To find the probability of more than 2 motor vehicle deaths in a given day in Dutchess County, New York, we can use the Poisson distribution. The Poisson distribution is suitable here because we are dealing with the number of events occurring in a fixed interval of time (in this case, a day).
The mean number of motor vehicle deaths per year is given as 35.4. To find the mean per day, we divide this number by the number of days in a year (approximately 365):
\[ \lambda = \frac{35.4}{365} \approx 0.09726027397 \]
For the Poisson distribution, the probability of observing \( k \) events when the mean is \( \lambda \) is given by the formula:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
We are interested in finding the probability of more than 2 motor vehicle deaths in a day, which is calculated as:
\[ P(X > 2) = 1 - P(X \leq 2) \]
where
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
Now we calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) using the Poisson formula:
-
Calculate \( P(X = 0) \): \[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-0.09726027397} \cdot 1 = e^{-0.09726027397} \approx 0.9062182657 \]
-
Calculate \( P(X = 1) \): \[ P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-0.09726027397} \cdot 0.09726027397 \approx 0.9062182657 \cdot 0.09726027397 \approx 0.08811886017 \]
-
Calculate \( P(X = 2) \): \[ P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-0.09726027397} \cdot (0.09726027397)^2}{2} \approx 0.9062182657 \cdot \frac{0.009463511692}{2} \approx 0.004295366982 \]
Now we can calculate \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.9062182657 + 0.08811886017 + 0.004295366982 \approx 0.99863249285 \]
Finally, we find \( P(X > 2) \):
\[ P(X > 2) = 1 - P(X \leq 2) \approx 1 - 0.99863249285 \approx 0.00136750715 \]
Now rounding the answer to the nearest ten-thousandth:
\[ P(X > 2) \approx 0.0014 \]
Thus, the probability that there are more than 2 motor vehicle deaths on a given day in Dutchess County is approximately 0.0014.