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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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08 Jul 2018, 05:27
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10% (02:12) correct 90% (01:56) wrong based on 156 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. $$\frac{\sqrt{2}}{2}$$

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

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08 Jul 2018, 06:03
2
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 $$x > \sqrt{2} = 0$$.

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

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27 Sep 2018, 11:53
1
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:

If Prob($$x < 0$$) = Prob($$x > \sqrt{2}$$) = 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. $$\frac{\sqrt{2}}{2}$$

We are talking about the median of a continuous random variable with a given probability density function... and all this wording is absolutely out-of-GMAT´s-scope!

On the other hand, the same problem could be put into GMAT´s reality (for instance) like that:

Quote:
In the figure above, what is the value of x such that the areas under the line from 0 to x, and from x to $$\sqrt{2}$$, coincide?

$$? = x\,\,\,\,:\,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = {S_{\,{\text{III}}}}\,\,\,\,\,\,\left( * \right)$$

$$\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{{S_{\,{\text{total}}}}}}{2} = \frac{1}{2}\left( {\frac{{\sqrt 2 \cdot \sqrt 2 }}{2}} \right) = \frac{1}{2}$$

$$\frac{1}{2} = {S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{x \cdot x}}{2} + x\left( {\sqrt 2 - 1} \right) = x\sqrt 2 - \frac{{{x^2}}}{2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,2x\sqrt 2 - {x^2} = 1\,$$

Now we have at least two possible continuations:

First: test alternatives (easy but not so quick).

Second: solve the second-degree equation (need "insight", but VERY quick), as follows (for instance):

$$2x\sqrt 2 - {x^2} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} - 2x\sqrt 2 + 1 = 0$$

$${\text{Sum}}\,\, = \,\,\,2\sqrt 2 \,\,\,\,\,,\,\,\,\,{\text{Product}}\,\, = \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x_1} = \sqrt 2 - 1\,\,,\,\,\,{x_2} = \sqrt 2 + 1$$

$${x_2} > \sqrt 2 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {x_1} = \sqrt 2 - 1$$

This solution follows the notations and rationale taught in the GMATH method.

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

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27 Sep 2018, 12:19
workout wrote:
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 $$0 ≤ x <\sqrt{2}$$ ?

Hi, workout!

I have taken the liberty of answering your question. I hope you both (Bunuel and yourself) don´t mind!

When dealing with continuous random variables, the probability of choosing ANY particular value is ALWAYS zero, therefore it really doesn´t matter if you consider to include, or not, any of the interval extremities...

When dealing with continuous random variables, it is useful to consider intervals, not "points", for example(s):

What is the probability that you will buy a light bulb that will not burn-out in less than 2h?

In this case, you are interested in the interval [0,2] , or ]0,2] , or [0,2[ , or ]0,2[ ...

Curiosity: in this situation, it is commonly considered an exponential distribution (see image below).

(The graphs are related to the density and accumulated functions.)

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

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18 Oct 2018, 14:19
Hi fskilnik

I've a problem with the usage of the term "median" here. For e.g. the median of set A (1,2,100) is 2 although 100 not equal 1. So in our case, the median of x must be the value corresponding to x= root2/2. What's wrong here?
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The random variable x has the following continuous probability distrib  [#permalink]

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18 Oct 2018, 17:00
1
HisHo wrote:
Hi fskilnik

I've a problem with the usage of the term "median" here. For e.g. the median of set A (1,2,100) is 2 although 100 not equal 1. So in our case, the median of x must be the value corresponding to x= root2/2. What's wrong here?

Hi, HisHo.

Thank you for your interest in the subject, even with my explicit mention that it is out of GMAT´s scope.

It is NOT as if x denotes the "position" in a list and y denotes the corresponding value at that position.

In this context, the vertical axis offers probabilities associated with values (intervals of values) in the horizontal ("x") axis.

When we are asked the median value of x, it is really the median value of the possible values of x, taking into account the "frequency" of these values that is given (shown) in the vertical axis. (This sentence will be explained more explicitly below.)

If you understood my last sentence, now it must be clear the answer is not:

(1) y=f(a) , where a is (square root of 2)/2
(2) y=f(b) , for any other b

In both cases because y does not have this "value" property associated with the position that would be given in the x axis.
(This is not what is going on, so to speak!)

(3) (square root of 2)/2 , because the "distribution of frequencies" is "irregular", I mean, the smaller (positive) values of x are more frequent...
Consequence: the value x=m (the median) that divides the "list" of x´s (the whole interval from 0 to the square root of 2) such that,
the probability of randomly choosing a number in the interval [0,m] is 50% , is less the (square root of 2)/2 ... got it?

I hope you understood (and liked) my INFORMAL comments.

Regards and success in your studies,
Fabio.
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Re: The random variable x has the following continuous probability distrib  [#permalink]

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16 Oct 2019, 18:49

Kind regards!
Re: The random variable x has the following continuous probability distrib   [#permalink] 16 Oct 2019, 18:49
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