I am having difficulty with one particular item in my homework. Its in the section of the text on "Continuous Random Variables" The homework question reads as follows:

"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.

The provided function for this is $$f(x)=\int\frac{k}{x^4}, x>0\:and\:k=3 $$

I have calculated the mean to equal 1.500 and the standard deviation to equal 0.866. The question reads:

(e) What is the probability that headway is within 1 standard deviation of the mean value?

I them proceeded to calculate the following integral limits

1.500-0.866= 0.634, and 1.500+0.866=2.366

I then use these in the integral $$f(x)=\int_{0.634}^{2.366} \frac{3}{x^4} dx$$

The answer I receive is 3.849 which is not the answer that is listed for this problem. The listed answer is: 0.925

What am I doing wrong?

What is the probability that headway is within 1 standard deviation of the mean value?

How do I calculate this? I would really like to know.


There's a typo in the density. The domain is not $x>0$, as the integral for $0<x<1$ is divergent. $$f(x)=\int\frac3{x^4}~, \qquad x\geq1$$

You probably have been aware of this since the mean $\mu = \frac32$ and standard deviation $\sigma = \frac{ \sqrt{3} }2$ are correct.

For the desired probability, it's just a tiny misstep in the lower limit. The correct integral is

$$f(x) = \int_{ \max(\mu-\sigma, 1)}^{\mu+\sigma} \frac{3}{x^4} dx = \int_1^{(3+\sqrt{3})/2} \frac{3}{x^4} dx \approx 0.924501 $$

again since $x \geq 1$. Note that $\mu - \sigma = \frac{ 3 - \sqrt{3}}2 \approx 0.634 < 1$ is out of bounds.


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