a fun integration question!! (1 Viewer)

EvoRevolution · Apr 19, 2009

same question somethin was wrong with the other 1

$\120dpi The\ "normal\ distribution\ with\ mean\ \mu\ and\ standard\ deviation\ \sigma" \\is\ used\ in\ the\ HSC\ to\ scale\ marks.\ Its\ shape\ is\ given\ by\ the\ "density\ function\": \\ \\ \Phi(x) = \frac{1}{\sqrt{2}\pi \sigma }e^{\frac{-(x-\mu)^2}{2\sigma^2}} \\ \\ (a) find\ \Psi (x) \, the\ density\ function\for\ a\ standard\ deviation\ of\ unity\ and\ a\ mean\ of\ zero\\ \\ (b) Evaluate \int_{0}^{a}x\Psi (x)dx, where\ a \in R, a > 0.$

shuning · Apr 19, 2009

回复: a fun integration question!!

so i was right LOL the last 1 didnt seem right xD

Trebla · Apr 19, 2009

LOL, this is so stats! Stats is fun!
First one just sub in σ = 1 and μ = 0.
For the second one, it's:
$\dfrac{1}{\sqrt{2\pi}}\left(1-e^{-\frac{a^2}{2}}\right)$

Aside: For the integral in the second part, if the limits were from negative infinity to positive infinity, you get what is called the expected value of the normal density which is the mean μ (in this case it is zero) since the integrand would be an odd function...

gurmies · Apr 19, 2009

$\Psi (x) = \frac{1}{\sqrt{2}\pi}e^{\frac{-x^{2}}{2}}$

$\int_{0}^{a}x\Psi (x)dx = \frac{1}{\sqrt{2}\pi }\int_{0}^{a}xe^{\frac{-x^{2}}{2}}dx$

$= \frac{1}{\sqrt{2}\pi }\left [e^{\frac{-x^{2}}{2}} \right ]^{0}_{a}$

$= \frac{1}{\sqrt{2}\pi }\left [e^{\frac{-x^{2}}{2}} \right ]^{0}_{a} =\frac{1}{\sqrt{2}\pi} -\frac{e^{\frac{-a^{2}}{2}}}{\sqrt{2}\pi }$

Note: I switched limits because of a negative that comes out when you use reverse chain rule.

EvoRevolution · Apr 19, 2009

ok u guys win, ill post another one

gurmies · Apr 19, 2009

Am I correct?

EvoRevolution · Apr 19, 2009

gurmies said:
$\Psi (x) = \frac{1}{\sqrt{2}\pi}e^{\frac{-x^{2}}{2}}$

$\int_{0}^{a}x\Psi (x)dx = \frac{1}{\sqrt{2}\pi }\int_{0}^{a}xe^{\frac{-x^{2}}{2}}dx$

$= \frac{1}{\sqrt{2}\pi }\left [xe^{\frac{-x^{2}}{2}} \right ]^{0}_{a}$

$= \frac{1}{\sqrt{2}\pi }\left [xe^{\frac{-x^{2}}{2}} \right ]^{0}_{a} = -\frac{ae^{\frac{-a^{2}}{2}}}{\sqrt{2}\pi }$

Note: I switched limits because of a negative that comes out when you use reverse chain rule.

u integrated wrong, $\120dpi \int f'(x)e^{f(x)}=e^{f(x)}$

don't need the x

gurmies · Apr 19, 2009

Ah yes, pardon me - very silly.

Trebla · Apr 19, 2009

gurmies said:
$\Psi (x) = \frac{1}{\sqrt{2}\pi}e^{\frac{-x^{2}}{2}}$

$\int_{0}^{a}x\Psi (x)dx = \frac{1}{\sqrt{2}\pi }\int_{0}^{a}xe^{\frac{-x^{2}}{2}}dx$

$= \frac{1}{\sqrt{2}\pi }\left [e^{\frac{-x^{2}}{2}} \right ]^{0}_{a}$

$= \frac{1}{\sqrt{2}\pi }\left [e^{\frac{-x^{2}}{2}} \right ]^{0}_{a} = -\frac{e^{\frac{-a^{2}}{2}}}{\sqrt{2}\pi }$

Note: I switched limits because of a negative that comes out when you use reverse chain rule.

An exponential of 0 gives 1, not 0...

Timothy.Siu · Apr 19, 2009

Trebla said:
An exponential of 0 gives 1, not 0...

lol

gurmies · Apr 19, 2009

Jeezus -.- I've been at home all day, i'm going nuts.

EDIT: In retrospect, I see why I neglected that. My initial wrong integral was xe^.. and when x = 0, whole thing = 0. When I edited it, I forgot to check for other things as a result =)

shaon0 · Apr 20, 2009

gurmies said:
Jeezus -.- I've been at home all day, i'm going nuts.

dw about it.

a fun integration question!! (1 Viewer)

EvoRevolution

Member

shuning

Member

Trebla

Administrator

gurmies

Drover

EvoRevolution

Member

gurmies

Drover

EvoRevolution

Member

gurmies

Drover

Trebla

Administrator

Timothy.Siu

Prophet 9

gurmies

Drover

shaon0

...

Users Who Are Viewing This Thread (Users: 0, Guests: 1)