2

I need to understand an explanation from a swirl() lesson in a course on "Statistical Inference", menu option "Probability2". At 76% of completion, there is a question:

The quantile v of a CDF is the point x_v at which the CDF has the value v. More precisely, F(x_v)=v. A percentile is a quantile in which v is expressed as a percentage.

What percentile is the median?
The answer is "50th"... ok...

Right after, there is another question:

What is the 50th percentile of the CDF F(x)=(x^2)/4 from the example above?

I know from here, the answer is $1.414214$, but I cannot find the reasoning to reach the result. I'm not sure if it is just a simple semantics catch, but I'm finding hard to understand how to reach the answer. Any idea or help?

Community
  • 1
Pedro Carneiro Jr.
  • 21
  • 2
  • 1
    Welcome to CV. There's a good discussion of this on CV here ... http://stats.stackexchange.com/questions/137931/when-would-we-use-tantiles-and-the-medial-rather-than-quantiles-and-the-median Basically, the CDF is a weighted value where the 50% "tantile" is called a *medial*. – Mike Hunter Nov 20 '15 at 17:17
  • It's easy to understand why you haven't attained an answer yet: the question is ill-posed, because $x^2/4$ clearly is not a CDF. Your source likely intends this definition to apply only for $x$ between $0$ and $2$, setting the value to $1$ for $x\gt 2$ and $0$ for $x\le 0$, but this is only a supposition. – whuber Dec 14 '15 at 19:49

2 Answers2

2

Median is when F(x)=(x^2)/4 = 0.5, that would be the 50th percentile. So we want to solve this:
(x^2)/4 = 1/2
We go like this:
x^2/2 = 1
And we soon arrive at this:
x^2 = 2.
And that means we need sqrt(2) to obtain the x which is 1.41...

Techwych
  • 21
  • 2
1

I just did this. They are looking for sqrt(2), but it should be entered as equiv_val(sqrt(2)) which gives the result you found.

gung - Reinstate Monica
  • 132,789
  • 81
  • 357
  • 650
David
  • 11
  • 1
  • Could you please explain what assumptions you made and calculations you carried out to arrive at this answer? – whuber Dec 14 '15 at 19:51