Question

PROB&STAT ANYONE????

"Time headway" in traffic flow is the elapsed time between the time that one car finishes passing a fixed point and the instant that the next car begins to pass that point. Let X = the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy flow. The pdf of X is the following.


f(x) =
0.15e−0.15(x − 0.5) x ≥ 0.5
0 otherwise

(a) What is the probability that time headway is at most 6 sec?

Answer 1

hey nottim, are u helping me out?

Answer 2

don't you just want
\[\int\limits_{.5}^{6}0.15e^{−0.15(x − 0.5)}dx\]

Answer 3

i dont know how to use it

Answer 4

you don't know how to integrat it ?

Answer 5

nope
:(

Answer 6

Do a substitution...let \(u=-.15(x-.5)\)

Answer 7

and for x i plug in my numbers?

Answer 8

you find \(du\)

then you can change the limits of integration based on the subtitution \(u\)

Answer 9

\[\int\limits_{.5}^{6}0.15e^{−0.15(x − 0.5)}dx\]

\[u=-.15(x-.5)\]
\[du=-.15dx\]
\[\frac{du}{-.15}=dx\]

\[\int\limits_{.5}^{6}0.15e^{−0.15(x − 0.5)}dx=\int\limits_{0}^{-.15(6-.5)}0.15e^{u}\frac{du}{-.15}\]

Answer 10

what would du be?

Answer 11

\[\int\limits_{0}^{-.15(6-.5)}0.15e^{u}\frac{du}{-.15}=-\int\limits_{0}^{-.825}e^{u}du\]

\[-\left.e^u\right|_{0}^{-.825}=-e^{-.825}+e^{0}\approx.561765\]

Answer 12

thank youuuu so much

Answer 13

I hope you understand what I did.

Answer 14

how'd u get -.825?

Answer 15

forget i figured it out, thanks!