How to Solve Multivariate Probability — Multivariate — maxima and minima from a known CDF Questions on Exam P

Exam P Topic: Multivariate Probability — Multivariate — maxima and minima from a known CDF Verified Procedural Question
Sample Practice Problem ID: #6KW4O
A materials lab records 4 independent peak-load readings. Each has CDF \(F(x)=\tfrac12[1+\sin(\pi x/4)]\) for \(-2\le x\le 2\). Let \(M=\max(X_{1},\ldots,X_{4})\) and \(m=\min(X_{1},\ldots,X_{4})\), where \(X_{i}\) is the \(i\)th reading. Calculate the expected largest reading.
(A)0.4531
(B)0.6796
(C)0.6041
(D)0.9062
(E)0.75
📖 Worked Solution & Strategy
The \(4\) readings \(X_{1}, \ldots, X_{4}\) are independent and identically distributed on \([-2,2]\) with CDF \(F(x) = \frac12[1+\sin(\pi x/4)]\). For the maximum \(M = \max(X_{1},\ldots,X_{4})\), independence gives \(F_{M}(x) = \Pr(X_{1} \le x, \ldots, X_{4} \le x) = F(x)^{4}\). For a continuous random variable supported on \([a,b]\) with \(b=2\), the expectation tail identity gives \(\operatorname{E}[M] = b - \int_{a}^b F_{M}(x)\,dx = 2 - \int_{-2}^2 F(x)^{4}\,dx = 2 - \int_{-2}^2 \left[\frac12(1+\sin(\pi x/4))\right]^{4}\,dx = \frac{29}{32}\).
Therefore, the result is \(\frac{29}{32}\approx 0.9062\), so the correct answer is option (D).

Final Answer: Option (D)

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