| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - large lambda direct |
| Difficulty | Moderate -0.3 This is a standard S2 Poisson distribution question covering routine bookwork (stating conditions), straightforward probability calculations using tables, and a normal approximation with continuity correction. While part (iii)(a) requires knowing when to apply the approximation (λ>15 rule), all techniques are textbook exercises with no novel problem-solving required. Slightly easier than average due to the structured guidance and standard content. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Cars pass independently of one another and at constant average rate | B1 | "Independently", refer to cars. |
| B1 | Not "constant rate", "constant probability". No extra conditions. (2 marks) | |
| Ignore all references to "singly" (which is \(\textit{wrong}\) in this context!) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(< 7) - P(< 3) = 0.6728 - 0.1118 = \mathbf{0.561(0)}\) | M1 | 0.680 or 0.681: M1A0. Allow from calculator, no working |
| A2 | 0.4491 or 0.5679: M1A1. Allow from calculator, no working | |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(4) + P(5) + P(6) + P(7) = 0.1118 + 0.1454 + 0.1575 + 0.1462 = \mathbf{0.561(0)}\) | M1 | Correct formula for \(\geq 3\) probabilities from Po(6.5) added, can be implied |
| A1 | 3, 4 or 5 correct terms (e.g. \(P(3) = 0.06880\)), can be algebraic or implied | |
| (3 marks) | Answer, a.r.t. 0.561 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Po}(30) = N(30, 30)\) | M1 | Normal, mean 30, stated or implied |
| \(1 - \Phi\left(\frac{35.5 - 30}{\sqrt{30}}\right) = 1 - \Phi(1.004)\) | A1 | Variance or SD 30 |
| M1 | Standardise, their \(\lambda\), \(\lambda\), allow wrong/no cc, var/SD errors | |
| A1 | cc, \(\sqrt{\lambda}\) correct | |
| \(= 1 - 0.8422 = \mathbf{0.1578}\) | A1 | Answer, a.r.t. 0.158 [NB: 0.157 may be from exact. See below] |
| B1 | Or "\(\lambda\) large", etc., but no other conditions. (6 marks) | |
| Normal suitable as \(30 > 15\) | ||
| If numerical comparison, must involve 15. | ||
| SC: Exact Poisson, 0.1574, max B1 | ||
| SC: Po(30), N(15, 15): M0B1 M1A1A0 B1, max 4/6 |
| Answer | Marks | Guidance |
|---|---|---|
| Cars do not pass independently/randomly, as one may be immediately followed by another | B1 | Any plausible relevant explanation in context, needn't be connected to (1 mark) |
| Allow explanations that might also hold for smaller \(\lambda\) | ||
| Do not allow comment on size of \(\lambda\) unless explained in valid way, e.g. "\(\lambda\) too large so cars follow one another", but not "\(\lambda\) too large for Poisson" |
## (i)
Cars pass independently of one another and at constant average rate | B1 | "Independently", refer to cars.
| B1 | Not "constant rate", "constant probability". No extra conditions. (2 marks)
| | Ignore all references to "singly" (which is $\textit{wrong}$ in this context!)
## (ii) $\alpha$
$P(< 7) - P(< 3) = 0.6728 - 0.1118 = \mathbf{0.561(0)}$ | M1 | 0.680 or 0.681: M1A0. Allow from calculator, no working
| A2 | 0.4491 or 0.5679: M1A1. Allow from calculator, no working
| | | (3 marks)
## or $\beta$
$P(4) + P(5) + P(6) + P(7) = 0.1118 + 0.1454 + 0.1575 + 0.1462 = \mathbf{0.561(0)}$ | M1 | Correct formula for $\geq 3$ probabilities from Po(6.5) added, can be implied
| A1 | 3, 4 or 5 correct terms (e.g. $P(3) = 0.06880$), can be algebraic or implied
| | (3 marks) | Answer, a.r.t. 0.561
## (iii)(a)
$\text{Po}(30) = N(30, 30)$ | M1 | Normal, mean 30, stated or implied
$1 - \Phi\left(\frac{35.5 - 30}{\sqrt{30}}\right) = 1 - \Phi(1.004)$ | A1 | Variance or SD 30
| M1 | Standardise, their $\lambda$, $\lambda$, allow wrong/no cc, var/SD errors
| A1 | cc, $\sqrt{\lambda}$ correct
$= 1 - 0.8422 = \mathbf{0.1578}$ | A1 | Answer, a.r.t. 0.158 [NB: 0.157 may be from exact. See below]
| B1 | Or "$\lambda$ large", etc., but no other conditions. (6 marks)
| | Normal suitable as $30 > 15$
| | If numerical comparison, must involve 15.
| | SC: Exact Poisson, 0.1574, max B1
| | SC: Po(30), N(15, 15): M0B1 M1A1A0 B1, max 4/6
## (b)
Cars do not pass independently/randomly, as one may be immediately followed by another | B1 | Any plausible relevant explanation in context, needn't be connected to (1 mark) | conditions, e.g. "steady stream". $\textit{Not}$ "several cars might pass at once".
| | Allow explanations that might also hold for smaller $\lambda$
| | Do not allow comment on size of $\lambda$ unless explained in valid way, e.g. "$\lambda$ too large so cars follow one another", but not "$\lambda$ too large for Poisson"The number of cars passing a point on a single-track one-way road during a one-minute period is denoted by $X$. Cars pass the point at random intervals and the expected value of $X$ is denoted by $\lambda$.
\begin{enumerate}[label=(\roman*)]
\item State, in the context of the question, two conditions needed for $X$ to be well modelled by a Poisson distribution. [2]
\item At a quiet time of the day, $\lambda = 6.50$. Assuming that a Poisson distribution is valid, calculate P$(4 \leq X < 8)$. [3]
\item At a busy time of the day, $\lambda = 30$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that a Poisson distribution is valid, use a suitable approximation to find P$(X > 35)$. Justify your approximation. [6]
\item Give a reason why a Poisson distribution might not be valid in this context when $\lambda = 30$. [1]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2016 Q6 [12]}}