Question

9. + 5/7 points Previous Answers DevoreStat9 4.E.013. My Notes + Ask Your Teacher '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 (sec). Suppose that in a particular traffic environment, the distribution of time headway has the following form. f(x) = (a) Determine the value of k for which fix) is a lcgitimate pdf. (b) Obtain the cumulative distribution function. x1 x 1 (c) Use the rdf from () ta determine the probability that headway exceeds 2 ser. (Round your answer to three decimal places.) .0625 Use the cdf from (b) to determine the probability that headway is between 2 and 3 sec. (Raund your answer to three decimal places.) .0501543 (d) Obtain the mean value of headway and the standard deviation of headway. (Round your answers to three decimal places.) mean 1.335 standard deviation (c) What is the probability that headway is within 1 standard deviation of the mean valuc? (Round your answer to three decimal places.) Need Help? Rendit Watch Talk to a Tutor Submit Answer Practice Another Version

Answer 1

The distribution of X has the following form :

fx(x) = 5 x > 1

( a ) It is a legitimate PDF if it evaluates over to 1 when integrated over entire range .

= fx(x)dx = JT=1

$\Rightarrow k\int_{x=1}^{\infty}\frac{1}{x^5} \ \ dx=1$

kl -5 dx = 1 Jr=1

= =

I= [_21

Substituting limit of x from 1 to $\infty$

1-x -1 = 1

$\Rightarrow \frac{k}{4}=1$

k=4

.. fx() = 4.1-5; 1>1

( b ) CDF is obtained by integrating PDF from 1 to x

$\Rightarrow F_X(x)=P(X\leq x)=\int_{t=1}^{x}4t^{-5}dt$

$=4[\frac{t^{-5+1}}{-5+1}]$

Substituting limit of t from 1 to x

=Fx(x) = 4

= -1-4+1

$\Rightarrow F_X(x)=1-\frac{1}{x^4} \ \ ; \ \ x>1$

and, Fx(2) = 0 ; Otherwise

( c )  $P(X>2)=1-P(X\leq 2)=1-F_X(2)$

$=1-(1=\frac{1}{2^4})=1-\frac{15}{16}=\frac{1}{16}=0.0625$

P(2 < X <3) = P(X <3) - P(X<2) = F(3) - F(2)

$=(1-\frac{1}{3^4})-(1-\frac{1}{2^4})=\frac{1}{16}-\frac{1}{81}=0.0502$

( d )   $E(X)=\int_{x=1}^{\infty}x.f_X(x)dx$

$=\int_{x=1}^{\infty}x.4x^{-5}dx$

$=4\int_{x=1}^{\infty}x^{-4}dx$

$=4[\frac{x^{-4+1}}{-4+1}]=-\frac{4}{3}[x^{-3}]$

Substituting limit of x from 1 to $\infty$

$\Rightarrow E(X)=-\frac{4}{3}[0-1]=\frac{4}{3}=1.333$

$E(X^2)=\int_{x=1}^{\infty}x^2.f_X(x)dx$

$=\int_{x=1}^{\infty}x^2.4x^{-5}dx$

$=4\int_{x=1}^{\infty}x^{-3}dx$

$=4[\frac{x^{-3+1}}{-3+1}]=-2[x^{-2}]$

Substituting limit of x from 1 to $\infty$

$\Rightarrow E(X^2)=-2[0-1]=2$

$\because V(X)=E(X^2)-[E(X)]^2$

$\Rightarrow V(X)=2-(\frac{4}{3})^2=2-\frac{16}{9}=\frac{2}{9}$

...S.D(X) = VV(X) = 1 = 0.4714

( e ) Probability that headway is within one Standard Deviation of Mean Value i.e,

$=P(\frac{4}{3}-0.4714<X<\frac{4}{3}+0.4714)$

$=P(0.862<X<1.805)$

$=P(X<1.805)-P(X<0.862)$

$=F(1.805)-F(0.862)$

$=F(1.805)$ ...............Since , F ( x ) is 0 for x < 1

$=1-\frac{1}{1.805^4}$

= 0906