From a recent statistical analysis for the last five years, on an average there
are 4 (major) air accidents per month in the world. Let X be the number of air accidents occurred in a randomly selected month. It is known that X ~ Poisson() approximately, where the intensity 4 accidents (monthly average number of accidents). Find the probability that there will be 4 or more air accidents
Sooo...I see you are having problems with this too! LOL
Try the Poisson Distribution.
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Values:
x = 0,1,2,3
m = 4
Calculate P(0), P(1), P(2), P(3). Add together, then subtract that value from 1. This will be your probability.
I hope this will help get you started.
To find the probability that there will be 4 or more air accidents in a randomly selected month, we can use the Poisson distribution. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space.
In this case, X follows a Poisson distribution with an intensity (or average rate) of λ ≈ 4 accidents per month. The probability mass function (PMF) of the Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
where e is the base of the natural logarithm and k! represents the factorial of k.
To find the probability P(X ≥ 4), we need to calculate the sum of the probabilities for X = 4, 5, 6, 7, and so on, up to infinity. However, calculating an infinite sum is not practical. Thankfully, we can use the complement rule to simplify the calculation.
The complement rule states that:
P(X ≥ k) = 1 - P(X < k)
Therefore, in our case:
P(X ≥ 4) = 1 - P(X < 4)
To find P(X < 4), we can calculate the sum of the probabilities for X = 0, 1, 2, and 3. Let's calculate it step by step:
P(X = 0) = (e^(-λ) * λ^0) / 0! = e^(-4) ≈ 0.0183
P(X = 1) = (e^(-λ) * λ^1) / 1! = e^(-4) * 4 / 1 ≈ 0.0733
P(X = 2) = (e^(-λ) * λ^2) / 2! = e^(-4) * 4^2 / 2 ≈ 0.1465
P(X = 3) = (e^(-λ) * λ^3) / 3! = e^(-4) * 4^3 / 6 ≈ 0.1953
Now, we can calculate P(X < 4) by summing up these probabilities:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) ≈ 0.0183 + 0.0733 + 0.1465 + 0.1953 ≈ 0.4334
Finally, we can calculate P(X ≥ 4) using the complement rule:
P(X ≥ 4) = 1 - P(X < 4) ≈ 1 - 0.4334 ≈ 0.5666
Therefore, the probability that there will be 4 or more air accidents in a randomly selected month is approximately 0.5666, or 56.66%.