IB Mathematics AHL 4.17 Poisson distribution -AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
Total weeks \(n = 50\). Mean \(\mu = 1.76\).
\(s^2 = \frac{\sum f(x – \bar{x})^2}{n – 1} = \frac{83.12}{49} \approx 1.69632\).
Unbiased variance estimate = \(1.70\) (3 s.f.).

(b)
In a Poisson distribution, the mean and variance are equal. Here, the mean (\(1.76\)) and unbiased variance (\(1.70\)) are very similar, suggesting the model is appropriate.

(c)
(i) \(P(X=4) = \frac{e^{-1.76} \times 1.76^4}{4!} \approx 0.06879\). Expected \(j = 50 \times 0.06879 = \mathbf{3.44}\).
(ii) \(P(X \geq 5) = 1 – P(X \leq 4) \approx 0.0319\). Expected \(k = 50 \times 0.0319 = \mathbf{1.68}\).

(d)
The test requires that all expected frequencies are at least \(5\)[cite: 1415]. Since \(j < 5\) and \(k < 5\), these categories must be combined.

(e)
Degrees of freedom \(df = (\text{number of cells}) – 1 – (\text{parameters estimated})\).
Cells after combining (\(0, 1, 2, 3, \geq 4\)) \(= 5\). Estimated \(\lambda = 1\).
\(df = 5 – 1 – 1 = \mathbf{3}\).

(f)
Using technology with combined observed values \(\{8, 16, 13, 8, 5\}\) and expected values \(\{8.62, 15.17, 13.35, 7.83, 5.03\}\):
\(p\)-value \(\approx 0.991\).

(g)
Since \(0.991 > 0.05\), do not reject the null hypothesis. There is insufficient evidence to suggest the data does not follow a Poisson distribution.