| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Find minimum n for probability condition |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson approximation to binomial with standard probability calculations. Part (a) requires recognizing n=10000, p=1/75000 gives λ=10000/75000≈0.133, then computing P(X≥1)=1-P(X=0). Part (b) involves solving e^(-n/75000)>0.9 for n, requiring logarithms. Both parts are routine applications of a standard approximation technique 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 probabilities |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | M1 | SOI. Allow Po(0.133). |
| Answer | Marks | Guidance |
|---|---|---|
| P(X ⩾ 1) = 1 – e 15 | M1 | Allow incorrect λ allow one end error |
| = 0.125 (3 sf) | A1 | SC Partially unsupported final answer: |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(b) | n |
| Answer | Marks |
|---|---|
| 75000 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| e 75000 > 0.9 | M1 | Allow ‘=’ |
| Answer | Marks | Guidance |
|---|---|---|
| 75000 | M1 | Attempt ln both sides |
| Largest value of n is 7902 | A1 | CWO. Must be an integer. |
| Answer | Marks | Guidance |
|---|---|---|
| > 0.9 | M1 | Allow ‘=’ |
| –μ > ln 0.9 [μ < 0.10536] | M1 | Attempt ln both sides |
| n = μ × 75000 | B1 | |
| Largest value of n is 7902 | A1 | CWO. Must be an integer. |
| Answer | Marks |
|---|---|
| 75000 | B1 |
| Answer | Marks |
|---|---|
| 75000 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 75000 | M1 | Attempt ln or log both sides |
| Largest value of n is 7901 | A1 | CWO Must be an integer |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | 2
Po
15 | M1 | SOI. Allow Po(0.133).
−2
P(X ⩾ 1) = 1 – e 15 | M1 | Allow incorrect λ allow one end error
= 0.125 (3 sf) | A1 | SC Partially unsupported final answer:
2
Po stated B1 then unsupported 0.125 B1
15
SC Use of Binomial (0.1248) B1 only
Use of Normal scores M0
3
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | n
λ =
75000 | B1
− n
e 75000 > 0.9 | M1 | Allow ‘=’
Allow incorrect λ
n
– > ln 0.9 [n < 7902.04]
75000 | M1 | Attempt ln both sides
Largest value of n is 7902 | A1 | CWO. Must be an integer.
Alternative method for Question 5(b)
e-μ
> 0.9 | M1 | Allow ‘=’
–μ > ln 0.9 [μ < 0.10536] | M1 | Attempt ln both sides
n = μ × 75000 | B1
Largest value of n is 7902 | A1 | CWO. Must be an integer.
Alternative method for Question 5(b)
74999
75000 | B1
n
74999
> 0.9
75000 | M1
74999
nln > ln 0.9
75000 | M1 | Attempt ln or log both sides
Largest value of n is 7901 | A1 | CWO Must be an integer
4
Question | Answer | Marks | GuidanceOn average, 1 in 75000 adults has a certain genetic disorder.
\begin{enumerate}[label=(\alph*)]
\item Use a suitable approximating distribution to find the probability that, in a random sample of 10000 people, at least 1 has the genetic disorder. [3]
\item In a random sample of $n$ people, where $n$ is large, the probability that no-one has the genetic disorder is more than 0.9.
Find the largest possible value of $n$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q5 [7]}}