I got a set of data Time (s) and Size (Bytes) for each I calculate a Rate (s/Bytes). I'm trying to predict a Time knowing a Size and using the Mean and the of the Rates. Something like 11 minutes 48 seconds ±3σ 2 minutes 30 seconds

For now, to calculate the mean I Sum the Times and the Sizes and calculate the Mean using these two values. But to calculate de Standard Deviation I need to perform the calculation on the calculated Rates of my set of data.

  • Does that make any sense to calculate the Mean on Times - Sizes sums and the Standard Deviation on each calculated Rates?

  • Or should I calculate the Mean of each calculated Rates knowing that I'm using a Standard Deviation on each calculated Rates to give ±3σ estimation?

Are both method correct or should I apply the second solution even if the Mean calculation of the first solution is more accurate?

  • $\begingroup$ A rate would be bytes per second, not seconds per bytes. $\endgroup$ – joriki Sep 6 '18 at 8:13
  • $\begingroup$ @joriki no a rate is a ratio between two quantities Cf en.wikipedia.org/wiki/Rate_(mathematics) $\endgroup$ – cz3ch Sep 6 '18 at 12:05
  • $\begingroup$ @joriki I got bytes and seek seconds so my rate is effectively s/bytes, and not bytes/s or I would invert the result which is not really effective. In any case a rate is a ratio between two quantities $\endgroup$ – cz3ch Sep 6 '18 at 12:09
  • $\begingroup$ Of course you're free to use language in any way you find convenient. I doubt you'll find many examples of ratios with time in the numerator being called a "rate", and consequently the resulting text will be confusing to people who are used to this convention, but if you think that in your case there's an advantage in doing so that outweighs this disadvantage, sure, why not. By the way, that Wikipedia article mentions that commonly the denominator is time and also provides some examples of rates with non-time denominators, but none with time numerators. $\endgroup$ – joriki Sep 6 '18 at 12:46
  • $\begingroup$ @joriki all I need is the Expected Time of Arrival to the end of processing data, I know the amount of data to process in Bytes and seek the Time to process this data. What should be you suggestion to obtain the Expected Time of Arrival? $\endgroup$ – cz3ch Sep 6 '18 at 14:30

The seeking Mean and Standard Deviation are on a ratio, and we can't Mean all Ratio directly. So to calculate the Mean the only way is by summing Times and Sizes and to calculate the Ratio of these two values. And to calculate the estimator of the Standard Deviation, the only ways is to use all calculated Ratios. I have no other way accomplished these two tasks.


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