A sample of size 10 has density 2(x−μ)e^(−(x−μ)²) for x>μ, zero otherwise. Its observed minimum is 1. Using min X_i−μ as pivot, a 95% confidence interval for μ is (c,1). Find c to two decimals.
For T=min X_i−μ≥0, P(T>t)=e^(−10t²). Choose a with P(0<T<a)=0.95: e^(−10a²)=0.05, so a=√(−ln0.05/10). The event is min X_i−a<μ<min X_i. Thus c=1−a≈**0.45**.
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