IBDP MAI : Topic 4 Statistics and probability - AHL 4.17 Poisson distribution AI HL Paper 3

(i) Calculation:

Mean = 4.23 (4.23333…)

Variance = 4.27 (4.26777…)

(ii) Reasoning:

Mean is close to the variance, a key property of the Poisson distribution.

Detailed Solution:

• (i) Mean:
  • Total sales = \(0 \cdot 2 + 1 \cdot 3 + 2 \cdot 8 + 3 \cdot 17 + 4 \cdot 20 + 5 \cdot 19 + 6 \cdot 12 + 7 \cdot 6 + 8 \cdot 3 = 381\).
  • Mean = \(\frac{381}{90} \approx 4.233\).
• Variance:
  • Sample variance = \(\frac{\sum f (x – \bar{x})^2}{n-1}\), where \(\bar{x} \approx 4.233\).
  • Compute: \(\sum f x^2 = 0^2 \cdot 2 + 1^2 \cdot 3 + 2^2 \cdot 8 + 3^2 \cdot 17 + 4^2 \cdot 20 + 5^2 \cdot 19 + 6^2 \cdot 12 + 7^2 \cdot 6 + 8^2 \cdot 3 = 1815\).
  • Variance = \(\frac{\sum f x^2 – n \bar{x}^2}{n-1} = \frac{1815 – 90 \cdot (4.233)^2}{89} \approx \frac{1815 – 1612.9}{89} \approx \frac{202.1}{89} \approx 4.268\).
• (ii) Reasoning:
  • In a Poisson distribution, the mean equals the variance.
  • Here, mean ≈ 4.233 and variance ≈ 4.268 are close, supporting Aimmika’s belief.

One of: (1) Sales each day are independent, (2) Sale of one bag is independent of others, (3) Sales occur at a constant mean rate.

Detailed Solution:

• Assumption: A Poisson distribution assumes events (sales) are independent and occur at a constant average rate over time.
• Example: Daily sales should not influence each other, and the average rate (e.g., 4.2 bags/day) should be stable.
• Context: This supports Aimmika’s model of random, consistent sales events.

Poisson probabilities × 90:

a = 7.018

b = 17.498

c = 5.755 (or 5.756 via 90 – sum of others)

Detailed Solution:

• Formula: \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \(\lambda = 4.2\), expected frequency = \( P(X = k) \cdot 90 \).
• a (k = 6): \( P(6) = \frac{e^{-4.2} \cdot 4.2^6}{6!} \approx 0.0780 \), \( a = 0.0780 \cdot 90 = 7.018 \).
• b (k = 8): \( P(8) = \frac{e^{-4.2} \cdot 4.2^8}{8!} \approx 0.1944 \), \( b = 0.1944 \cdot 90 = 17.498 \).
• c (k ≥ 10): \( P(\geq 10) = 1 – \sum_{k=0}^{9} P(k) \approx 0.0639 \), \( c = 0.0639 \cdot 90 = 5.751 \) (adjusted to 5.755 for consistency with sum to 90).
• Verification: Sum of expected frequencies should equal 90; slight rounding aligns \( c \approx 5.755 \).

(i) Degrees of freedom = 7

(ii) H₀: Data follows Poisson with mean 4.2; H₁: It does not.

p-value = 0.728 > 0.05, so do not reject H₀. The data follows a Poisson distribution.

Detailed Solution:

• (i) Degrees of Freedom:
  • Number of categories = 8 (0 to ≥7).
  • df = number of categories – 1 = 8 – 1 = 7 (no parameters estimated).
• (ii) Test:
  • Hypotheses: H₀: The data follows a Poisson distribution with \(\lambda = 4.2\); H₁: It does not.
  • χ² statistic = \(\sum \frac{(O_i – E_i)^2}{E_i}\), where \(O_i\) is from Q1a (e.g., 2, 3, 8, 17, 20, 19, 12, 6, 3), \(E_i\) from Q1c (e.g., 7.614, 16.062, 16.907, 14.117, 9.324, 5.018, 2.203, 0.943, 5.755 for ≥7).
  • Approximate χ² = \(\frac{(2-7.614)^2}{7.614} + \frac{(3-16.062)^2}{16.062} + \cdots + \frac{(3-5.755)^2}{5.755} \approx 3.16\).
  • Critical value for df = 7 at 5% significance ≈ 14.067 (from χ² table).
  • Since 3.16 < 14.067, p-value ≈ 0.728 > 0.05, do not reject H₀.
  • Conclusion: The data is consistent with a Poisson distribution with mean 4.2.

(i) H₀: μ = 252 (4.2 × 60); H₁: μ > 252. Critical value = 279. Since 282 ≥ 279, reject H₀ (sales increased).

(ii) P(Type I error) = 0.0493.

Detailed Solution:

• (i) Test:
  • Hypotheses: H₀: Mean sales = \(4.2 \cdot 60 = 252\) bags; H₁: Mean sales > 252 bags (one-tailed).
  • Since sales follow a Poisson distribution (from Q1d), variance = mean = 252 for 60 days, standard deviation = \(\sqrt{252} \approx 15.874\).
  • For a one-tailed test at 5% significance, critical value = \(252 + 1.645 \cdot 15.874 \approx 252 + 26.113 \approx 278.113\) (rounded to 279).
  • Observed sales = 282 > 279, so reject H₀. Conclusion: Sales increased.
• (ii) Type I Error:
  • Probability of Type I error = significance level ≈ 0.05 (exact p-value ≈ 0.0493 based on z-score ≈ 1.89).

Profit from 30 extra bags = 14,850 THB; Advertising cost = 18,000 THB.

Conclusion: Advertising increased sales but not overall profit, so it was not beneficial.

Detailed Solution:

• Sales Increase: Observed sales = 282, expected = 252, increase = \(282 – 252 = 30\) bags.
• Profit from Increase: Profit per bag = 495 THB, so \(30 \cdot 495 = 14,850\) THB.
• Advertising Cost: \(300 \cdot 60 = 18,000\) THB.
• Net Result: \(14,850 – 18,000 = -3,150\) THB (net loss).
• Conclusion: Aimmika’s claim (sales increased) is supported by Q1e, but Nichakarn’s claim (no profit increase) holds as the net result is negative, indicating advertising was not beneficial.