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

See detailed explanation.

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) (k)/(x^7) dx=\left.(-1)/(6) (k)/(x^6) \right|_1^(\infty) = (1)/(6)((k)/(1^6)-0) = (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) = (1)/(6)\left.((6)/(t^6))\right|_(1)^x = 1- (1)/(x^6)`

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

The case
`\text{Pr}(2\leqX\leq 3) = F(3)-F(2) = 1-(1)/(3^6)-(1-(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)(x^k)/(x^7) dx=6\int_(1)^(\infty)x^(k-7) dx = (6)/(k-6) x^(k-6)`

If k<6, then

`=(6)/(k-6)(\left.(1)/(x^(6-k)))\right|_1^(\infty) = (6)/(k-6)(0-1) = (6)/(6-k)`

So, the mean (k=1) is 6/(6-1) = 6/5= 1.2 . And the standard deviation is
`\sqrt[]{(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- (1)/(1.445^6) = 0.89`