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The random variable x has the following continuous probabil
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26 Aug 2017, 01:55
26 Aug 2017, 01:55
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25% (02:38) correct
74% (01:58) wrong based on 138 sessions
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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. \(\frac{\sqrt{2}}{2}\)
Re: The random variable x has the following continuous probabil
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03 Mar 2018, 04:34
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.
Skewed.png [ 131.58 KiB | Viewed 15034 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://gre.myprepclub.com/forum/download/ ... iew&id=923
Area 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.
Inkedtriangle_LI.jpg [ 841.95 KiB | Viewed 15054 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)
Best understood with following image.
Attachment:
Skewed.png [ 131.58 KiB | Viewed 15034 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://gre.myprepclub.com/forum/download/ ... iew&id=923
Area 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 15054 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)
Re: The random variable x has the following continuous probabil
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18 Sep 2017, 02:25
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. \(\frac{\sqrt{2}}{2}\)
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.
