| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Scaled time period sums |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties requiring scaling parameters for different time periods and basic probability calculations. Part (a) involves simple scaling and complement probability, part (b) uses the sum of independent Poisson variables, and part (c) requires solving P(X≥1)=0.9 using logarithms. All techniques are standard for S2 level with no novel insight required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks |
|---|---|
| 6(a) | [λ = 2.7] 1 − e−2.7(1 + 2.7 + 2.72 ) or 1 − e−2.7(1 + 2.7 +3.645) |
| Answer | Marks | Guidance |
|---|---|---|
| or 1- ( 0.06721 + 0.1815 + 0.2450 ) | M1 | Any λ. Allow one end error. |
| Answer | Marks | Guidance |
|---|---|---|
| = 0.506 (3 sf) | A1 | SC unsupported answer 0.506 scores B1. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | λ = 1.95 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| or 0.1423 + 0.2774+ 0.2705 + 0.1758 | M1 | Any λ. Allow one end error. |
| Answer | Marks | Guidance |
|---|---|---|
| = 0.866 | A1 | SC unsupported answer 0.866 scores B1B1. |
| Answer | Marks | Guidance |
|---|---|---|
| 6(c) | 1 – e-2.1x ⩾ 0.90 or 1 – e−λ ⩾ 0.90 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| −2.1x < ln0.1 or −λ < ln0.1 [ λ > 2.3026, 2.3026/2.1 ] | M1 | Rearrange and attempt take logs of relevant form. |
| 1.096 or 10.96 accept 1.097 or 10.97 | *A1 | Seen. |
| She must wait for at least 11 minutes | A1 dep |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(a) ---
6(a) | [λ = 2.7] 1 − e−2.7(1 + 2.7 + 2.72 ) or 1 − e−2.7(1 + 2.7 +3.645)
2
or 1- ( 0.06721 + 0.1815 + 0.2450 ) | M1 | Any λ. Allow one end error.
Must see expression.
= 0.506 (3 sf) | A1 | SC unsupported answer 0.506 scores B1.
2
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | λ = 1.95 | B1
e−1.95(1 + 1.95 + 1.952 + 1.953 ) or e−1.95(1 + 1.95 + 1.90125 +1.2358)
2 3!
or 0.1423 + 0.2774+ 0.2705 + 0.1758 | M1 | Any λ. Allow one end error.
Must see expression.
= 0.866 | A1 | SC unsupported answer 0.866 scores B1B1.
3
--- 6(c) ---
6(c) | 1 – e-2.1x ⩾ 0.90 or 1 – e−λ ⩾ 0.90 | M1 | OE Condone use of ‘=’ throughout.
[e−2.1x < 0.1] or e−λ < 0.1
−2.1x < ln0.1 or −λ < ln0.1 [ λ > 2.3026, 2.3026/2.1 ] | M1 | Rearrange and attempt take logs of relevant form.
1.096 or 10.96 accept 1.097 or 10.97 | *A1 | Seen.
She must wait for at least 11 minutes | A1 dep
SC Use of trial and improvement.
Use of 1–e-λ any numerical λ (not 2.1) ie one trial M1.
Use of enough trials to give an answer of 0.90 (2sf) M1.
λ=2.30 i.e. 3sf accuracy AND 1.09… or 10.9 … A1.
Then 11 A1 dep.
4
Question | Answer | Marks | GuidanceThe numbers of customers arriving at service desks $A$ and $B$ during a $10$-minute period have the independent distributions $\text{Po}(1.8)$ and $\text{Po}(2.1)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that during a randomly chosen $15$-minute period more than $2$ customers will arrive at desk $A$. [2]
\item Find the probability that during a randomly chosen $5$-minute period the total number of customers arriving at both desks is less than $4$. [3]
\item An inspector waits at desk $B$. She wants to wait long enough to be $90\%$ certain of seeing at least one customer arrive at the desk.
Find the minimum time for which she should wait, giving your answer correct to the nearest minute. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q6 [9]}}