Question

"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) = k/(x^7) ; x > 1

0 ; x<1
a. Determine the value of K for which f(x) is a legitimate pdf
b. Obtain the cumulative distribution function

F(x) = [ Blank ] ; x > 1

0 ; x < 1

c. Use the cdf from (b) to determine the probability that headway exceeds 2 sex. (Round your answer to three decimal places.)
Use the cdf from (b) to determine the probability that headway is between 2 and 3 sec (round your answer to three decimal places)
d. Obtain the mean value of headway and the standard deviation of headway. (Roun to nearest 3 decimal places)
e. What is the probability that headway is within 1 standard deviation of the mean value?

Answer 1

Answer:

See detailed explanation.

Step-by-step explanation:

a) Recall that if a function is a pdf, then it must happen that $\int_{-\infty}^{\infty} f(x) dx = 1$. Then,

$\int_{-\infty}^{\infty} f(x) dx=\int_{1}^{\infty} \frac{k}{x^7} dx=\left.\frac{-1}{6} \frac{k}{x^6} \right|_1^{\infty} = \frac{1}{6}(\frac{k}{1^6}-0) = \frac{k}{6}=1$

which gives as k = 6.

b). Recall that $F(x) = \text{Pr}(X\leq x ) = \int_{-\infty}^x f(t) dt$ Then, F(x) =0 if $x\leq 1$. Then if x>1 we have from the previous point that

$F(x) = \frac{1}{6}\left.(\frac{6}{t^6})\right|_{1}^x = 1- \frac{1}{x^6}$

c). We want to know $\text{Pr}(X>2) = 1- \text{Pr}(X\leq 2) = 1- (1-\frac{1}{2^6}) = 0.016.$

The case $\text{Pr}(2\leqX\leq 3) = F(3)-F(2) = 1-\frac{1}{3^6}-(1-\frac{1}{2^6}) =0.014$

d). We have that $\mu = \text{E}(X), \sigma = \sqrt[]{\text{E}(X^2)-\text{E}(X)^2}$. Where, $\text{E}(X^k)=\int_{-\infty}^{\infty}x^kf(x) dx$

Then,

$\text{E}(X^k)=6\int_{1}^{\infty}\frac{x^k}{x^7} dx=6\int_{1}^{\infty}x^{k-7} dx = \frac{6}{k-6} x^{k-6}$

If k<6, then

$=\frac{6}{k-6}(\left.\frac{1}{x^{6-k}})\right|_1^{\infty} = \frac{6}{k-6}(0-1) = \frac{6}{6-k}$

So, the mean (k=1) is 6/(6-1) = 6/5= 1.2 . And the standard deviation is $\sqrt[]{\frac{6}{6-2}-1.2^2}= \sqrt[]{1.5-1.2^2} = 0.245$.

e. We are asked for $\text{Pr}(|X-\mu|\leq \sigma)$ Which is equivalent to

$\text{Pr}(-\sigma+\mu \leq X \leq \sigma+\mu ) =\text{Pr}(0.955 \leq X \leq 1.445 ) = F(1.445) - F(0.955) = F(1.445) = 1- \frac{1}{1.445^6} = 0.89$