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The random variable x has the following continuous probabil [#permalink]
26 Aug 2017, 01:55
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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: The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\). 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. \(\frac{\sqrt{2}}{2}\)
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Re: The random variable x has the following continuous probabil [#permalink]
18 Sep 2017, 02:25
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Carcass 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: The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\). 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. \(\frac{\sqrt{2}}{2}\) \(\sqrt{2}^{1}\) and \(\frac{\sqrt{2}}{2}\) are the same expression. One is the rationalized form of the other. Thus they should be both right.



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Re: The random variable x has the following continuous probabil [#permalink]
18 Sep 2017, 05:47
Yes. True. They are equal. Regards
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Re: The random variable x has the following continuous probabil [#permalink]
28 Feb 2018, 01:47
Hey Carcass, can you explain the answer to this question?



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Re: The random variable x has the following continuous probabil [#permalink]
28 Feb 2018, 12:51
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Attachment:
triangle.jpg [ 28.16 KiB  Viewed 1649 times ]
Actually, the question boils down to this question. which point halves the triangle into to part ?' because the max probability is always 1. In this case, you have equal probability that x is on the right part of the triangle or in the left. If the coordinates of C are (\(\sqrt{2}\), 0 ), then the answer is \(\sqrt{2}\) minus the only point below this, which means 1. C is the answer., Hope now is clear. Regards
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Re: The random variable x has the following continuous probabil [#permalink]
03 Mar 2018, 03:55
IlCreatore wrote: Carcass 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: The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\). 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. \(\frac{\sqrt{2}}{2}\) \(\sqrt{2}^{1}\) and \(\frac{\sqrt{2}}{2}\) are the same expression. One is the rationalized form of the other. Thus they should be both right. why not give full explanation? why GMATclub's rules doesnt applied here



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Re: The random variable x has the following continuous probabil [#permalink]
03 Mar 2018, 04:02
The same applies here. Above is a FULL explanation. The more the questions are tricky the more they boil down in few concepts to solve them. Do not know why you said that above is not a full explanation
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Re: The random variable x has the following continuous probabil [#permalink]
03 Mar 2018, 04:34
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A median value of any probability distribution divides the area under the probabaility distribution in two equal parts. Best understood with following image. Attachment:
Skewed.png [ 131.58 KiB  Viewed 1574 times ]
Now in our case we need to split the triangular distribution along a line \(x=?\) (parallel to y axis) such that area of the right half is same as the left half. https://greprepclub.com/forum/download/ ... iew&id=923Area of right half = \(\frac{1}{2} \times\) Area of the larger triangle. Now look at the figure below and we have marked out the hight and length of the triangle as x. Now height = length for this triangle because the given line has slope 1. Attachment:
Inkedtriangle_LI.jpg [ 841.95 KiB  Viewed 1575 times ]
Area of right half = \(\frac{1}{2} \times x^2\)= \(\frac{1}{2} \times \frac{1}{2}\times \sqrt{2}^2\). Solving for x we get x =1. So median value has to be \(\sqrt{2}1\) (refer to the figure above)
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Re: The random variable x has the following continuous probabil [#permalink]
29 Sep 2018, 04:49
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@Sandy @Carcass can you explain why the smaller triangle is definitively an isosceles triangle?



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Re: The random variable x has the following continuous probabil [#permalink]
29 Sep 2018, 05:35
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yashkanoongo wrote: @Sandy @Carcass can you explain why the smaller triangle is definitively an isosceles triangle? Because it is given in the question that the slope of the line is 1. So any triangle made from the line and line parallel to y axis will be an isosceles right triangle.
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Re: The random variable x has the following continuous probabil [#permalink]
30 Sep 2018, 04:02
sandy wrote: yashkanoongo wrote: @Sandy @Carcass can you explain why the smaller triangle is definitively an isosceles triangle? Because it is given in the question that the slope of the line is 1. So any triangle made from the line and line parallel to y axis will be an isosceles right triangle. Thanks for replying man but can you elaborate on your explanation or point me in the direction of a source where I can read up more about this. I dont entirely understand the current explanation. Thanks for the help!



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Re: The random variable x has the following continuous probabil [#permalink]
30 Sep 2018, 04:35
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Attachment:
InkedInkedtriangle_LI.jpg [ 856.62 KiB  Viewed 803 times ]
Equation of line making 45 degrees as shown in the figure above is x+y=100 (say, it can be any number) what is the value of y at a point x=10? y=90. Distance between the point (10,0) and (100, 0) is also 90. This this makes the triangle isosceles. Hope this clears up the doubt. This is not exactly some concept just basic geometry, so id ont know excatly which resource would be the correct recommendation for this.
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Re: The random variable x has the following continuous probabil [#permalink]
30 Sep 2018, 04:52
sandy wrote: Attachment: InkedInkedtriangle_LI.jpg Equation of line making 45 degrees as shown in the figure above is x+y=100 (say, it can be any number) what is the value of y at a point x=10? y=90. Distance between the point (10,0) and (100, 0) is also 90. This this makes the triangle isosceles. Hope this clears up the doubt. This is not exactly some concept just basic geometry, so id ont know excatly which resource would be the correct recommendation for this. This definitely clears the doubt, thanks a lot!



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Re: The random variable x has the following continuous probabil [#permalink]
02 Nov 2018, 09:50
I used an equation of this sort 1/2 * (root(2)  x) * y = x * y + 1/2 * (root(2)  y) * x, provided (x,y) divide the area into two halves Not sure how to proceed



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Re: The random variable x has the following continuous probabil [#permalink]
19 Nov 2018, 14:21
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indiragre18 wrote: I used an equation of this sort 1/2 * (root(2)  x) * y = x * y + 1/2 * (root(2)  y) * x, provided (x,y) divide the area into two halves Not sure how to proceed Get the value of y in terms of x from the equation of the line \(x+y=\sqrt{2}\). You would get a quadratic equation in x then solve for x.
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Re: The random variable x has the following continuous probabil
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