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

Question

Hank sets up a bird table in his garden to provide the local birds with some food. Hank notices that a specific bird, a large magpie, visits several times per month and he names him Bill. Hank models the number of times per month that Bill visits his garden as a Poisson distribution with mean \(3.1\).

(a) Using Hank’s model, find the probability that Bill visits the garden on exactly \(4\) occasions during one particular month.

(b) Over the course of \(3\) consecutive months, find the probability that Bill visits the garden:

(i) on exactly \(12\) occasions.

(ii) during the first and third month only.

(c) Find the probability that over a \(12\)-month period, there will be exactly \(3\) months when Bill does not visit the garden.

After the first year, a number of baby magpies start to visit Hank’s garden. It may be assumed that each of these baby magpies visits the garden randomly and independently, and that the number of times each baby magpie visits the garden per month is modelled by a Poisson distribution with mean \(2.1\).

(d) Determine the least number of magpies required, including Bill, in order that the probability of Hank’s garden having at least \(30\) magpie visits per month is greater than \(0.2\).

▶️ Answer/Explanation

Markscheme

a
\( X_1 \sim \text{Poisson}(3.1) \), \( P(X_1 = 4) = \frac{e^{-3.1} \times 3.1^4}{4!} \).
Calculate: \( 3.1^4 \approx 92.3521 \), \( 4! = 24 \), \( e^{-3.1} \approx 0.045049 \),
\( P(X_1 = 4) \approx \frac{0.045049 \times 92.3521}{24} \approx 0.1733 \).
Explanation:
Use the Poisson probability formula with mean \(3.1\).
Result:
\(0.173\) (\(0.173349\ldots\))

b (i)
Over \(3\) months, \( X_2 \sim \text{Poisson}(3 \times 3.1) = \text{Poisson}(9.3) \),
\( P(X_2 = 12) = \frac{e^{-9.3} \times 9.3^{12}}{12!} \).
Calculate: \( 9.3^{12} \approx 1809846.7 \), \( 12! = 479001600 \), \( e^{-9.3} \approx 0.00009957 \),
\( P(X_2 = 12) \approx \frac{0.00009957 \times 1809846.7}{479001600} \approx 0.0799 \).
Explanation:
Use the Poisson distribution for the total visits over \(3\) months.
Result:
\(0.0799\) (\(0.0798950\ldots\))

b (ii)
\( P(X_1 > 0) = 1 – P(X_1 = 0) = 1 – e^{-3.1} \approx 1 – 0.045049 = 0.954951 \),
\( P(X_1 = 0) = e^{-3.1} \approx 0.045049 \),
Probability for first and third month only = \( P(X_1 > 0)^2 \times P(X_1 = 0) \),
\( = 0.954951^2 \times 0.045049 \approx 0.91189 \times 0.045049 \approx 0.0411 \).
Explanation:
Calculate the probability for each month independently and combine.
Result:
\(0.0411\) (\(0.0410817\ldots\))

c
\( P(X_1 = 0) = e^{-3.1} \approx 0.045049 \),
Number of months with no visits ~ \( \text{Binomial}(12, 0.045049) \),
\( P(\text{exactly 3 months}) = \binom{12}{3} \times (0.045049)^3 \times (1 – 0.045049)^9 \),
\( \binom{12}{3} = 220 \), \( (0.045049)^3 \approx 0.00009138 \), \( (0.954951)^9 \approx 0.636 \),
\( P \approx 220 \times 0.00009138 \times 0.636 \approx 0.0133 \).
Explanation:
Use the binomial distribution for \(12\) months with the probability of no visits.
Result:
\(0.0133\) (\(0.013283\ldots\))

d
METHOD ONE
\[ \begin{array}{|c|c|c|} \hline n & \lambda & P(X \geq 30) \\ \hline \ldots & \ldots & \ldots \\ 10 & 24.1 & 0.136705 \\ 11 & 26.2 & 0.253384 \\ \hline \end{array} \]
Note: Award \(M1\) for evidence of a cumulative Poisson with \(\lambda\).
\(A1\) for \(0.136705\) and \(A1\) for \(0.253384\).
so require \(12\) magpies (including Bill)

METHOD TWO
evidence of a cumulative Poisson with \(\lambda = 3.1 + 2.1n\)
sketch of curve and \(y = 0.2\)
(intersect at) \(10.5810\ldots\)
rounding up gives \(n = 11\)
so require \(12\) magpies (including Bill)
Explanation:
Use cumulative Poisson probabilities and graphical intersection to determine the minimum number of magpies.
Result:
\(12\) magpies (including Bill)

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

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