Question

The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.8. Answer the following two questions: a) What is the probability that more than two hurricanes will hit the Florida coast in a year? b) What is the probability that exactly one hurricane will hit the coast of Florida in a year?

Answer 1

Answer:

a) $P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

b) $P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

Step-by-step explanation:

Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that $X \sim Poisson(\lambda=0.8)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda=0.8$

$E(X)=\mu =\lambda=0.8$

Part a

For this case we want this probability: $P(X>2)$

And for this case we can use the complement rule like this:

$P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

Part b

Using the probability mass function we have:

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$