To address your inquiry about the distribution of these events over the specified timeframes, we can approach this problem using basic probability and statistics.
Step 1: Establish the Framework
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Timeframes:
- Total observation period: 23 months.
- Period with events: 6 months.
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Event count: Total of 5 events (4 car break-ins and mailbox thefts).
Step 2: Probability Distribution
Assuming the events are independent and randomly occurring at a constant rate, we can model this with a Poisson distribution which is often suitable for counting the number of events in a fixed interval of time.
For simplicity, we'll model the average rate of occurrence for the past 23 months.
- Let \( \lambda \) be the average rate of events (events/months):
- In 23 months with 5 events, the average rate \( \lambda = \frac{5 \text{ events}}{23 \text{ months}} \approx 0.217 \text{ events/month} \).
Step 3: Calculate the Expected Number of Events in 6 Months
To find the expected number of events in just the past 6 months, we multiply \( \lambda \) by 6:
\[ E(6 \text{ months}) = \lambda \times 6 \approx 0.217 \text{ events/month} \times 6 \text{ months} = 1.302 \text{ events} \]
Step 4: Calculate the Probability of Observing 5 Events in 6 Months
Using the Poisson probability mass function (PMF):
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
where \( k \) is the number of events (5 in your case);
We can compute the probability of observing 5 events in 6 months:
\[ P(X = 5) = \frac{e^{-1.302} \cdot 1.302^5}{5!} \]
Calculating the components:
- \( e^{-1.302} \approx 0.273 \)
- \( 1.302^5 \approx 4.855 \)
- \( 5! = 120 \)
So:
\[ P(X = 5) \approx \frac{0.273 \cdot 4.855}{120} \approx \frac{1.325}{120} \approx 0.01104 \]
Step 5: Compare with the Expected Distribution Over 23 Months
Next, we calculate the probability of having the same 5 events occur within a 23-month period. This is essentially examining how likely 5 events is in a broader time frame.
Using the same approach with \( \lambda \) for 23 months, we examine:
\[ P(X = 5) = \frac{e^{-5} \cdot 5^5}{5!} \]
Calculating components:
- \( e^{-5} \approx 0.007 \)
- \( 5^5 = 3125 \)
Thus:
\[ P(X = 5) \approx \frac{0.007 \cdot 3125}{120} \approx \frac{21.875}{120} \approx 0.18229 \]
Conclusion
To answer your question regarding the likelihood of the events occurring:
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Probability of 5 events in 6 months: Approximately 1.104%.
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Probability of 5 events in 23 months: Approximately 18.229%.
Given these estimates:
Conclusion on Hypotheses
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The probability that 5 events occurred in 6 months under the assumption of a uniform distribution suggests those events are highly unlikely random occurrences (only about 1.1% chance).
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Comparatively, there's an approximately 18.23% chance that similar events could occur randomly over 23 months.
Variance Conclusion
This makes it about 17 times more likely (or 18.23% / 1.1%) that something changed post-08/16/24, supporting your second hypothesis more than the first.
This suggests that there may indeed be a factor influencing these events that arose around the time of the first incident. Further investigation would be advisable to determine if there are specific changes in the neighborhood that correlate with the increase in incidents.
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