Question

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

Answer 1

Answer:

The answer is "0.6227 and 0.5971".

Step-by-step explanation:

$\to p(x\leq 2)=p(0)+p(1)+p(2)$

                  $=e^{-2.2}(1+2.2^{1}+\frac{2.2^2}{2!})\\\\=0.6227$

$\to x \sim emp(\frac{1}{2.2})\\\\\to f(x)=\frac{1}{2.2} emp1-\frac{x}{2.2}$

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

                  $=1-emp1-2|2.2\\\\=0.5971$