Answer :
Final answer:
Using the Poisson distribution process, Jack can expect to hear the door slam at least once during the one hour from midnight to 1am with a probability of 88.52%.
Explanation:
The subject of your question falls under the category of Mathematics, particularly statistics. You're looking to apply the Poisson distribution in a practical scenario, which is used to model the number of times an event happens in a fixed interval of time or space.
The average rate of sound occurrence is 13 times in a 6-hour interval. However, to find the probability of the sound occurring in a 1 hour interval, you need to adjust this rate proportionally. So, the 1-hour rate will be 13 (sounds per 6 hours) divided by 6 which is approx 2.1667 sounds per hour.
The probability of hearing a slam at least once is just 1 minus the probability of hearing no slams. The formula for Poisson distribution is P(k; λ) = (λ^k * e^-λ)/k!, where λ is the mean rate (2.1667), k is the number of occurrences, and e is Euler's number (approx 2.71828).
For k=0 (no occurrence), it becomes P(0; 2.1667) = (2.1667^0 * e^-2.1667)/0! = e^-2.1667. After calculating we get approximately 0.1148.
So, the probability of hearing a door slam at least once during the one hour from midnight to 1am is 1-0.1148 = 0.8852, or 88.52%.
Learn more about Poisson Distribution here:
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