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

The answer is "0.6227 and 0.5971".

Explanation:

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

`=e^(-2.2)(1+2.2^(1)+(2.2^2)/(2!))\\\\=0.6227`

`\to x \sim emp((1)/(2.2))\\\\\to f(x)=(1)/(2.2) emp1-(x)/(2.2)`

`\to p(x \leq 2)=\int^(2)_(0) (1)/(2.2) emp1(-(x)/(2.2)\ dx`

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