Augustine S. answered • 12/03/23
PhD Candidate at Georgia Tech. 6+ years of Tutoring Experience.
X is a discrete random variable with Poisson(λ) distribution. Note that we use Poisson distributions to express "number of times something is going to happen", so of course X≥0. Its pmf (probability mass function) is given by:
f(k) = P(X=k) = (λ^k * exp(-λ)) / k!
Now that we know that, let's tackle a and b together (the rest then should be straightforward). Let's start with b which is a little easier.
b) P(X≤1) = P(X=0) + P(X=1) = f(0) + f(1) = 0.5^0 * exp(-0.5) / 0! + 0.5^1 * exp(-λ) / 1! = exp(-0.5) + 0.5 * exp(-0.5) = 0.9098
So this is a pretty high probability. This is because Poisson pmf with small λ are heavily concentrated close to 0 (see wiki also Wikipedia page about Poisson distribution).
a) P(X≥2) = 1 - P(X=0) - P(X=1) = 1 - f(0) - f(1) = 1 - 5^0 * exp(-5) / 0! - 5^1 * exp(-5) / 1! = 1- exp(-5) - 5*exp(-5) = 1 - 0.0404 = 0.9596
Now that we have a high λ, the pmf is going to be more shifted to the right, so it makes sense that this is a high probability (note that in the previous question with λ=0.5, we would get P(X≥2) = 1 - P(X≤1) = 0.0902 which is completely different. So λ plays a great role here.
Similarly, you should be able to answer all questions. If you need any additional help, please feel free to contact me!
