If η1, η2, η3, η4, η5 are independent and identically distributed exponential random variables with the parameter λ.
So far I've done this:
P (max(η1, …, η5) ≤ a) = P (η1 ≤ a ∩ η2 ≤ a ∩ η3 ≤ a ∩ η4 ≤ a ∩ η5 ≤ a) = P (η1 ≤ a) * P (η2 ≤ a) ... P (η5 ≤ a) = [F(x)]^5
Let $W = \max(X_1, \dots, X_{5}),$ where $X_i$ are a random sample from $\mathsf{Exp}(\lambda)$, where $\lambda = 1/\mu$ is a rate parameter.
$$F_W(a) = P(W \le a) = P(X_1 \le a) \times \cdots \times P(X_5 \le a) = (1- e^{-\lambda a})^5.$$
PDF can be found by differentiating CDF, but that gets messy with five random variables. $E(W) = 1/5\lambda + 1/4\lambda + 1/3\lambda + 1/2\lambda + 1/\lambda = 2.283333/\lambda.$ [Intuitive argument: Suppose $W$ is the waiting for the failure of a parallel system with 5 components, each with failure rate $\lambda.$ The waiting time for the first failure is $1/5\lambda.$ By the no-memory property of exponentials, the additional waiting time for the second component to fail is $1/4\lambda,$ and so on.]