Question 7 - A-Level Maths

CAIE FP2 2011 November — Question 7 11 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2011
Session	November
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Distribution
Type	Multiple independent components
Difficulty	Standard +0.3 This is a straightforward application of the exponential distribution with standard techniques: finding probabilities using the CDF, applying binomial distribution to multiple trials, and solving an inequality involving the exponential function. All steps are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average.
Spec	5.02i Poisson distribution: random events model 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities 5.02l Poisson conditions: for modelling 5.02m Poisson: mean = variance = lambda

7 The lifetime, in hours, of a 'Trulite' light bulb is a random variable $T$. The probability density function f of $T$ is given by $$\mathrm { f } ( t ) = \begin{cases} 0 & t < 0 \\ \lambda \mathrm { e } ^ { - \lambda t } & t \geqslant 0 \end{cases}$$ where $\lambda$ is a positive constant. Given that the mean lifetime of Trulite bulbs is 2000 hours, find the probability that a randomly chosen Trulite bulb has a lifetime of at least 1000 hours. A particular light fitting has 6 randomly chosen Trulite bulbs. Find the probability that no more than one of these bulbs has a lifetime less than 1000 hours. By using new technology, the proportion of Trulite bulbs with very short lifetimes is to be reduced. Find the least value of the new mean lifetime that will ensure that the probability that a randomly chosen Trulite bulb has a lifetime of no more than 4 hours is less than 0.001 .

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Question 7:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
State or find $\lambda$: $\lambda = 1/2000$ or $0.0005$	B1
Find $p = P(T \geq 1000)$: $1 - \int_0^{1000}\lambda e^{-\lambda t}\,dt = 1 + [e^{-\lambda t}]_0^{1000} = e^{-0.5} = 0.607$	M1 A1
Find $P(N=1)$ where $N$ of 6 bulbs have $T < 1000$: $P(N=1) = 6p^5(1-p)\,[= 0.194]$	B1
Find $P(N \leq 1)$: $P(N \leq 1) = P(N=1) + p^6 = 0.244$	M1 A1
Formulate inequality for new $\lambda$: $0.001 > \int_0^4 \lambda e^{-\lambda t}\,dt = [-e^{-\lambda t}]_0^4 = 1 - e^{-4\lambda}$	M1
$-4\lambda > \ln 0.999$	A1
Find minimum mean $1/\lambda$: $1/\lambda > -4/\ln 0.999$, minimum is $4000$	A1 M1 A1

Total: 11 marks

## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| State or find $\lambda$: $\lambda = 1/2000$ or $0.0005$ | B1 | |
| Find $p = P(T \geq 1000)$: $1 - \int_0^{1000}\lambda e^{-\lambda t}\,dt = 1 + [e^{-\lambda t}]_0^{1000} = e^{-0.5} = 0.607$ | M1 A1 | |
| Find $P(N=1)$ where $N$ of 6 bulbs have $T < 1000$: $P(N=1) = 6p^5(1-p)\,[= 0.194]$ | B1 | |
| Find $P(N \leq 1)$: $P(N \leq 1) = P(N=1) + p^6 = 0.244$ | M1 A1 | |
| Formulate inequality for new $\lambda$: $0.001 > \int_0^4 \lambda e^{-\lambda t}\,dt = [-e^{-\lambda t}]_0^4 = 1 - e^{-4\lambda}$ | M1 | |
| $-4\lambda > \ln 0.999$ | A1 | |
| Find minimum mean $1/\lambda$: $1/\lambda > -4/\ln 0.999$, minimum is $4000$ | A1 M1 A1 | |

**Total: 11 marks**

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7 The lifetime, in hours, of a 'Trulite' light bulb is a random variable $T$. The probability density function f of $T$ is given by

$$\mathrm { f } ( t ) = \begin{cases} 0 & t < 0 \\ \lambda \mathrm { e } ^ { - \lambda t } & t \geqslant 0 \end{cases}$$

where $\lambda$ is a positive constant. Given that the mean lifetime of Trulite bulbs is 2000 hours, find the probability that a randomly chosen Trulite bulb has a lifetime of at least 1000 hours.

A particular light fitting has 6 randomly chosen Trulite bulbs. Find the probability that no more than one of these bulbs has a lifetime less than 1000 hours.

By using new technology, the proportion of Trulite bulbs with very short lifetimes is to be reduced. Find the least value of the new mean lifetime that will ensure that the probability that a randomly chosen Trulite bulb has a lifetime of no more than 4 hours is less than 0.001 .

\hfill \mbox{\textit{CAIE FP2 2011 Q7 [11]}}

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