IB Mathematics AHL 4.18: Test for proportion AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
A Poisson model requires random, independent events occurring at a constant average rate. The gardener might have observed:
– Snails are scattered randomly across the park without forming clusters.
– The number of snails increases proportionally with the area inspected, and occurrences appear independent.
Additionally, noting that the mean number of snails approximates the variance in counts could further support this model.

(b)
Calculate the mean for the \( 12 \, \text{m}^2 \) area: \( \lambda = 12 \times 0.2 = 2.4 \). This follows a Poisson distribution \( X \sim \text{Po}(2.4) \).
To find \( P(X = 4) \), use the Poisson formula: \( P(X = 4) = \frac{e^{-2.4} \times 2.4^4}{4!} \).
Compute step-by-step: \( 2.4^4 = 33.1776 \), \( 4! = 24 \), and \( e^{-2.4} \approx 0.0907 \).
Thus, \( P(X = 4) \approx \frac{0.0907 \times 33.1776}{24} \approx 0.1254 \).
Probability \( \approx \mathbf{0.125} \, (12.5\%) \).

(c)
For \( P(X < 3) \), compute the cumulative probability up to 2 snails, i.e., \( P(X \leq 2) \).
Using \( \lambda = 2.4 \), calculate \( P(X = 0) + P(X = 1) + P(X = 2) \):
\( P(X \leq 2) = e^{-2.4} \times (1 + 2.4 + \frac{2.4^2}{2}) \).
Here, \( 2.4^2 = 5.76 \), \( \frac{5.76}{2} = 2.88 \), so \( 1 + 2.4 + 2.88 = 6.28 \).
Then, \( P(X \leq 2) \approx 0.0907 \times 6.28 \approx 0.570 \).
Probability \( \approx \mathbf{0.570} \, (57.0\%) \).

(d)
First, find the probability of at least one snail in a single \( 12 \, \text{m}^2 \) inspection:
\( P(X \geq 1) = 1 – P(X = 0) = 1 – e^{-2.4} \).
With \( e^{-2.4} \approx 0.0907 \), \( P(X \geq 1) \approx 1 – 0.0907 = 0.9093 \).
For three independent inspections, the probability of at least one snail each time is \( (0.9093)^3 \).
Compute: \( 0.9093^2 \approx 0.8269 \), then \( 0.8269 \times 0.9093 \approx 0.752 \).
Probability \( \approx \mathbf{0.752} \, (75.2\%) \).

(e)
To test if the mean number of snails increased after rain, define the hypotheses:
– \( H_0: \mu = 2.4 \) (mean remains \( 2.4 \) per \( 12 \, \text{m}^2 \)).
– \( H_1: \mu > 2.4 \) (mean has increased due to rain).

(f)
For a \( 1\% \) significance level, determine the critical region where \( P(X \geq c) \leq 0.01 \) when \( \lambda = 2.4 \).
Calculate: \( P(X \geq 7) \approx 0.0116 \), \( P(X \geq 8) \approx 0.00334 \).
Since \( 0.00334 < 0.01 < 0.0116 \), the critical region is \( X \geq 8 \).

(g)
After heavy rain, the mean rises to \( 0.75 \) per \( \text{m}^2 \), so for \( 12 \, \text{m}^2 \), \( \lambda = 12 \times 0.75 = 9 \).
A Type II error occurs if \( H_0 \) is not rejected when \( H_1 \) is true, i.e., \( P(X \leq 7 \mid \lambda = 9) \).
Sum probabilities from 0 to 7: \( P(X \leq 7) = e^{-9} \times (1 + 9 + \frac{9^2}{2} + \frac{9^3}{6} + \frac{9^4}{24} + \frac{9^5}{120} + \frac{9^6}{720} + \frac{9^7}{5040}) \).
Approximate: \( e^{-9} \approx 0.000123 \), terms sum to about 6643, so \( 0.000123 \times 6643 \approx 0.324 \).
Probability \( \approx \mathbf{0.324} \, (32.4\%) \).