An observer counts 240 vehicles per hour at a | vedclass

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Question

(A) The average arrival rate, $\lambda$, is $240 	ext{ vehicles/hour} = \frac{240}{3600} 	ext{ vehicles/second} = \frac{1}{15} 	ext{ vehicles/secon

Vedclass · Accepted Answer

$\frac{e^2-5}{e^2}$

Vedclass · Answer

(A) The average arrival rate, $\lambda$, is $240 	ext{ vehicles/hour} = \frac{240}{3600} 	ext{ vehicles/second} = \frac{1}{15} 	ext{ vehicles/second}$.For a time interval $t = 30 	ext{ seconds}$, the expected number of arrivals is $\mu = \lambda t = \frac{1}{15} 	imes 30 = 2$.According to the Poisson distribution, the probability of $n$ arrivals is $P(n) = \frac{\mu^n e^{-\mu}}{n!}$.We need to find the probability of more than two vehicles arriving, i.e.,$P(n > 2) = 1 - [P(0) + P(1) + P(2)]$.Calculating individual probabilities:$P(0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$$P(1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$$P(2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$Summing these probabilities: $P(n \leq 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$.Therefore, $P(n > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2} = \frac{e^2 - 5}{e^2}$.

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

An observer counts $240$ vehicles per hour at a specific location on a highway. Assuming that the arrival of vehicles at the location follows a Poisson distribution, the probability that more than two vehicles arrive over a $30 \text{ sec}$ time interval is

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