OCR S2 2012 June — Question 4 11 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2012
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle period normal approximation - scaled period (normal approximation only)
DifficultyModerate -0.3 This is a straightforward application of standard Poisson distribution techniques. Part (i) requires recalling a textbook condition (independence/randomness), parts (ii)-(iii) are direct calculator work with given parameters, and part (iv) is a routine normal approximation. All steps are standard S2 material with no problem-solving or novel insight required.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)

4 In a rock, small crystal formations occur at a constant average rate of 3.2 per cubic metre.
  1. State a further assumption needed to model the number of crystal formations in a fixed volume of rock by a Poisson distribution. In the remainder of the question, you should assume that a Poisson model is appropriate.
  2. Calculate the probability that in one cubic metre of rock there are exactly 5 crystal formations.
  3. Calculate the probability that in 0.74 cubic metres of rock there are at least 3 crystal formations.
  4. Use a suitable approximation to calculate the probability that in 10 cubic metres of rock there are at least 36 crystal formations.

Question 4(i):
AnswerMarks Guidance
AnswerMarks Guidance
Crystals must occur independently of one anotherB1 Allow interpreted, or "randomly" but nothing else. Must be contextualised; no other answers included. Ignore "singly". But allow "probability… is independent"
[1]
Question 4(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(e^{-3.2}\dfrac{3.2^5}{5!} = \mathbf{0.114(0)}\)M1 Formula, or .0608 or .1781 or .1075 or .1203 (tables)
A1Answer a.r.t. 0.114, implies both marks
[2]
Question 4(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(Po(2.368)\)M1 \(Po(0.74 \times 3.2)\) stated or implied. Allow for \(0.75 \times 3.2\) etc, e.g. \(Po(2.4)\)
\(1 - e^{-2.368}\!\left(1 + 2.368 + \dfrac{2.368^2}{2}\right)\)M1 \(1 -\) correct Poisson terms, their \(\lambda\), allow \(\pm 1\) term. Don't allow second M1 from \(\lambda\) in tables, e.g. if MR, treat as E–1.
\(= \mathbf{0.4219}\)A1 Answer, a.r.t. 0.422, implies all 3 marks. If no working: don't give M1A0
[3]
Question 4(iv):
AnswerMarks Guidance
AnswerMarks Guidance
\(Po(32) \approx N(32, 32)\)M1 \(N(\lambda, \lambda)\) stated or implied, allow \(\sqrt{\lambda}\) or \(\lambda^2\) for var. Needs \(\lambda \geq 15\)
A1\(N(32, 32)\), allow \(\sqrt{32}\) or \(32^2\) for var
\(1 - \Phi\!\left(\dfrac{35.5 - 32}{\sqrt{32}}\right)\)M1 Standardise with \(\lambda\) and \(\sqrt{\lambda}\) or \(\lambda\), allow cc errors but not \(\sqrt{n}\); both cc and \(\sqrt{}\) correct. Can get (M0A0) M1A1 from \(\lambda < 15\). Typically, no cc \(\to 0.2203\), or \(32 \to 0.4565\), 3/5 (but needs evidence, not just answer)
A1both cc and \(\sqrt{}\) correct
\(= 1 - \Phi(.619) = \mathbf{0.2681}\)A1 Final answer, a.r.t. 0.268
[5]
## Question 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Crystals must occur independently of one another | B1 | Allow interpreted, or "randomly" but nothing else. Must be contextualised; no other answers included. Ignore "singly". But allow "probability… is independent" |
| | **[1]** | |

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## Question 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-3.2}\dfrac{3.2^5}{5!} = \mathbf{0.114(0)}$ | M1 | Formula, or .0608 or .1781 or .1075 or .1203 (tables) |
| | A1 | Answer a.r.t. 0.114, implies both marks |
| | **[2]** | |

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## Question 4(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $Po(2.368)$ | M1 | $Po(0.74 \times 3.2)$ stated or implied. Allow for $0.75 \times 3.2$ etc, e.g. $Po(2.4)$ |
| $1 - e^{-2.368}\!\left(1 + 2.368 + \dfrac{2.368^2}{2}\right)$ | M1 | $1 -$ correct Poisson terms, their $\lambda$, allow $\pm 1$ term. Don't allow second M1 from $\lambda$ in tables, e.g. if MR, treat as E–1. |
| $= \mathbf{0.4219}$ | A1 | Answer, a.r.t. 0.422, implies all 3 marks. If no working: don't give M1A0 |
| | **[3]** | |

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## Question 4(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $Po(32) \approx N(32, 32)$ | M1 | $N(\lambda, \lambda)$ stated or implied, allow $\sqrt{\lambda}$ or $\lambda^2$ for var. Needs $\lambda \geq 15$ |
| | A1 | $N(32, 32)$, allow $\sqrt{32}$ or $32^2$ for var |
| $1 - \Phi\!\left(\dfrac{35.5 - 32}{\sqrt{32}}\right)$ | M1 | Standardise with $\lambda$ and $\sqrt{\lambda}$ or $\lambda$, allow cc errors but not $\sqrt{n}$; both cc and $\sqrt{}$ correct. Can get (M0A0) M1A1 from $\lambda < 15$. Typically, no cc $\to 0.2203$, or $32 \to 0.4565$, 3/5 (but needs evidence, not just answer) |
| | A1 | both cc and $\sqrt{}$ correct |
| $= 1 - \Phi(.619) = \mathbf{0.2681}$ | A1 | Final answer, a.r.t. 0.268 |
| | **[5]** | |

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4 In a rock, small crystal formations occur at a constant average rate of 3.2 per cubic metre.\\
(i) State a further assumption needed to model the number of crystal formations in a fixed volume of rock by a Poisson distribution.

In the remainder of the question, you should assume that a Poisson model is appropriate.\\
(ii) Calculate the probability that in one cubic metre of rock there are exactly 5 crystal formations.\\
(iii) Calculate the probability that in 0.74 cubic metres of rock there are at least 3 crystal formations.\\
(iv) Use a suitable approximation to calculate the probability that in 10 cubic metres of rock there are at least 36 crystal formations.

\hfill \mbox{\textit{OCR S2 2012 Q4 [11]}}