) and variance σ² (spread). Written X ~ N(μ, σ²). The total area under the curve equals 1. It is used to model many naturally occurring phenomena — heights, weights, exam marks, measurement errors.
Key Properties of the Normal Distribution
Shape Properties
- Symmetric about the mean μ
- Mean = Median = Mode = μ
- Bell-shaped curve — single peak at μ
- Asymptotic to x-axis (tails never touch)
- Total area under curve = 1
- 68% of data within μ±σ
- 95% within μ±2σ
- 99.7% within μ±3σ
Effect of μ and σ
- Increasing μ shifts the curve right
- Decreasing μ shifts the curve left
- Larger σ → flatter, wider curve
- Smaller σ → taller, narrower curve
- The area under the curve is always 1 regardless of μ and σ
Standardisation — The Z-Score
Standardisation Formula
Z = (X − μ)/σ where Z ~ N(0, 1)
P(X < x) = P(Z < (x−μ)/σ) = Φ(z)
Φ(−z) = 1 − Φ(z) (symmetry of Standard Normal)
P(a < X < b) = Φ((b−μ)/σ) − Φ((a−μ)/σ)
Reading the Normal Distribution Table (Φ table):
The table gives Φ(z) = P(Z < z) for z > 0. For negative z: use Φ(−z) = 1−Φ(z).
Key values: Φ(1.645) ≈ 0.95, Φ(1.960) ≈ 0.975, Φ(2.326) ≈ 0.99, Φ(2.576) ≈ 0.995
These give critical values for common significance levels in hypothesis testing.
The table gives Φ(z) = P(Z < z) for z > 0. For negative z: use Φ(−z) = 1−Φ(z).
Key values: Φ(1.645) ≈ 0.95, Φ(1.960) ≈ 0.975, Φ(2.326) ≈ 0.99, Φ(2.576) ≈ 0.995
These give critical values for common significance levels in hypothesis testing.
📐 Worked Example 1 — Normal Probabilities
X ~ N(50, 16). Find: (a) P(X < 54) (b) P(X > 46) (c) P(44 < X < 58) (d) P(|X−50| < 3)
1
(a) z=(54−50)/4=1. P(X<54)=Φ(1)=0.8413
2
(b) z=(46−50)/4=−1. P(X>46)=P(Z>−1)=Φ(1)=0.8413
3
(c) z₁=(44−50)/4=−1.5; z₂=(58−50)/4=2
P=Φ(2)−Φ(−1.5)=0.9772−(1−0.9332)=0.9772−0.0668=0.9104
P=Φ(2)−Φ(−1.5)=0.9772−(1−0.9332)=0.9772−0.0668=0.9104
4
(d) P(47<X<53)=P(−0.75<Z<0.75)=2Φ(0.75)−1=2(0.7734)−1=0.5468
📐 Worked Example 2 — Finding μ or σ from a Probability
(a) X~N(μ,25) and P(X<30)=0.8. Find μ. (b) X~N(60,σ²) and P(X>70)=0.1. Find σ.
1
(a) P(X<30)=0.8 → Φ(z)=0.8 → z=0.842
(30−μ)/5=0.842 → 30−μ=4.21 → μ=25.79
(30−μ)/5=0.842 → 30−μ=4.21 → μ=25.79
2
(b) P(X>70)=0.1 → P(X<70)=0.9 → Φ(z)=0.9 → z=1.282
(70−60)/σ=1.282 → σ=10/1.282=7.80
(70−60)/σ=1.282 → σ=10/1.282=7.80
📐 Worked Example 3 — Finding Both μ and σ
X~N(μ,σ²). P(X<20)=0.2 and P(X<30)=0.8. Find μ and σ.
1
P(X<20)=0.2 → z₁=−0.842 (since Φ(0.842)=0.8, so Φ(−0.842)=0.2)
P(X<30)=0.8 → z₂=0.842
P(X<30)=0.8 → z₂=0.842
2
(20−μ)/σ=−0.842 ...(1) (30−μ)/σ=0.842 ...(2)
Subtract: 10/σ=1.684 → σ=5.94
Subtract: 10/σ=1.684 → σ=5.94
3
From (2): 30−μ=0.842×5.94=5 → μ=25
2. Normal Approximations S1
Normal Approximation to the Binomial
If X~B(n,p) with n large and p not too extreme (rule: np>5 and nq>5), then X is approximately N(np, npq). Apply continuity correction — a half-unit adjustment to convert a discrete inequality to a continuous one.
Continuity Correction Table:
P(X=k) → P(k−½ < Y < k+½)
P(X ≤ k) → P(Y < k+½) P(X < k) → P(Y < k−½)
P(X ≥ k) → P(Y > k−½) P(X > k) → P(Y > k+½)
where Y ~ N(np, npq)
P(X=k) → P(k−½ < Y < k+½)
P(X ≤ k) → P(Y < k+½) P(X < k) → P(Y < k−½)
P(X ≥ k) → P(Y > k−½) P(X > k) → P(Y > k+½)
where Y ~ N(np, npq)
📐 Worked Example 4 — Normal Approximation to Binomial
X~B(100, 0.4). Use Normal approximation to find P(X ≤ 35).
1
Check: np=40>5, nq=60>5 ✓. Y~N(40, 24) (μ=np=40, σ²=npq=24)
2
P(X≤35) ≈ P(Y<35.5) (continuity correction)
z=(35.5−40)/√24=(35.5−40)/4.899=−0.918
z=(35.5−40)/√24=(35.5−40)/4.899=−0.918
3
P(Z<−0.918)=1−Φ(0.918)=1−0.8206=0.1794
Normal Approximation to the Poisson
If X~Po(λ) with λ large (rule: λ>15), then X is approximately N(λ, λ). Again apply continuity correction.
📐 Worked Example 5 — Normal Approximation to Poisson
X~Po(25). Use Normal approximation to find P(X ≥ 30).
1
λ=25>15 ✓. Y~N(25,25), σ=5.
2
P(X≥30) ≈ P(Y>29.5). z=(29.5−25)/5=0.9
3
P(Z>0.9)=1−Φ(0.9)=1−0.8159=0.1841
3. Sampling Distributions and the Central Limit Theorem S1
Sampling Distribution of X̄: When a random sample of size n is taken from a population with mean μ and variance σ², the sample mean X̄ has: E(X̄) = μ and Var(X̄) = σ²/n. The standard deviation of X̄ is σ/√n — called the standard error.
Sampling Distribution Formulae
If X~N(μ,σ²) then X̄~N(μ, σ²/n) exactly
Central Limit Theorem: For large n (n≥30), X̄~N(μ, σ²/n) approximately, regardless of population distribution
Standard Error: SE = σ/√n
Standardise X̄: Z = (X̄−μ)/(σ/√n)
📐 Worked Example 6 — Sample Mean Distribution
The weight of apples is distributed with mean 120g and standard deviation 15g. A sample of 25 apples is selected. Find: (a) the distribution of the sample mean (b) P(X̄ > 125) (c) P(115 < X̄ < 122).
1
(a) X̄~N(120, 15²/25)=N(120, 9). SE=3.
2
(b) z=(125−120)/3=5/3≈1.667
P(X̄>125)=1−Φ(1.667)=1−0.9525=0.0475
P(X̄>125)=1−Φ(1.667)=1−0.9525=0.0475
3
(c) z₁=(115−120)/3=−5/3≈−1.667; z₂=(122−120)/3=0.667
P=Φ(0.667)−Φ(−1.667)=0.7475−0.0475=0.7000
P=Φ(0.667)−Φ(−1.667)=0.7475−0.0475=0.7000
4. Hypothesis Testing — Population Mean S1
Hypothesis Test: A formal statistical procedure to decide whether sample evidence supports or contradicts a claim about a population parameter. The procedure uses a test statistic and compares it to a critical value (or computes a p-value).
Seven-Step Hypothesis Testing Procedure:
1. State H₀ and H₁ — Null hypothesis (no change/claimed value) and Alternative (what you suspect).
2. State significance level α (usually 5% or 1%).
3. State the test statistic and its distribution under H₀.
4. Calculate the test statistic from the sample data.
5. Find the critical value or calculate the p-value.
6. Compare test statistic to critical region (or p-value to α).
7. Conclude in context — use words about the original claim.
1. State H₀ and H₁ — Null hypothesis (no change/claimed value) and Alternative (what you suspect).
2. State significance level α (usually 5% or 1%).
3. State the test statistic and its distribution under H₀.
4. Calculate the test statistic from the sample data.
5. Find the critical value or calculate the p-value.
6. Compare test statistic to critical region (or p-value to α).
7. Conclude in context — use words about the original claim.
One-Tailed and Two-Tailed Tests
| Test Type | H₁ | Critical Region | Critical Value (5%) |
|---|---|---|---|
| One-tailed (upper) | μ > μ₀ | Z > z_α | Z > 1.645 |
| One-tailed (lower) | μ < μ₀ | Z < −z_α | Z < −1.645 |
| Two-tailed | μ ≠ μ₀ | |Z| > z_{α/2} | |Z| > 1.960 |
📐 Worked Example 7 — Z-Test for Population Mean
The weight of cereal boxes is claimed to be μ=500g with σ=8g. A sample of 36 boxes has mean 497g. Test at 5% significance level whether the population mean has decreased.
1
H₀: μ=500 H₁: μ<500 (one-tailed, lower)
Significance level: α=5%=0.05
Significance level: α=5%=0.05
2
Under H₀: X̄~N(500, 64/36)=N(500, 16/9). SE=4/3≈1.333
Test statistic: Z=(497−500)/(8/√36)=(−3)/(8/6)=−3×6/8=−2.25
Test statistic: Z=(497−500)/(8/√36)=(−3)/(8/6)=−3×6/8=−2.25
3
Critical value (one-tailed, 5%): z = −1.645
Z=−2.25 < −1.645 → falls in the critical region.
Z=−2.25 < −1.645 → falls in the critical region.
4
Conclusion: Reject H₀. There is sufficient evidence at the 5% significance level to conclude that the mean weight of cereal boxes has decreased below 500g.
📐 Worked Example 8 — Two-Tailed Z-Test
A machine produces bolts with diameter N(10mm, 0.04mm²). After maintenance, a sample of 20 bolts has mean 10.08mm. Test at 1% level whether the mean diameter has changed.
1
H₀: μ=10 H₁: μ≠10 (two-tailed)
α=1%, two-tailed → critical values: Z=±2.576
α=1%, two-tailed → critical values: Z=±2.576
2
SE=√(0.04/20)=√0.002=0.04472
Z=(10.08−10)/0.04472=0.08/0.04472=1.789
Z=(10.08−10)/0.04472=0.08/0.04472=1.789
3
|Z|=1.789 < 2.576 → NOT in critical region.
Conclusion: Do not reject H₀. Insufficient evidence at the 1% level to conclude the mean diameter has changed after maintenance.
Conclusion: Do not reject H₀. Insufficient evidence at the 1% level to conclude the mean diameter has changed after maintenance.
5. Hypothesis Testing — Binomial Proportion S1
Proportion Test: When testing a claim about a probability p, the test statistic is the number of successes X in a binomial experiment. Under H₀: X~B(n, p₀). Find the probability of getting the observed result or more extreme — this is the p-value.
📐 Worked Example 9 — Binomial Hypothesis Test
A manufacturer claims 20% of items are defective. A quality inspector finds 9 defectives in a random sample of 30. Test at 5% level whether the proportion of defectives has increased.
1
H₀: p=0.2 H₁: p>0.2 (one-tailed, upper)
Under H₀: X~B(30, 0.2). α=5%.
Under H₀: X~B(30, 0.2). α=5%.
2
P-value = P(X≥9 | X~B(30,0.2)) = 1−P(X≤8)
Using tables: P(X≤8)≈0.8867
P-value = 1−0.8867 = 0.1133
Using tables: P(X≤8)≈0.8867
P-value = 1−0.8867 = 0.1133
3
P-value=0.1133 > 0.05=α. NOT in critical region.
Conclusion: Do not reject H₀. There is insufficient evidence at the 5% level to conclude that the proportion of defectives has increased above 20%.
Conclusion: Do not reject H₀. There is insufficient evidence at the 5% level to conclude that the proportion of defectives has increased above 20%.
📐 Worked Example 10 — Critical Region for Binomial Test
X~B(20, p). Test H₀: p=0.3 vs H₁: p<0.3 at 5%. Find the critical region.
1
Need the largest c such that P(X≤c | p=0.3) < 0.05.
Calculate cumulative probabilities under B(20, 0.3):
Calculate cumulative probabilities under B(20, 0.3):
2
P(X≤2)≈0.0355 < 0.05 ✓
P(X≤3)≈0.1071 > 0.05 ✗
Critical region: X ≤ 2 (actual significance level ≈ 3.55%)
P(X≤3)≈0.1071 > 0.05 ✗
Critical region: X ≤ 2 (actual significance level ≈ 3.55%)
3
If the observed value of X is ≤ 2, reject H₀ and conclude there is evidence that p < 0.3.
Critical Region vs P-value Approach:
Critical region: Find the boundary value c such that P(X ≤ c) < α (for lower tail). Reject H₀ if observation falls in critical region.
P-value: Calculate P(result as extreme or more extreme | H₀ true). Reject H₀ if p-value < α.
Both give the same conclusion — Cambridge questions may ask for either or both.
Critical region: Find the boundary value c such that P(X ≤ c) < α (for lower tail). Reject H₀ if observation falls in critical region.
P-value: Calculate P(result as extreme or more extreme | H₀ true). Reject H₀ if p-value < α.
Both give the same conclusion — Cambridge questions may ask for either or both.
6. Type I and Type II Errors S1
Type I Error: Rejecting H₀ when it is actually true — a false positive. The probability of a Type I error equals the significance level α. Cannot be reduced without increasing the sample size.
Type II Error: Failing to reject H₀ when it is actually false — a false negative. Probability denoted β. Depends on the true value of the parameter.
Type II Error: Failing to reject H₀ when it is actually false — a false negative. Probability denoted β. Depends on the true value of the parameter.
| H₀ True | H₀ False | |
|---|---|---|
| Reject H₀ | Type I Error (probability=α) | Correct decision (Power=1−β) |
| Do not reject H₀ | Correct decision (probability=1−α) | Type II Error (probability=β) |
📐 Worked Example 11 — Type I and Type II Error Probabilities
X~B(20,p). Test H₀:p=0.4 vs H₁:p≠0.4 at 5%. Critical region is X≤4 or X≥13. (a) Find P(Type I Error). (b) Find P(Type II Error) when p=0.6.
1
(a) P(Type I) = P(X≤4 or X≥13 | p=0.4)
= P(X≤4|B(20,0.4)) + P(X≥13|B(20,0.4))
≈ 0.0510 + P(X≥13)≈0.0510+0.0210=0.0210+0.0510=0.0210
More precisely: P(X≤4)≈0.0510, P(X≥13)≈1−P(X≤12)≈1−0.9790=0.0210
P(Type I)≈0.0510+0.0210=0.0720
= P(X≤4|B(20,0.4)) + P(X≥13|B(20,0.4))
≈ 0.0510 + P(X≥13)≈0.0510+0.0210=0.0210+0.0510=0.0210
More precisely: P(X≤4)≈0.0510, P(X≥13)≈1−P(X≤12)≈1−0.9790=0.0210
P(Type I)≈0.0510+0.0210=0.0720
2
(b) P(Type II | p=0.6) = P(not in critical region | p=0.6)
= P(5≤X≤12 | B(20,0.6))
= P(X≤12|B(20,0.6)) − P(X≤4|B(20,0.6))
≈ 0.5841 − 0.0003 ≈ 0.5838
= P(5≤X≤12 | B(20,0.6))
= P(X≤12|B(20,0.6)) − P(X≤4|B(20,0.6))
≈ 0.5841 − 0.0003 ≈ 0.5838
📝 Exam Practice Questions
Q1 [4 marks] — X~N(72, 100). Find: (a) P(X<80) (b) P(60<X<75) (c) The value of a such that P(X>a)=0.3.
σ=10.
(a) z=(80−72)/10=0.8. P(X<80)=Φ(0.8)=0.7881
(b) z₁=(60−72)/10=−1.2; z₂=(75−72)/10=0.3
P=Φ(0.3)−Φ(−1.2)=0.6179−0.1151=0.5028
(c) P(X>a)=0.3 → P(X<a)=0.7 → z=0.524
a=72+0.524×10=77.24
(a) z=(80−72)/10=0.8. P(X<80)=Φ(0.8)=0.7881
(b) z₁=(60−72)/10=−1.2; z₂=(75−72)/10=0.3
P=Φ(0.3)−Φ(−1.2)=0.6179−0.1151=0.5028
(c) P(X>a)=0.3 → P(X<a)=0.7 → z=0.524
a=72+0.524×10=77.24
Q2 [4 marks] — X~N(μ,σ²). P(X<15)=0.25 and P(X<25)=0.75. Find μ and σ.
P(X<15)=0.25 → z₁=−0.674 (since Φ(0.674)=0.75)
P(X<25)=0.75 → z₂=0.674
(15−μ)/σ=−0.674 ...(1); (25−μ)/σ=0.674 ...(2)
Subtract: 10/σ=1.348 → σ=7.42
μ=(15+25)/2=20 (by symmetry, since z-scores are equal and opposite)
P(X<25)=0.75 → z₂=0.674
(15−μ)/σ=−0.674 ...(1); (25−μ)/σ=0.674 ...(2)
Subtract: 10/σ=1.348 → σ=7.42
μ=(15+25)/2=20 (by symmetry, since z-scores are equal and opposite)
Exam Tip: When the two probabilities add to 1 (0.25+0.75=1), the two x-values are symmetric about the mean. The mean is simply their average: μ=(15+25)/2=20. This quick check saves time.
Q3 [4 marks] — A sample of 50 is taken from a population with σ=12. The sample mean is 43.6. Test at 5% whether the population mean has increased from a claimed value of 41.
H₀: μ=41 H₁: μ>41 (one-tailed, upper). α=5%.
Under H₀: X̄~N(41, 144/50). SE=12/√50=1.697
Z=(43.6−41)/1.697=2.6/1.697=1.532
Critical value (5%, one-tail)=1.645. Z=1.532<1.645 → not in critical region.
Conclusion: Do not reject H₀. Insufficient evidence at 5% level that the mean has increased above 41.
Under H₀: X̄~N(41, 144/50). SE=12/√50=1.697
Z=(43.6−41)/1.697=2.6/1.697=1.532
Critical value (5%, one-tail)=1.645. Z=1.532<1.645 → not in critical region.
Conclusion: Do not reject H₀. Insufficient evidence at 5% level that the mean has increased above 41.
Q4 [4 marks] — A six-sided die is suspected to be biased towards 6. In 60 rolls, a 6 appears 16 times. Test at 5% significance whether the die is biased. Use the Normal approximation with continuity correction.
H₀: p=1/6 H₁: p>1/6. Under H₀: X~B(60,1/6).
np=10, nq=50 — Normal approx valid. Y~N(10, 60×1/6×5/6)=N(10, 25/3)
SE=√(25/3)=2.887
P(X≥16) ≈ P(Y>15.5)=(15.5−10)/2.887=5.5/2.887=1.905
P(Z>1.905)=1−Φ(1.905)=1−0.9717=0.0283<0.05
Reject H₀. Evidence at 5% that the die is biased towards 6.
np=10, nq=50 — Normal approx valid. Y~N(10, 60×1/6×5/6)=N(10, 25/3)
SE=√(25/3)=2.887
P(X≥16) ≈ P(Y>15.5)=(15.5−10)/2.887=5.5/2.887=1.905
P(Z>1.905)=1−Φ(1.905)=1−0.9717=0.0283<0.05
Reject H₀. Evidence at 5% that the die is biased towards 6.
Q5 [4 marks] — Records show that 30% of patients recover within a week. A new treatment is given to 15 patients and 8 recover within a week. (a) Test at 10% level whether the new treatment improves recovery. (b) What would be the conclusion at 5%?
H₀: p=0.3 H₁: p>0.3. X~B(15,0.3) under H₀.
P-value = P(X≥8) = 1−P(X≤7).
P(X≤7)=ΣP: using B(15,0.3): P(X≤7)≈0.9500
P-value=1−0.9500=0.0500
(a) p-value=0.05≤0.10 → Reject H₀ at 10%. Evidence that new treatment improves recovery.
(b) p-value=0.05≤0.05 → Also reject at 5% — borderline case.
P-value = P(X≥8) = 1−P(X≤7).
P(X≤7)=ΣP: using B(15,0.3): P(X≤7)≈0.9500
P-value=1−0.9500=0.0500
(a) p-value=0.05≤0.10 → Reject H₀ at 10%. Evidence that new treatment improves recovery.
(b) p-value=0.05≤0.05 → Also reject at 5% — borderline case.
Exam Tip: When the p-value equals exactly the significance level (p-value=α=0.05), reject H₀. The rejection condition is p-value ≤ α, so equality leads to rejection. In practice, Cambridge marks accept either conclusion with clear justification at the boundary.
Q6 [3 marks] — Explain the meaning of a Type I error in the context of Q5. State the probability of making a Type I error if the test is conducted at 5% significance.
A Type I error in this context means concluding that the new treatment improves the recovery rate (rejecting H₀) when in fact the recovery rate is still 30% (H₀ is true) — incorrectly concluding the treatment is effective when it is not.
The probability of a Type I error = the significance level = 0.05 (5%).
This is the probability of observing X≥8 when p=0.3 (i.e. getting a result this extreme by chance alone when the null hypothesis is true).
The probability of a Type I error = the significance level = 0.05 (5%).
This is the probability of observing X≥8 when p=0.3 (i.e. getting a result this extreme by chance alone when the null hypothesis is true).