convert SD of travel time to SD of speed (reciprocal of SD?)

I am trying to get the standard deviation of the reciprocal values of a dataset. Unfortunately, I don't have the original data, only the mean, standard deviation, and n.

Specifically, the mean is 281 seconds, SD is 22 seconds, and n is 39. I need to get speed - the reciprocal of distance. Knowing the distance traveled is 5.9 miles, 281 seconds works out to 75.6 mph easily enough. But how can I get the standard deviation of speed?

The speed calculated from time one standard deviation above the mean (303 seconds) is 70.1 mph, and from one standard deviation below the mean (259 seconds) is 82.0 mph. One is 5.5 mph below the mean speed, the other is 6.4 mph above the mean speed.

I understand why the two differences are different, I just don't know what to do about it. They can't both be one standard deviation!

How can I calculate the standard deviation of speed from the standard deviation of time? Can a reciprocal standard deviation be calculated? I plan on using the standard deviation of speed to calculate the 85th percentile speed, which assumes that the distribution curve is normal (I think) - does getting the reciprocal distort the curve, making it non-normal, and therefore the whole excercise moot?

Thank you all.


TS Contributor
\( V = \frac {S} {T} \)

where \( V \) is the (velocity) speed, \( S \) is the distance traveled which is assumed to be deterministic here, and \( T \) is the time.

If you are willing to make a distributional assumption on \( V \) which is complete determined by its mean and variance, e.g. normal distribution, then actually you can calculate the theoretical SD:

\( SD[V] = S \times SD\left[\frac {1} {T} \right] \)

I have not calculate about the theoretical SD of the reciprocal of normal. But once you can estimate the parameters of the distribution of \( T \) by the sample mean and variance estimates, then \( SD\left[\frac {1} {T} \right] \) is just a function of the parameters and you can plug in the parameter estimates to get \( \hat{SD}[V] \). In the worse case scenario, if the SD is not analytically tractable, you can do numerical integration or simulation.

If you do not want to make any distributional assumption, you can use the Delta method, which is just Taylor series expansion about the mean:

\( SD[V] = S \times SD\left[\frac {1} {T} \right]
\approx S \times SD\left[E[T] + \frac {-1} {E[T]^2}(T - E[T]) \right]
= \frac {S} {E[T]^2} \times SD[T] \)

in which you can again plug in the estimates. The approximation is good when \( n \) is large and the distribution is concentrated around the mean. If you are interested you can also refer a simple example here:
Thank you very much for your answer. I'm in way over my head. Delta method, Taylor series, numerical integration ... all in explody head territory for me.

I am willing/forced to assume that the distribution is normal, and completely determined by the mean and variance, but I only know SD[T], not SD[1/T]. I guess that is my question. If I had the original data, I could calculate it, but I don't. (It's not simply 1/SD[T], is it? I think I tried that.)

Since what I really need is the 85th percentile speed, which is 1.04 SDs above the mean speed, would it make sense to assume that the time 1.04 SDs below the mean will convert to the speed 1.04 SDs above the mean? That would shortcut all this.
I've been reading up on Taylor series and moments, and I think I've got my head around it, at least enough to make it work. That last formula you posted is the key!

Thank you very much BGM!
I have a related question:

I have a distribution, X, (more or less Normal) of about 500 samples, mostly with values between 0 and 1, but there are outliers and even a few negative samples too. I am interested in a measure of the variance (SD) of the reciprocal (1/X) distribution.
When I first compute the reciprocal of each value, and then compute the SD, it is heavily affected by the outliers near zero, and especially by the few negative samples. That is why I prefer to base my measure of variance on X, rather than on 1/X.
Would it be correct to estimate the SD of 1/X by computing SD(X)/mean(X)^2 ??