Question
Dutchess County, New York, has been experiencing a mean of 35.4 motor vehicle deaths each year. Find the probability that on a given day, there are more than 2 motor vehicle deaths.
Round your answer to the nearest ten thousandth.
Round your answer to the nearest ten thousandth.
Answers
Answer
0.0001
Answered by
GPT-4o mini
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:
1. **Calculate \( P(X = 0) \)**:
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-0.09726027397} \cdot 1 = e^{-0.09726027397} \approx 0.9062182657
\]
2. **Calculate \( P(X = 1) \)**:
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-0.09726027397} \cdot 0.09726027397 \approx 0.9062182657 \cdot 0.09726027397 \approx 0.08811886017
\]
3. **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**.
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:
1. **Calculate \( P(X = 0) \)**:
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-0.09726027397} \cdot 1 = e^{-0.09726027397} \approx 0.9062182657
\]
2. **Calculate \( P(X = 1) \)**:
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-0.09726027397} \cdot 0.09726027397 \approx 0.9062182657 \cdot 0.09726027397 \approx 0.08811886017
\]
3. **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**.
Answered by
GPT-4o mini
Let's go through the calculations again to find the probability that on a given day in Dutchess County, there are more than 2 motor vehicle deaths using the Poisson distribution.
1. **Find the mean deaths per day:**
The annual mean is given as 35.4 deaths per year. To find the mean per day (\(\lambda\)):
\[
\lambda = \frac{35.4}{365} \approx 0.09726027
\]
2. **Calculate \(P(X \leq 2)\)**, which is the cumulative probability of having 0, 1, or 2 deaths in a day.
We need to compute:
\[
P(X = 0), P(X = 1), P(X = 2)
\]
Using the Poisson probability mass function:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
**Calculating** \(P(X = 0)\):
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-0.09726027}
\]
Calculating this value:
\[
P(X = 0) \approx e^{-0.09726027} \approx 0.9060
\]
**Calculating** \(P(X = 1)\):
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-0.09726027} \cdot 0.09726027
\]
\[
P(X = 1) \approx 0.9060 \cdot 0.09726027 \approx 0.0881
\]
**Calculating** \(P(X = 2)\):
\[
P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-0.09726027} \cdot (\lambda^2)}{2}
\]
\[
P(X = 2) \approx \frac{0.9060 \cdot (0.09726027^2)}{2} \approx \frac{0.9060 \cdot 0.009463}{2} \approx 0.0043
\]
Now sum these probabilities to find \(P(X \leq 2)\):
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.9060 + 0.0881 + 0.0043 \approx 0.9984
\]
3. **Calculate \(P(X > 2)\)**:
\[
P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9984 \approx 0.0016
\]
After reviewing the entire process, it seems my previous calculation of ending as 0.0014 was slightly off. The corrected final answer is approximately:
\[
P(X > 2) \approx 0.0016
\]
**Rounded to the nearest ten-thousandth:**
If again verified, \(P(X > 2) \approx 0.0016\).
If you are referring to a specific result you expect (0.0001), please clarify, as otherwise, the probability based on the given Poisson distribution model calculation estimates for more than 2 deaths in a day are reasonably concluded at approximately **0.0016**.
1. **Find the mean deaths per day:**
The annual mean is given as 35.4 deaths per year. To find the mean per day (\(\lambda\)):
\[
\lambda = \frac{35.4}{365} \approx 0.09726027
\]
2. **Calculate \(P(X \leq 2)\)**, which is the cumulative probability of having 0, 1, or 2 deaths in a day.
We need to compute:
\[
P(X = 0), P(X = 1), P(X = 2)
\]
Using the Poisson probability mass function:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
**Calculating** \(P(X = 0)\):
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-0.09726027}
\]
Calculating this value:
\[
P(X = 0) \approx e^{-0.09726027} \approx 0.9060
\]
**Calculating** \(P(X = 1)\):
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-0.09726027} \cdot 0.09726027
\]
\[
P(X = 1) \approx 0.9060 \cdot 0.09726027 \approx 0.0881
\]
**Calculating** \(P(X = 2)\):
\[
P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-0.09726027} \cdot (\lambda^2)}{2}
\]
\[
P(X = 2) \approx \frac{0.9060 \cdot (0.09726027^2)}{2} \approx \frac{0.9060 \cdot 0.009463}{2} \approx 0.0043
\]
Now sum these probabilities to find \(P(X \leq 2)\):
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.9060 + 0.0881 + 0.0043 \approx 0.9984
\]
3. **Calculate \(P(X > 2)\)**:
\[
P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9984 \approx 0.0016
\]
After reviewing the entire process, it seems my previous calculation of ending as 0.0014 was slightly off. The corrected final answer is approximately:
\[
P(X > 2) \approx 0.0016
\]
**Rounded to the nearest ten-thousandth:**
If again verified, \(P(X > 2) \approx 0.0016\).
If you are referring to a specific result you expect (0.0001), please clarify, as otherwise, the probability based on the given Poisson distribution model calculation estimates for more than 2 deaths in a day are reasonably concluded at approximately **0.0016**.
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