Question

a

Answer 1

Sorry, I'm not familiar with Poisson distributions

Answer 2

The Meteorology Dept. of U of Hawaii modeled the number of hurricanes coming within 250 nautical miles of Honolulu during a year using a Poisson distribution with a mean of 0.45. Using this model, determine the probabilities of the following events: a) at least one hurricane will come within 250 nautical miles of Honolulu during the next year b) at most 4 hurricanes will come within 250 nautical miles of Honolulu during the next year

Let \(X\) denote the random variable for number of hurricanes. \(X\) follows a Poisson dist. with mean \(\lambda=0.45\), so the probability density function is
\[f(x)=\dfrac{e^{-\lambda}\lambda^x}{x!}~~~~\text{for }x=0,1,2,...\]
For part (a), you must find \(P(X\ge1)\), and for part (b), \(P(X\le4)\).
\[\begin{align*}P(X\ge1)&=1-P(X<1)\\
&=1-P(X=0)\\
&=1-\dfrac{e^{-0.45}0.45^0}{0!}\\
&=1-e^{-0.45}
\end{align*}\]
First answer is correct!
\[\begin{align*}P(X\le4)&=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\\
&=\frac{e^{-0.45}0.45^0}{0!}+\frac{e^{-0.45}0.45^1}{1!}+\frac{e^{-0.45}0.45^2}{2!}+\frac{e^{-0.45}0.45^3}{3!}\\
&~~~~~~+\frac{e^{-0.45}0.45^4}{4!}
\end{align*}\]
Your last term (which would be my first term) is off by a bit. The answer in part (a) is for \(1-P(X=0)\), and not \(P(X=0)\) itself.