# How to find something that is 1 standard deviation of the mean value?

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 $$f(x)=\int_{0.634}^{2.366} \frac{3}{x^4} dx$$

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 $$0 is divergent. $$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$$ are correct.
$$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.