Question

Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway between two randomly selected consecutive cars (sec). Suppose that in a different traffic environment, the distribution of time headway has the form f(x) 5 5 k x4 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. c. Use the cdf from (b) to determine the probability that headway exceeds 2 sec and also the probability that headway is between 2 and 3 sec. d. Obtain the mean value of headway and the standard deviation of headway. e. What is the probability that headway is within 1 standard deviation of the mean value?

Answer 1

Answer:

a. k = 3

b. Cumulative distribution function X, $F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c.  Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e.  Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

$f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.$

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

$\int\limits^{\infty}_{-\infty} f(x)dx=1$

So, the equation becomes,

$\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3$

b. For this problem, the cumulative distribution function is defined as :

$F(x) = \int\limits^1_{\infty} f(x)dx + \int\limits^x_1 f(x)dx$

Now,

$F(x) = 0 + \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 + \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}$

Therefore the cumulative distribution function X is,

$F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

$P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125$

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

$P(2$

≅ 0.088

d. Mean value of headway,

$E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5$

And,

$E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3$

The standard deviation of headway is,

$= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254$

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

$P(\alpha - \beta < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X$

From part b, F(x) = 0, if x ≤ 1

$=1-(2.366)^{-3}\\=0.9245$