Final answer:
The probability that more than 3 users visit a website in a given minute, given an average rate of 2.6 visitors per minute and using a Poisson distribution, is approximately 0.2645 or 26.45%.
Step-by-step explanation:
We need to find the probability that X, the number of website visitors per minute, is greater than 3. Given the average rate of 2.6 visitors per minute, the situation can be modeled using a Poisson distribution. The probability mass function (PMF) for a Poisson distribution is:
P(X = k) = (e-λλk)/k!,
where λ is the average rate (2.6 in this case), k is the actual number of successes (number of users visiting the website), and e is approximately equal to 2.71828 (Euler's number). To find the probability that more than 3 users visit the website, we need to calculate 1 minus the sum of the probabilities for 0, 1, 2, and 3 visitors, i.e., 1 - P(X ≤ 3).
Here's the calculation:
- P(X = 0) = (2.71828-2.6 × 2.60) / 0! = 0.074273578
- P(X = 1) = (2.71828-2.6 × 2.61) / 1! = 0.193110904
- P(X = 2) = (2.71828-2.6 × 2.62) / 2! = 0.250844175
- P(X = 3) = (2.71828-2.6 × 2.63) / 3! = 0.217233052
Add these probabilities together:
P(X ≤ 3) = 0.074273578 + 0.193110904 + 0.250844175 + 0.217233052 = 0.735461709
Now, to find P(X > 3), subtract P(X ≤ 3) from 1:
P(X > 3) = 1 - P(X ≤ 3) = 1 - 0.735461709 = 0.264538291.
Therefore, the probability that more than 3 users visit the website in a given minute is approximately 0.2645 or 26.45%.