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 3 independent peak-load readings. Each has CDF \(F(x)=\tfrac12[1+\sin(\pi x/4)]\) for \(-2\le x\le 2\). Let \(M=\max(X_1,\ldots,X_3)\) and \(m=\min(X_1,\ldots,X_3)\), where \(X_i\) is the \(i\)th reading. Calculate the expected largest reading.
(A)0.1875
(B)0.3750
(C)0.5625
(D)0.7500
(E)0.5000
📖 Worked Solution & Strategy
The \(3\) readings \(X_1, \ldots, X_{3}\) are independent and identically distributed on \([-2,2]\) with CDF \(F(x) = \frac12[1+\sin(\pi x/4)]\). For the maximum \(M = \max(X_1,\ldots,X_{3})\), independence gives \(F_M(x) = \Pr(X_1 \le x, \ldots, X_{3} \le x) = F(x)^{3}\). For a continuous random variable supported on \([a,b]\) with \(b=2\), the expectation tail identity gives \(\operatorname{E}[M] = b - \int_a^b F_M(x)\,dx = 2 - \int_{-2}^2 F(x)^{3}\,dx = 2 - \int_{-2}^2 \left[\frac12(1+\sin(\pi x/4))\right]^{3}\,dx = \frac{3}{4}\). The evaluated result is \(\frac{3}{4}\), which matches the correct option.
Therefore, the result is \(\frac{3}{4}\approx 0.7500\), so the correct answer is option (D).

Final Answer: Option (D)

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