Probability Functions of Random Variable Question

[a_chak09@hotmail.com]

Suppose X has density function x^(-1/2)/2 for 0 < x < 1 and 0 otherwise. Find:
(a) the distribution function
(b) P(X > 3/4)
(c) P(1/9 < X > 1/4)

Attempted Answer:
(a) Distribution Function:

. . .\(\displaystyle \dfrac{1}{2}\,\)\(\displaystyle \displaystyle \int\, x^{-1/2}\, dx\, =\, \sqrt{\strut x\,}\)

Answer: F(X) = x^(1/2) <== CORRECT

(b) Probability:

. . .\(\displaystyle \displaystyle \begin{align}P\,\left(X\, >\, \dfrac{3}{4}\right)\, &=\, \int_{3/4}^1\, x^{-1/2}\, dx

\\ \\ &=\, \sqrt{\strut x\,}\, \bigg|_{3/4}^1

\\ \\ &=\, 1\, -\, \sqrt{\strut \dfrac{3}{4}\,} \end{align}\)

Answer: 1 - 1/(3^(1/2)). .<== WRONG

(Answer: 0.13)

(c) Probability:

. . .\(\displaystyle |displaystyle \begin{align} P\, \left(\dfrac{1}{9}\, <\, X\, <\, \dfrac{1}{4}\right)\, &=\, \int_0^{1/4}\, \dfrac{x^{-1/2}}{2}\, dx\, -\, \int_0^{1/9}\, \dfrac{x^{-1/2}}{2}\, dx

\\ \\ &=\, \sqrt{\strut x\,}\,\bigg|_0^{1/4}\, -\, \sqrt{\strut x\,}\,\bigg|_0^{1/9}

\\ \\ &=\, \sqrt{\strut \dfrac{1}{4}\,}\, -\, \sqrt{\strut \dfrac{1}{9}\,}

\\ \\ &=\, \dfrac{1}{6} \end{align}\) Answer: 1/6 . .<== CORRECT

I HAVE NO IDEAS WHAT B IS!!

 

[Ishuda]

Suppose X has density function x^(-1/2)/2 for 0 < x < 1 and 0 otherwise. Find:
(a) the distribution function
(b) P(X > 3/4)
(c) P(1/9 < X > 1/4)

Attempted Answer:
(a) Distribution Function: F(X) = x^(1/2) <-- CORRECT
(b) P(X > 3/4) = integral[x=3/4 to x=1] (x^(-1/2)/2)dx = 1 - sqrt(3/4) <-- WRONG (Answer: 0.13)
(c) P(1/9 < X > 1/4) = integral[x=1/4 to x=0] (x^(-1/2)/2)dx - integral[x=1/9 to x=0] (x^(-1/2)/2)dx = 1/6 <-- CORRECT

I HAVE NO IDEAS WHAT B IS!!

For (b), what is the value of \(\displaystyle 1\, -\, \sqrt{\frac{3}{4}}\). Seem to me it is pretty close to 0.13

 

[pka]

Suppose X has density function x^(-1/2)/2 for 0 < x < 1 and 0 otherwise. Find:
(a) the distribution function
(b) P(X > 3/4)
(c) P(1/9 < X > 1/4)

c) THAT means absolutely nothing. The question in written incorrectly.
It could be \(\displaystyle \mathcal{P}\left(\frac{1}{9}<X<\frac{1}{4}\right)=\displaystyle\int_{1/9}^{1/4} {{{(2\sqrt x )}^{ - 1}}dx} \) It cannot be as written.

b) \(\displaystyle \mathcal{P}\left(X>\frac{3}{4}\right)= \displaystyle \int_{3/4}^1 {{{(2\sqrt x )}^{ - 1}}dx} \).

 

[Ishuda]

Suppose X has density function x^(-1/2)/2 for 0 < x < 1 and 0 otherwise. Find:
(a) the distribution function
(b) P(X > 3/4)
(c) P(1/9 < X > 1/4)

c) THAT means absolutely nothing. The question in written incorrectly.
It could be \(\displaystyle \mathcal{P}\left(\frac{1}{9}<X<\frac{1}{4}\right)=\displaystyle\int_{1/9}^{1/4} {{{(2\sqrt x )}^{ - 1}}dx} \) It cannot be as written.

b) \(\displaystyle \mathcal{P}\left(X>\frac{3}{4}\right)= \displaystyle \int_{3/4}^1 {{{(2\sqrt x )}^{ - 1}}dx} \).

Isn't that obviously a typo? But then I guess I've sort of turned into a nit-picker also as you can see in one of my recent posts. Thank you for pointing that out though.

 

[stapel]

(b) Probability:

. . .\(\displaystyle \displaystyle \begin{align}P\,\left(X\, >\, \dfrac{3}{4}\right)\, &=\, \int_{3/4}^1\, x^{-1/2}\, dx

\\ \\ &=\, \sqrt{\strut x\,}\, \bigg|_{3/4}^1

\\ \\ &=\, 1\, -\, \sqrt{\strut \dfrac{3}{4}\,} \end{align}\)

Answer: 1 - 1/(3^(1/2)). .<== WRONG

(Answer: 0.13)

How did you go from:

. . . . .\(\displaystyle 1\, -\, \sqrt{\strut \dfrac{3}{4}\,}\, =\, 1\, -\, \dfrac{\sqrt{\strut 3\,}}{\sqrt{\strut 4\,}}\, =\, 1\, -\, \dfrac{\sqrt{\strut 3\,}}{2}\)

...to:

. . . . .\(\displaystyle 1\, -\, \dfrac{1}{\sqrt{\strut 3\,}}\)

What were your steps? What happened to the 4?

Note to others: The erroneous "greater than" symbol was in the post, but not in the attached image. The post has been corrected through inclusion of the text from that image.