Respuesta :
Answer:
Step-by-step explanation:
Given
Poisson Distribution with mean of 9 per hour
Mean of Process[tex]=\lambda t[/tex]
Probability that over the next hour , only one car will arrive
Probability is given by
[tex]P\left ( x\left ( t \right )=k\right )=\frac{e^{-\lambda t}\left ( \lambda t\right )^k}{k!}[/tex]
For k=1 and t=1
[tex]P\left ( x\left ( 1\right )=1\right )=\frac{e^{-\lambda 1}\left ( \lambda \cdot 1\right )^1}{1!}[/tex]
[tex]P\left ( x\left ( 1\right )=1\right )=\frac{e^{9\cdot 1}\left ( \lambda \cdot 1\right )^1}{1!}=0.0011[/tex]
(b) For next 6 hours , more than 22 cars will arrive will be given by
[tex]\lambda =9[/tex]
[tex]\lambda \times t=9\times 6=54[/tex]
k=22, t=6 hr
[tex]P\left ( x\left ( 4\right )> 22\right )=1-P\left ( x\left ( 4\right )\leq 22\right )[/tex]
[tex]=1-\sum_{k=22}^{k=0}\frac{e^{54}\left ( 54\right )^k}{k!}[/tex]
