Bunuel wrote:
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:
If Prob(\(x < 0\)) = Prob(\(x > \sqrt{2}\)) = 0, what is the median of x?
A. \(\frac{\sqrt{2} - 1}{2}\) \(\,\,\,\,\,\,\,\) B. \(\frac{\sqrt{2}}{4}\) \(\,\,\,\,\,\,\,\) C. \(\sqrt{2} - 1\) \(\,\,\,\,\,\,\,\) D. \(\frac{\sqrt{2} + 1}{4}\) \(\,\,\,\,\,\,\,\) E. \(\frac{\sqrt{2}}{2}\)
We are talking about the median of a continuous random variable with a given probability density function... and all this wording is absolutely out-of-GMAT´s-scope!
On the other hand, the same problem could be put into GMAT´s reality (for instance) like that:
Quote:
In the figure above, what is the value of x such that the areas under the line from 0 to x, and from x to \(\sqrt{2}\), coincide?
\(? = x\,\,\,\,:\,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = {S_{\,{\text{III}}}}\,\,\,\,\,\,\left( * \right)\)
\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{{S_{\,{\text{total}}}}}}{2} = \frac{1}{2}\left( {\frac{{\sqrt 2 \cdot \sqrt 2 }}{2}} \right) = \frac{1}{2}\)
\(\frac{1}{2} = {S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{x \cdot x}}{2} + x\left( {\sqrt 2 - 1} \right) = x\sqrt 2 - \frac{{{x^2}}}{2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,2x\sqrt 2 - {x^2} = 1\,\)
Now we have at least two possible continuations:
First: test alternatives (easy but not so quick).
Second: solve the second-degree equation (need "insight", but VERY quick), as follows (for instance):
\(2x\sqrt 2 - {x^2} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} - 2x\sqrt 2 + 1 = 0\)
\({\text{Sum}}\,\, = \,\,\,2\sqrt 2 \,\,\,\,\,,\,\,\,\,{\text{Product}}\,\, = \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x_1} = \sqrt 2 - 1\,\,,\,\,\,{x_2} = \sqrt 2 + 1\)
\({x_2} > \sqrt 2 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {x_1} = \sqrt 2 - 1\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.