Discussion
Let \(x\) and \(y\) be two independent normal random variables with mean \(\mu_{1}\) and \(\mu_{2}\) respectively and common variance \(\sigma^{2}\). For \(0 \lt \alpha \lt 1\) if the \(100\alpha\)th quantile of \(x\) is larger than \(100\alpha\)th quantile of \(y\), then :
Answer: B
The answer is \(\mu_{1} \gt \mu_{2}\).
The \(100\alpha\)th quantile of \(N(\mu, \sigma^{2})\) is \(\mu + \sigma z_{\alpha}\), where \(z_{\alpha}\) is the standard normal quantile. With a common \(\sigma\),
\[ \begin{aligned} &\mu_{1} + \sigma z_{\alpha} \gt \mu_{2} + \sigma z_{\alpha} \\ &\iff \mu_{1} \gt \mu_{2} \end{aligned} \]
Because the two distributions differ only by a shift, every quantile of \(x\) exceeds the corresponding quantile of \(y\) by \(\mu_{1} - \mu_{2}\). So the means cannot be equal or reversed, and something definite can be said.
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