Citat:
Ursprungligen postat av
Jonmax
Tjena! Pluggar matte på distans å sitter fast med en upg gällande triangelfördelning
Ser ut så här:
http://postimg.org/image/cnokvrmit/
Har hittat denna förklaringen men förstår inte så mycket...
"This is about a different distribution from the uniform one; and one that you can deal with by calculation.
For a function f to be a probability density function on an interval of values x, then we require that f(x) >= 0 on the interval and that the area under the pdf from the left to the right hand end of the interval is 1. You can find the area of the triangle by using half base times height, or by doing an integration.
(a) You calculate the probability of a value between 4.2 and 4.4 as the area under the sloping line from 4.2 to 4.4. This is the difference of two triangular areas; or it is the integral of f from 4.2 to 4.4 ."
Någon som är insatt som schysst skulle kunna hjälpa lite?
Hur börjar jag?
Tack!
f(x) = m(x-p)
I figuren ser du att f(4) = 1 och f(6) = 0
f(6) = m(6-p) = 0 => p = 6
f(4) = m(4-p) = m(4-6) = -2m = 1 -> m = -1/2
så f(x) = -1/2(x-6) = 3 - x/2
Den rosa ytan beräknar du genom att integrera f(x) mellan gränserna 4,2 och 4,4
[;
P(4,2\le X \le 4,4) = \int_{4,2}^{4,4} f(x) \mathrm{d}x = \int_{4,2}^{4,4} (3-\frac{x}{2})\mathrm{d}x
;]