Final answer:
Given the variance of the exponential random variable Y, its probability density function (pdf) is determined, then E[Y²] is calculated using the properties of exponential distribution, and the probability P[Y > 5] is found using the survival function.
Step-by-step explanation:
The variance of an exponential random variable Y is equal to the square of its mean (μ2).
Therefore, given that var[Y] = 25, we can deduce that μ = √25 = 5.
This leads us to the probability density function (pdf) of Y, which, for an exponential distribution, is given by
f(y) = (1/μ)e-y/μ for y ≥ 0.
Hence, the pdf of Y is f(y) = (1/5)e-y/5 for y ≥ 0.
To find E[Y2], we use the fact that for an exponential random variable, E[Y2] = 2μ2.
Thus, E[Y2] = 2(5)2 = 50.
For the probability that Y > 5, P[Y > 5], we integrate the pdf from 5 to infinity: P[Y > 5] = ∫5∞ f(y)dy which equals e-5/5. This is simply the exponential survival function evaluated at 5, which gives us P[Y > 5] = e-1 ≈ 0.3679.