Question 10 - A-Level Maths

CAIE FP2 2009 June — Question 10 12 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2009
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Distribution
Type	Link Poisson to exponential
Difficulty	Challenging +1.2 This question requires understanding the link between Poisson and exponential distributions (a key Further Maths concept), then applying the Central Limit Theorem to sum of exponentials. While it involves multiple statistical concepts and a two-part proof structure, the steps are fairly standard for Further Maths students: deriving P(T>t) from Poisson, recognizing the exponential distribution, and using normal approximation for sum of exponentials. The justification requirement adds modest difficulty, but this is a well-rehearsed connection in FP2/S2 syllabi.
Spec	5.02i Poisson distribution: random events model 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.03a Continuous random variables: pdf and cdf

10 The number of hits per minute on a particular website has a Poisson distribution with mean 0.8. The time between successive hits is denoted by \(T\) minutes. Show that \(\mathrm { P } ( T > t ) = \mathrm { e } ^ { - 0.8 t }\) and hence show that \(T\) has a negative exponential distribution. Using a suitable approximation, which should be justified, find the probability that the time interval between the 1st hit and the 51st hit exceeds one hour.

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Question 10:

Part 1 (5 marks)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(P(T > t) = P(\text{hits in } t \text{ mins} = 0)\)	M1	Relate \(P(T>t)\) to number of hits
\(= P_0(0.8t) = e^{-0.8t}\)	M1 A1	Relate to Poisson distribution. A.G.
\(F(t) = P(T < t) = 1 - e^{-0.8t}\)	B1	Find distribution \(F(t)\) of \(T\)
\(f(t) = 0.8\, e^{-0.8t}\)	B1	Differentiate to find \(f(t)\) in required form

Part 2 - EITHER route (7 marks)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(S = \sum_{i=1}^{50} T_i\)	M1	State or imply required probability
\(\mu_S = 50(1/0.8) = 62.5\)	A1	State or use mean of \(S\)
\(\sigma_S^2 = 50(1/0.8)^2 = 78.1[25]\)	A1	State or use variance of \(S\)
By Central Limit Theorem	B1	Justify use of Normal approximation (A.E.F.): or 50 is large or \(50 > \) e.g. 30
\(1 - \Phi((60 - \mu_S)/\sigma_S)\)	M1	Evaluate approximate probability (A.E.F.)
\(= \Phi(0.283) = 0.611\)	A1; A1

Part 2 - OR route (7 marks)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(P(\text{hits in 60 mins} < 50)\) [allow \(\leq 50\)]	(M1)	State or imply required probability
\(\mu = 60 \times 0.8 = 48\)	(A1)	State or use mean
\(\sigma^2 = 60 \times 0.8 = 48\)	(A1)	State or use variance
48 is large or \(48 > \) e.g. 15	(B1)	Justify use of Normal approximation (A.E.F.)
\(\Phi((49.5 - \mu)/\sigma)\)	(M1)	Evaluate approximate probability (A.E.F.)
\(= \Phi(0.216[5]) = 0.586\)	(A1; A1)
\(\Phi((50-\mu)/\sigma) = 0.614\) earns B1		S.R. Omission of continuity correction

# Question 10:

## Part 1 (5 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(T > t) = P(\text{hits in } t \text{ mins} = 0)$ | M1 | Relate $P(T>t)$ to number of hits |
| $= P_0(0.8t) = e^{-0.8t}$ | M1 A1 | Relate to Poisson distribution. **A.G.** |
| $F(t) = P(T < t) = 1 - e^{-0.8t}$ | B1 | Find distribution $F(t)$ of $T$ |
| $f(t) = 0.8\, e^{-0.8t}$ | B1 | Differentiate to find $f(t)$ in required form |

## Part 2 - EITHER route (7 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $S = \sum_{i=1}^{50} T_i$ | M1 | State or imply required probability |
| $\mu_S = 50(1/0.8) = 62.5$ | A1 | State or use mean of $S$ |
| $\sigma_S^2 = 50(1/0.8)^2 = 78.1[25]$ | A1 | State or use variance of $S$ |
| By Central Limit Theorem | B1 | Justify use of Normal approximation (A.E.F.): *or* 50 is large *or* $50 > $ e.g. 30 |
| $1 - \Phi((60 - \mu_S)/\sigma_S)$ | M1 | Evaluate approximate probability (A.E.F.) |
| $= \Phi(0.283) = 0.611$ | A1; A1 | |

## Part 2 - OR route (7 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{hits in 60 mins} < 50)$ [allow $\leq 50$] | (M1) | State or imply required probability |
| $\mu = 60 \times 0.8 = 48$ | (A1) | State or use mean |
| $\sigma^2 = 60 \times 0.8 = 48$ | (A1) | State or use variance |
| 48 is large *or* $48 > $ e.g. 15 | (B1) | Justify use of Normal approximation (A.E.F.) |
| $\Phi((49.5 - \mu)/\sigma)$ | (M1) | Evaluate approximate probability (A.E.F.) |
| $= \Phi(0.216[5]) = 0.586$ | (A1; A1) | |
| $\Phi((50-\mu)/\sigma) = 0.614$ earns B1 | | **S.R.** Omission of continuity correction |

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10 The number of hits per minute on a particular website has a Poisson distribution with mean 0.8. The time between successive hits is denoted by $T$ minutes. Show that $\mathrm { P } ( T > t ) = \mathrm { e } ^ { - 0.8 t }$ and hence show that $T$ has a negative exponential distribution.

Using a suitable approximation, which should be justified, find the probability that the time interval between the 1st hit and the 51st hit exceeds one hour.

\hfill \mbox{\textit{CAIE FP2 2009 Q10 [12]}}

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Q1 5 Q2 7 Q3 8 Q4 11 Q5 12 Q6 6 Q7 8 Q8 8 Q9 9 Q10 12 Q11 EITHER Q11 OR