| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Complementary event Poisson approximation |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson approximation to binomial with clearly stated parameters (n=94, p=0.1), requiring routine calculations of P(X=4), P(X≥4), and binomial probability raised to a power. Part (v) involves normal approximation which is also standard. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02c Linear coding: effects on mean and variance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Binomial (94,0.1) | B1 for binomial, B1 dep for parameters | 2 |
| (ii) \(n\) is large and \(p\) is small | B1, B1 Allow appropriate numerical ranges | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (B) Using tables: \(P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.0160 = 0.9840\) cao | M1 for calculation or use of tables, A1 for attempt to find \(P(X \geq 4)\), A1 cao | 5 |
| (iv) \(P(\text{sufficient rooms throughout August}) = 0.9840^{31} = 0.6065\) | M1, A1 FT | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Binomial (2914,0.1) | B1 for binomial, B1 dep. for parameters | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 300.5) = P\left(Z \leq \frac{300.5 - 291.4}{\sqrt{262.26}}\right) = P(Z \leq 0.5619) = \Phi(0.5619) = 0.7130\) | B1, B1, B1 for continuity corr., M1 for probability using correct tail, A1 cao (but FT wrong or omitted CC) | 5 |
**(i)** Binomial (94,0.1) | B1 for binomial, B1 dep for parameters | 2
**(ii)** $n$ is large and $p$ is small | B1, B1 Allow appropriate numerical ranges | 2
**(iii)** $\lambda = 94 \times 0.1 = 9.4$
**(A)** $P(X = 4) = e^{-9.4} \frac{9.4^4}{4!} = 0.0269$ (3 s.f.)
or from tables: 0.0429 - 0.0160 = 0.0269 cao
**(B)** Using tables: $P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.0160 = 0.9840$ cao | M1 for calculation or use of tables, A1 for attempt to find $P(X \geq 4)$, A1 cao | 5
**(iv)** $P(\text{sufficient rooms throughout August}) = 0.9840^{31} = 0.6065$ | M1, A1 FT | 2
**(v)** **(A)** $31 \times 94 = 2914$
Binomial (2914,0.1) | B1 for binomial, B1 dep. for parameters | 2
**(B)** Use Normal approx with
$\mu = np = 2914 \times 0.1 = 291.4$
$\sigma^2 = npq = 2914 \times 0.1 \times 0.9 = 262.26$
$P(X \leq 300.5) = P\left(Z \leq \frac{300.5 - 291.4}{\sqrt{262.26}}\right) = P(Z \leq 0.5619) = \Phi(0.5619) = 0.7130$ | B1, B1, B1 for continuity corr., M1 for probability using correct tail, A1 cao (but FT wrong or omitted CC) | 5
---2 A large hotel has 90 bedrooms. Sometimes a guest makes a booking for a room, but then does not arrive. This is called a 'no-show'. On average $10 \%$ of bookings are no-shows. The hotel manager accepts up to 94 bookings before saying that the hotel is full. If at least 4 of these bookings are no-shows then there will be enough rooms for all of the guests. 94 bookings have been made for each night in August. You should assume that all bookings are independent.
\begin{enumerate}[label=(\roman*)]
\item State the distribution of the number of no-shows on one night in August.
\item State the conditions under which the use of a Poisson distribution is appropriate as an approximation to a binomial distribution.
\item Use a Poisson approximating distribution to find the probability that, on one night in August,\\
(A) there are exactly 4 no-shows,\\
(B) there are enough rooms for all of the guests who do arrive.
\item Find the probability that, on all of the 31 nights in August, there are enough rooms for all of the guests who arrive.
\item (A) In August there are $31 \times 94 = 2914$ bookings altogether. State the exact distribution of the total number of no-shows during August.\\
(B) Use a suitable approximating distribution to find the probability that there are at most 300 no-shows altogether during August.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2008 Q2 [18]}}