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The random variable x has the following continuous probability distrib

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The random variable x has the following continuous probability distrib  [#permalink]

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New post 08 Jul 2018, 05:27
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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:

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. \(\frac{\sqrt{2}}{2}\)


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

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New post 08 Jul 2018, 06:03
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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:

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. \(\frac{\sqrt{2}}{2}\)


Attachment:
The attachment BK8gXRh.jpg is no longer available


An interesting question...

The frequency of x is given by the slanting line and x is between 0 and √2

Now the median will require knowing the centre point of this so formed triangle with x axis and y axis..

It will happen where a line drawn horizontally divides the so formed triangle in two equal areas..

Area of bigger triangle = 1/2 *√2*√2=1
So the are of smaller triangle should be 1/2
And let this happen at distance A from √2. Since the triangles - smaller and bigger - both are similar, the vertical line will also be A.

Area now =1/2 *A*A=1/2....A^2=1...A=1
The x axis is √2 and median lies 1 away from √2, so √2-1

C
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The random variable x has the following continuous probability distrib  [#permalink]

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New post 08 Jul 2018, 06:15
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:

The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).



Bunuel,

The question statement says \(0 ≤ x ≤\sqrt{2}\) and from the graph, the probability of x being \(\sqrt{2}\) is 0. So I guess the question statement should say \(0 ≤ x <\sqrt{2}\) ?
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Re: The random variable x has the following continuous probability distrib  [#permalink]

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New post 08 Jul 2018, 06:22
The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

Hi chetan2u Sir,
Could you please explain the significance of the highlighted information provided in the question?

And how this info is helpful in solving the question?

Thanking you.
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Re: The random variable x has the following continuous probability distrib  [#permalink]

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New post 08 Jul 2018, 06:32
PKN wrote:
The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

Hi chetan2u Sir,
Could you please explain the significance of the highlighted information provided in the question?

And how this info is helpful in solving the question?

Thanking you.


Same wine in different bottle
The point tells us that the left edge of x is 0 and right edge is √2.
Although same info is also stated by 0<=x<=√2, but we should be able to make out the redundant matter and not pay attention on it.
If someone starts spending time between the two statements pointing towards same thing, the Question / test makers aim is achieved.

So read the info as gi en and discard what is not important or what gives you the same wine in a different bottle.
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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: The random variable x has the following continuous probability distrib &nbs [#permalink] 08 Jul 2018, 06:32
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