Answer:
The probability that there will be between 1 and 3 is 0.4152.
Step-by-step explanation:
Let X denote the number of power outages at a nuclear power plant.
It is provided that X follows a Poisson distribution with a mean of 4 outages per year.
The probability mass function of X is:
[tex]p_{X}(x)=\frac{\lambda e^{-\lambda \cdot x}}{x!}[/tex]
Compute the probability that there will be between 1 and 3 as follows:
[tex]P(1\leq X\leq 3)=\sum\limits^{3}_{1}{\frac{4e^{-4 x}}{x!}}\\\\=0.07326+0.14653+0.19537\\\\=0.41516\\\\\approx 0.4152[/tex]
Thus, the probability that there will be between 1 and 3 is 0.4152.
