OCR Further Statistics AS 2018 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Statistics AS (Further Statistics AS)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeScaled time period sums
DifficultyModerate -0.8 This question tests basic recall of Poisson distribution conditions and straightforward applications of scaling and addition properties. Part (i) requires stating one standard condition (constant mean rate), while parts (ii) and (iii) involve routine calculations with no conceptual challenges—just scaling the parameter and using the additive property of independent Poisson variables.
Spec5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson

3 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by \(X\).
  1. State a further assumption needed for \(X\) to be well modelled by a Poisson distribution. Assume now that \(X\) can be well modelled by the distribution \(\operatorname { Po } ( 0.7 )\).
  2. Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws. The number of flaws in 100 m of FOC of a larger diameter has the distribution \(\mathrm { Po } ( 1.6 )\).
  3. Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4.

Question 3:
Part (i)
AnswerMarks Guidance
Flaws must occur at constant average rate (uniform rate)B1 [1] Context (e.g. "flaws") needed. Extra answers e.g. "singly": B0. Not "constant rate" or "average constant rate"
Part (ii)
AnswerMarks Guidance
\(\text{Po}(2.1)\) or \(e^{-\lambda}\dfrac{\lambda^3}{3!} = \mathbf{0.189}\)M1 A1 [2] M1: Po(2.1) stated or implied, or formula with \(\lambda = 2.1\) stated. A1: Awrt 0.189
Part (iii)
\(\text{Po}(3)\)
AnswerMarks Guidance
\(1 - P(\leq 3) = \mathbf{0.3528}\)M1 M1 A1 [3] M1: Po\((2\times0.7 + 1.6)\) stated or implied. M1: Allow \(1 - P(\leq 4) = 0.1847\), or from wrong \(\lambda\). A1: Awrt 0.353. Or all combinations \(\leq 3\); \(1-\) above, not just \(= 3\)
## Question 3:

### Part (i)
Flaws must occur at constant average rate (uniform rate) | **B1** [1] | Context (e.g. "flaws") needed. Extra answers e.g. "singly": B0. Not "constant rate" or "average constant rate"

### Part (ii)
$\text{Po}(2.1)$ or $e^{-\lambda}\dfrac{\lambda^3}{3!} = \mathbf{0.189}$ | **M1** A1 [2] | M1: Po(2.1) stated or implied, or formula with $\lambda = 2.1$ stated. A1: Awrt 0.189

### Part (iii)
$\text{Po}(3)$

$1 - P(\leq 3) = \mathbf{0.3528}$ | **M1** M1 A1 [3] | M1: Po$(2\times0.7 + 1.6)$ stated or implied. M1: Allow $1 - P(\leq 4) = 0.1847$, or from wrong $\lambda$. A1: Awrt 0.353. Or all combinations $\leq 3$; $1-$ above, not just $= 3$

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3 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by $X$.\\
(i) State a further assumption needed for $X$ to be well modelled by a Poisson distribution.

Assume now that $X$ can be well modelled by the distribution $\operatorname { Po } ( 0.7 )$.\\
(ii) Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws.

The number of flaws in 100 m of FOC of a larger diameter has the distribution $\mathrm { Po } ( 1.6 )$.\\
(iii) Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4.

\hfill \mbox{\textit{OCR Further Statistics AS 2018 Q3 [6]}}