Q#96: Random dog arrivals
Note: I believe it should be free and readily available for everyone to gain value from data, hence I pledge to keep this series free regardless of how large it grows.
Suppose dogs randomly arrive at a dog park at an average rate of 5.2 dogs per hour. Given this information, what is the probability of observing 6 dog arrivals in a given hour at the dog park?
TRY IT YOURSELF
ANSWER
Seems like a strange unaswerable question at first, but it actually is a statistics problem. It requires using the Poisson distribution to determine the probability of this event.
The Poisson distribution is a discrete probability distribution that models the number of events occurring within a fixed interval of time or space when events happen randomly and independently at a constant average rate.
In our case, we have an average rate of 5.2 dogs arriving per hour. The Poisson distribution can help us calculate the probability of observing a specific number of arrivals, such as 6 dogs, within that hour.
The Poisson Probability Formula
The probability mass function of the Poisson distribution is given by the formula:
P(X=k)=k!e−λ∗λk
Where:
- P(X=k) is the probability of observing exactly k arrivals.
- λ is the average rate at which events occur (in this case, the average number of dogs arriving per hour).
- e is Euler’s number, approximately equal to 2.71828.
- k is the number of events we want to find the probability for.
- k! is the factorial of k.
Calculating the Probability
To find the probability of observing 6 dog arrivals in a given hour at the dog park, we can plug our values into the formula. In this case, λ=5.2 and k=6.
P(X=6)=(e^−5.2∗5.2)/6!
Calculating this expression will give us the probability of exactly 6 dogs arriving in an hour, 0.1515.
Plug: Checkout all my digital products on Gumroad here. Please purchase ONLY if you have the means to do so. Use code: MEDSUB to get a 10% discount!
