Can you find the probability that there will be exactly 2 arrivals in a period of 5 minutes, assuming independent arrivals with a mean of 2,5 arrivals per minute?

To find the probability of exactly 2 arrivals in a period of 5 minutes, we can use the Poisson distribution formula:

P(X = k) = (e^(-λ) * λ^k) / k!

Where:
- P(X = k) is the probability of k arrivals
- e is the base of the natural logarithm (approximately 2.71828)
- λ is the mean number of arrivals per minute
- k is the number of arrivals we are interested in
- k! is the factorial of k

Given that the mean number of arrivals per minute is 2.5, we can calculate the mean number of arrivals in 5 minutes as λ = 2.5 * 5 = 12.5.

Now, we can plug in the values into the formula:

P(X = 2) = (e^(-12.5) * 12.5^2) / 2!
P(X = 2) = (e^(-12.5) * 156.25) / 2
P(X = 2) = (3.72665317208e-06 * 156.25) / 2
P(X = 2) = 0.000581413299

Therefore, the probability of exactly 2 arrivals in a period of 5 minutes is approximately 0.00058 or 0.058%.

Poisson Distribution:

The number of arrivals in a time interval follows a Poisson distribution with mean (λ) equal to the average number of arrivals per minute multiplied by the number of minutes in the interval:

λ = 2.5 arrivals/minute x 5 minutes = 12.5 arrivals

Probability of Exactly 2 Arrivals:

The probability of exactly k arrivals in a Poisson distribution is given by:

P(X = k) = (λ^k - e^-λ) / k!

where:

- k is the number of arrivals
- λ is the mean number of arrivals
- e is the mathematical constant approximately equal to 2.718

For k = 2, λ = 12.5, we get:

P(X = 2) = (12.5^2 - e^-12.5) / 2!
≈ 0.2153

Interpretation:

There is approximately a 21.53% probability that there will be exactly 2 arrivals in a period of 5 minutes, assuming independent arrivals with a mean of 2.5 arrivals per minute.