question 10\nthe number of magnitude 5 or larger earthquakes in alaska follow a poisson distribution with an…

Explanation:

Step1: Recall Poisson - distribution formula

The Poisson - distribution formula is $P(X = k)=\frac{\lambda^{k}e^{-\lambda}}{k!}$, where $X$ is the random variable representing the number of events, $k$ is the actual number of events that occur, $\lambda$ is the average number of events in the given time interval, and $e\approx2.71828$.

Step2: Calculate $\lambda$ for part a

Given that the average number of earthquakes per year is $\lambda_{year}=19$. For part a, we want to find the probability of $k = 24$ earthquakes in a year. $P(X = 24)=\frac{19^{24}e^{- 19}}{24!}$ $19^{24}=1.9018\times10^{30}$, $e^{-19}\approx4.0762\times10^{-9}$, $24! = 6.204484017\times10^{23}$ $P(X = 24)=\frac{1.9018\times10^{30}\times4.0762\times10^{-9}}{6.204484017\times10^{23}}\approx0.0397$

Step3: Calculate $\lambda$ for part b - d

The average number of earthquakes per year is $\lambda_{year}=19$. So the average number of earthquakes per month is $\lambda_{month}=\frac{19}{12}\approx1.5833$.

Step4: Calculate probability for part b

For $k = 1$ earthquake in a month, using the Poisson - distribution formula $P(X = 1)=\frac{(1.5833)^{1}e^{-1.5833}}{1!}$ $(1.5833)^{1}=1.5833$, $e^{-1.5833}\approx0.2059$ $P(X = 1)=1.5833\times0.2059\approx0.3260$

Step5: Calculate probability for part c

For $k = 0$ earthquake in a month, using the Poisson - distribution formula $P(X = 0)=\frac{(1.5833)^{0}e^{-1.5833}}{0!}$ Since $(1.5833)^{0}=1$ and $0! = 1$, $P(X = 0)=e^{-1.5833}\approx0.2059$

Step6: Calculate probability for part d

For $k = 2$ earthquakes in a month, using the Poisson - distribution formula $P(X = 2)=\frac{(1.5833)^{2}e^{-1.5833}}{2!}$ $(1.5833)^{2}=2.5068$, $e^{-1.5833}\approx0.2059$, $2! = 2$ $P(X = 2)=\frac{2.5068\times0.2059}{2}\approx0.2599$

Answer:

a) $0.0397$ b) $0.3260$ c) $0.2059$ d) $0.2599$