Answer:
(1)14.9% (2) 2.96% (3) 97.04%
Step-by-step explanation:
Formula for Poisson distribution: [tex]P(k) = \frac{\lambda^ke^{-k}}{k!}[/tex] where k is a number of guests coming in at a particular hour period.
(1) We can substitute k = 7 and [tex]\lambda = 7[/tex] into the formula:
[tex]P(k=7) = \frac{7^7e^{-7}}{7!}[/tex]
[tex]P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%[/tex]
(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours
[tex]P(k\leq2) = P(k=0)+P(k=1)+P(k=2)[/tex]
[tex]P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}[/tex]
[tex]P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}[/tex]
[tex]P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%[/tex]
(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:
[tex]P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%[/tex]
