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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 \(x > \sqrt{2} = 0\).
Re: The random variable x has the following continuous probabil
[#permalink]
03 Mar 2018, 04:34
12
Expert Reply
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 13898 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.
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 13922 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]
18 Sep 2017, 02:25
6
1
Bookmarks
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.
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
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!
Re: The random variable x has the following continuous probabil
[#permalink]
30 Sep 2018, 04:35
3
Expert Reply
Attachment:
InkedInkedtriangle_LI.jpg [ 856.62 KiB | Viewed 13087 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]
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
Re: The random variable x has the following continuous probabil
[#permalink]
19 Nov 2018, 14:21
1
Expert Reply
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.
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Re: The random variable x has the following continuous probabil
[#permalink]
11 Jun 2020, 19:33
sandy wrote:
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
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.
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
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)
Great explanation, but I wonder if the slope is -1 as opposed to 1? If we apply rise-over-run here, the slope would be negative. Am I getting it wrong somehow?
Re: The random variable x has the following continuous probabil
[#permalink]
20 Jun 2021, 04:53
1
A continuous probability distribution has a total area of 100%, or 1, underneath the entire curve. The median of such a distribution splits the area into two equal halves, with 50% of the area to the left of the median and the other 50% to the right of the median: In simpler terms, the random variable x has a 50% chance of being above the median and a 50% chance of being below the median. You can ignore the regions to the right or the left of this triangle, since the probability that x could fall in either of those regions is zero. So the question becomes this: what point on the x-axis will divide the large right triangle into two equal areas? One shortcut is to note that the area of the large isosceles right triangle must be 1, which equals the total area under any probability distribution curve. Confirm by means of the area formula for this right triangle: The quickest way to find the median is to consider the small isosceles right triangle. Triangle ABC must have an area of 1/2. So what must be the length of each of its legs, AB and BC? From the formula 1/2bh = 1/2, and noting that the base BC equals the height AB, the base BC must be 1 (the same as the height). Since the coordinates of point C are (\sqrt{2} , 0), the coordinates of point B must be (\sqrt{2} – 1, 0). That is, the median is \sqrt{2}– 1.
For normal distribution (we do not have a specific tag. And neither of GMATclub there is) you can use a trick: put the words normal distribution in the lens on top of every page