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W4_Tutorial_Q_2_8

# W4_Tutorial_Q_2_8 - Tutorial 3 2 The interarrival time...

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Unformatted text preview: Tutorial 3 2. The interarrival time follows an exponential distribution with mean 10 or parameter 0.1: so the counting process for the arrival of events is Poisson distributed with the same parameter 0.1. Let this process be N(t). We are looking for the probability that at least two customers arrive in a 10 minute period, i.e. ( 10 ¡ ≥ 2 = 1 − 10 ¡ = 0 ¡ − 10 ¡ = 1 ¡ And 10 ¡ = 0 ¡ = ? − 0.1×10 0.1 × 10 ¡ 0! = ? − 1 10 ¡ = 1 ¡ = ? − 0.1×10 0.1 × 10 ¡ 1 1! = ? − 1 ∴ ( 10 ≥ 2 ¡ = 1 − 2 ? − 1 = 0.2642 8. a) The number of storms N(t) up to time t depends on whether it is a good year¢or¢an¢“other” year. If it is a good year N(t) is Poisson rate 3, but if it is an other year N(t) is Poisson rate 5 i.e. in other years there are typically more storms. Because N(t) depends on which type of year it is, let us condition on which type of year it is: = £ 1 ݅? ݅¤ ?¥¥¦ ?§?...
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W4_Tutorial_Q_2_8 - Tutorial 3 2 The interarrival time...

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