Answer:
(a) 0.635
(b) 0.498
Explanation:
(a) The probability distribution is calculated using the cumulative distribution function of the Poisson as follows;
CDF =
![F(x)=e^(-\lambda t)\sum_(i = 0)^(\left | x \right |)(\lambda t^(i))/(i!)](https://img.qammunity.org/2021/formulas/mathematics/high-school/vyhvpqvkuhnydecsqycsnwwn59s5e2ctlu.png)
For x = three we have;
CDF = 0.781
For x = one we have;
CDF = 0.1465
Therefore, the probability that between one and three inclusive = 0.781 - 0.1465 = 0.635
(b) For x = 2 we have;
CDF = 0.4395
For x = one we have;
CDF = 0.9378
Therefore, the probability that between one and three inclusive = 0.9378- 0.4395= 0.498