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Re: The random variable x has the following continuous probabil [#permalink]
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Yes. True. They are equal.

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Re: The random variable x has the following continuous probabil [#permalink]
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]
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Attachment:
triangle.jpg
triangle.jpg [ 28.16 KiB | Viewed 14202 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.

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Re: The random variable x has the following continuous probabil [#permalink]
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:

Image

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]
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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]
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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]
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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]
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]
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Attachment:
InkedInkedtriangle_LI.jpg
InkedInkedtriangle_LI.jpg [ 856.62 KiB | Viewed 13247 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]
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]
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]
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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 [#permalink]
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.

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


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?
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Re: The random variable x has the following continuous probabil [#permalink]
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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.
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Re: The random variable x has the following continuous probabil [#permalink]
Hey Carcass where can I find more of this type of problem here? Probability combined with Normal distribution problems

Carcass wrote:
Yes. True. They are equal.

Regards
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Re: The random variable x has the following continuous probabil [#permalink]
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Chaithraln2499 wrote:
Hey Carcass where can I find more of this type of problem here? Probability combined with Normal distribution problems

Carcass wrote:
Yes. True. They are equal.

Regards


mmhhhhhhhh good question. Usually, the questions that covers more than one area of concepts are a bit difficult to find

For probability, you can use the specific tag https://gre.myprepclub.com/forum/search.ph ... tag_id=243

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

The consequences is you can have all the questions that contain in them the words above. It is basically a search by key words. Of course this is the result https://gre.myprepclub.com/forum/search.ph ... mit=Search

But you can pick up to search ONLY inside the sub forums: QCQC, PS, numeric entry , and so on

This is for PS https://gre.myprepclub.com/forum/search.ph ... mit=Search

Here is an example https://gre.myprepclub.com/forum/the-stand ... tml#p13490

Let me know if you do need further assistance
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Re: The random variable x has the following continuous probabil [#permalink]
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