Conditional Probability, Normal Approximation & PMCC Hypothesis Testing

1

Conditional Probability, Independence, and Mutually Exclusive Events

The conditional probability of event A occurring given that event B has already occurred is P(A | B) = P(A ∩ B) / P(B) (which rearranges to the multiplication rule P(A ∩ B) = P(A | B) P(B)). For any two events, the addition law is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Carefully distinguish the two special cases: 1. Mutually exclusive events cannot happen together: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B). 2. Independent events do not affect each other's probability: P(A | B) = P(A), which is equivalent to P(A ∩ B) = P(A) × P(B).

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2

Venn Diagrams and Tree Diagrams with Conditional Probability

When constructing a 3-circle Venn diagram for events A, B, and C, always start from the innermost central intersection P(A ∩ B ∩ C) and work outwards by subtracting overlaps so every region sums to 1. To find P(A | B') from a Venn diagram, restrict your sample space to the regions outside circle B (denominator = P(B') = 1 - P(B)) and put the region inside A but outside B in the numerator (P(A ∩ B')). In a tree diagram (such as medical diagnostic testing for a disease D), P(+ | D) and P(- | D') are the second-branch conditional probabilities; to find the 'reversed' conditional probability P(D | +) (Bayes' theorem), divide the 'Disease and Positive' branch product by the sum of all 'Positive' branches.

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3

The Normal Distribution and Standardising (Z-Scores)

A continuous random variable X ~ N(μ, σ²) has a bell-shaped probability density curve symmetrical about the mean μ, with points of inflection at x = μ - σ and x = μ + σ, and total area under the curve equal to 1.Note that for any continuous random variable, P(X = a) = 0, so P(X < a) = P(X ≤ a). Whenever μ or σ is unknown, you must standardise to the Standard Normal distribution Z ~ N(0, 1²) using Z = (X - μ) / σ, use your calculator's Inverse Normal function (with area to the left) to find the z-value, and solve simultaneous linear equations in μ and σ.

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4

Normal Approximation to the Binomial Distribution

If X ~ B(n, p) has a large number of trials n and a probability of success p close to 0.5 (so the distribution is roughly symmetrical, typically np > 5 and n(1 - p) > 5), calculating binomial probabilities directly can be approximated by a continuous Normal random variable Y ~ N(μ, σ²) with mean μ = np and variance σ² = np(1 - p) (so standard deviation σ = √[np(1 - p)]). Because X is discrete (integers only) and Y is continuous, you must apply a ±0.5 Continuity Correction: 1. P(X ≤ 12) becomes P(Y < 12.5); 2. P(X < 12) = P(X ≤ 11) becomes P(Y < 11.5); 3. P(X ≥ 12) becomes P(Y > 11.5); 4. P(X = 12) becomes P(11.5 < Y < 12.5).

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5

Product moment correlation coefficient (PMCC, r)

The Product Moment Correlation Coefficient r measures the strength and direction of the linear correlation between two variables x and y, always satisfying -1 ≤ r ≤ +1. A value of r = +1 indicates perfect positive linear correlation, r = -1 indicates perfect negative linear correlation, and r ≈ 0 indicates no linear correlation (though there could still be a strong non-linear curve). In particular, linear coding (adding constants or multiplying by positive scale factors, e.g., p = ax + b and q = cy + d with a, c > 0) has zero effect on r!

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6

Hypothesis Testing for Zero Correlation (ρ = 0)

To test whether a sample of n paired observations with sample PMCC r provides evidence of correlation in the underlying population (whose true correlation coefficient is denoted by the Greek letter ρ, never r!): 1. State the hypotheses in terms of ρ: Null hypothesis H₀: ρ = 0; Alternative hypothesis H₁: ρ > 0 (one-tailed positive correlation), H₁: ρ < 0 (one-tailed negative correlation), or H₁: ρ ≠ 0 (two-tailed correlation). 2. Look up the critical value c in the PMCC table for sample size n and significance level α (halving α in each tail if two-tailed). 3. Compare |r| with |c|: if |r| > |c| (r lies in the critical region), reject H₀ and conclude in context that there is sufficient evidence of correlation.

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1. Q1

   For two events A and B, P(A) = 0.45, P(B) = 0.40, and P(A | B) = 0.60. (a) Find P(A ∩ B), P(A ∪ B), and P(B | A'). (b) State with a numerical reason whether events A and B are independent. (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. (a) Use the conditional probability formula P(A ∩ B) = P(A | B) × P(B) = 0.60 × 0.40 = 0.24 [1 mark]. 2. Use the addition rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.45 + 0.40 - 0.24 = 0.61 [1 mark]. 3. Find P(A' ∩ B) (the region inside B but outside A): P(A' ∩ B) = P(B) - P(A ∩ B) = 0.40 - 0.24 = 0.16, and find P(A') = 1 - 0.45 = 0.55 [1 mark]. 4. Calculate P(B | A') = P(A' ∩ B) ÷ P(A') = 0.16 ÷ 0.55 = 16 / 55 (≈ 0.291) [1 mark]. 5. (b) Compare P(A | B) = 0.60 with P(A) = 0.45 (or compare P(A) × P(B) = 0.45 × 0.40 = 0.18 with P(A ∩ B) = 0.24) [1 mark]. 6. Since P(A | B) ≠ P(A) (0.60 ≠ 0.45), events A and B are not independent [1 mark].

2. Q2

   The masses X grams of bags of coffee produced by a machine are normally distributed with mean μ and standard deviation σ, so X ~ N(μ, σ²). Given that 10% of bags have a mass less than 243.6 g and 5% of bags have a mass greater than 258.2 g, find the values of μ and σ to 1 decimal place, and find the probability that out of 3 randomly chosen bags, all 3 have a mass greater than 250 g. (7 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. From P(X < 243.6) = 0.10, use Inverse Normal on Z ~ N(0, 1) with left-tail area 0.10 to get z₁ = -1.2816 [1 mark]. 2. From P(X > 258.2) = 0.05 → P(X < 258.2) = 0.95, use Inverse Normal with left-tail area 0.95 to get z₂ = +1.6449 [1 mark]. 3. Form two simultaneous linear equations using X = μ + zσ: μ - 1.2816σ = 243.6 (Equation 1) and μ + 1.6449σ = 258.2 (Equation 2) [1 mark]. 4. Subtract Equation 1 from Equation 2: 2.9265σ = 14.6 → σ = 4.99 g = 5.0 g (1 d.p) [1 mark]. 5. Substitute σ = 4.989 into Equation 1 to find μ: μ = 243.6 + 1.2816(4.989) = 250.0 g [1 mark]. 6. Since μ = 250.0 g, by symmetry of the Normal distribution P(X > 250.0) = 0.5 [1 mark]. 7. For 3 independent bags all exceeding 250 g: Probability = (0.5)³ = 0.125 (or 1/8) [1 mark].

3. Q3

   In a large city, 40% of commuters take the train to work. A random sample of 150 commuters is selected, and X is the number in the sample who take the train. (a) State the exact distribution of X and explain why a Normal approximation is appropriate. (b) Use a suitable Normal approximation (including a continuity correction) to estimate P(50 ≤ X < 68). (6 marks)

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   Click “Check Mark Scheme” to compare your answer, 1. (a) Exact distribution is X ~ B(150, 0.40) [1 mark]. 2. A Normal approximation is valid because n = 150 is large and p = 0.40 is close to 0.5 (with np = 60 > 5 and n(1 - p) = 90 > 5) [1 mark]. 3. (b) Calculate the mean and variance of the approximating Normal distribution Y ~ N(μ, σ²): μ = np = 150 × 0.40 = 60, and σ² = np(1 - p) = 60 × 0.60 = 36 (so σ = √36 = 6), giving Y ~ N(60, 6²) [2 marks]. 4. Apply the continuity correction to the discrete interval 50 ≤ X < 68 (which is 50 ≤ X ≤ 67): P(50 ≤ X ≤ 67) ≈ P(49.5 < Y < 67.5) [1 mark]. 5. Evaluate P(49.5 < Y < 67.5) for Y ~ N(60, 6²) using the calculator's Normal CD function: P(49.5 < Y < 67.5) = 0.8543 (or 0.854 to 3 s.f) [1 mark].

4. Q4

   A meteorologist investigates whether there is a negative correlation between altitude h metres and daily maximum temperature T °C across n = 12 weather stations. She calculates a Product Moment Correlation Coefficient of r = -0.542. Given that for n = 12, the one-tailed 5% critical value for the PMCC is 0.4973 and the one-tailed 2.5% critical value is 0.5760, test at the 5% level of significance whether there is evidence of negative correlation between altitude and daily maximum temperature. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. State the null and one-tailed alternative hypotheses using the population parameter ρ: H₀: ρ = 0 and H₁: ρ < 0 (one-tailed test at α = 0.05) [1 mark]. 2. State the critical value for n = 12 at the 5% one-tailed level in the negative tail: critical value = -0.4973 (critical region is r < -0.4973) [1 mark]. 3. Compare the observed test statistic with the critical value: since -0.542 < -0.4973 (or |-0.542| > 0.4973), r lies inside the critical region, so we reject H₀ [1 mark]. 4. Write a contextual conclusion: There is sufficient evidence at the 5% significance level to support the meteorologist's claim that there is a negative correlation between altitude and daily maximum temperature [1 mark].

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