Telephone calls arrive at the Global Airline reservation office in Louisville according to a Poisson distribution with a mean of 2.2 calls per minute What is the probability of receiving at least 2 calls during a one-minute interval? What is the probability that at most 2 minutes elapse between 1 call and the next

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Answer:

The answer is "0.6227 and 0.5971".

Step-by-step explanation:

[tex]\to p(x\leq 2)=p(0)+p(1)+p(2)[/tex]

                  [tex]=e^{-2.2}(1+2.2^{1}+\frac{2.2^2}{2!})\\\\=0.6227[/tex]

[tex]\to x \sim emp(\frac{1}{2.2})\\\\\to f(x)=\frac{1}{2.2} emp1-\frac{x}{2.2}[/tex]

[tex]\to p(x \leq 2)=\int^{2}_{0} \frac{1}{2.2} emp1(-\frac{x}{2.2}\ dx[/tex]

                  [tex]=1-emp1-2|2.2\\\\=0.5971[/tex]