Discussion
Suppose the number of accidents per week on a certain highway follows Poisson distribution with mean 2 then the percentage of weeks having no accidents is .................... (Given \(e^{-2}\) = 0.1353)
Answer: C
The percentage of weeks with no accidents is 13.53%.
For a Poisson distribution with mean m:
\[ \begin{aligned} P(X = x) &= \frac{e^{-m} m^{x}}{x!} \\ P(X = 0) &= e^{-2} = 0.1353 \end{aligned} \]
So 13.53% of weeks are expected to be accident-free.
27.06% is \(2e^{-2}\), the probability of exactly one (or exactly two) accidents, not of none. 6.77% is half of the correct figure, and 7.39% is \(e^{2}\) misread as a percentage.
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