Discussion
Given \((X, Y) \sim N_{2}(5, 10, 1, 25, \rho)\) and \(P[4 \lt Y \lt 16 | X = 5] = 0.954\), then \(\rho\) is equal to :
(It is given that \(\frac{1}{\sqrt{2\pi}}\int_{0}^{2} e^{-\frac{1}{2}z^{2}} dz = 0.477\))
Answer: C
\(\rho\) is \(\pm 0.8\).
Here \(\mu_{X} = 5\), \(\mu_{Y} = 10\), \(\sigma_{X}^{2} = 1\), \(\sigma_{Y}^{2} = 25\). Given \(X = x\), Y is normal with mean \(10 + 5\rho(x - 5)\) and variance \(25(1 - \rho^{2})\). At \(x = 5\) the mean is 10, and the interval (4, 16) is \(10 \pm 6\). Using the given value, \(P(|Z| \lt 2) = 2(0.477) = 0.954\), so
\[ \begin{aligned} \frac{6}{5\sqrt{1 - \rho^{2}}} &= 2 \\ \sqrt{1 - \rho^{2}} &= 0.6, \quad \rho^{2} = 0.64 \end{aligned} \]
Because \(x = 5\) is the mean of X, the sign of \(\rho\) does not move the conditional mean, so the data fix only \(\rho^{2}\): \(\rho = \pm 0.8\). Choosing +0.8 or -0.8 alone claims more than the information allows, and 0.96 is not a solution.
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