Question

Cars arriving for gasoline at a Shell station follow a Poisson distribution with a mean of 9 per hour. A. Determine the probability that over the next hour, only one car will arrive. Probability = B. Compute the probability that in the next 6 hours, more than 22 cars will arrive.

Answer 1

Answer:

Step-by-step explanation:

Given

Poisson Distribution with mean of 9 per hour

Mean of Process$=\lambda t$

Probability that over the next hour , only one car will arrive

Probability is given by

$P\left ( x\left ( t \right )=k\right )=\frac{e^{-\lambda t}\left ( \lambda t\right )^k}{k!}$

For k=1 and t=1

$P\left ( x\left ( 1\right )=1\right )=\frac{e^{-\lambda 1}\left ( \lambda \cdot 1\right )^1}{1!}$

$P\left ( x\left ( 1\right )=1\right )=\frac{e^{9\cdot 1}\left ( \lambda \cdot 1\right )^1}{1!}=0.0011$

(b) For next 6 hours , more than 22 cars will arrive will be given by

$\lambda =9$

$\lambda \times t=9\times 6=54$

k=22, t=6 hr

$P\left ( x\left ( 4\right )> 22\right )=1-P\left ( x\left ( 4\right )\leq 22\right )$

$=1-\sum_{k=22}^{k=0}\frac{e^{54}\left ( 54\right )^k}{k!}$