I am working on a statistics problem for my Engineering Statistics class.

The problem goes like this: You are measuring the communications latency between two processors. You take 6 million data points, all measured within .01 ms.

Using the given data set, I was able to find the following statistical parameters: mean = 3.6749 variance = 1.0822 std dev = 1.0403

90% confidence interval = 3.6749 +- .0007 95% confidence interval = 3.6749 +- .0008 99% confidence interval = 3.6749 +- .0011

Now I am tasked with finding the probability that the latency will exceed 10ms. I am having trouble figuring out how to compute that. I know it involves something with a z-score but every calculation I try leaves me with a z-score of like 15 thousand and I know that's definitely not right. Thank you for any help!

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    $\begingroup$ the z-sore is not that high - assuming that the mean/std are given in units of ms you should get the following $$z = \frac{10-3.6749}{1.04}$$ which is much less that quoted. Though the probability of exceeding the mean by $\approx 5\sigma$ is quite small. $\endgroup$ – Chinny84 Jul 28 '16 at 15:56

Essentially, you have your answer in the Comment by @Chinney84, but there seems to be some remaining confusion in the statement of your problem. The purpose of the 'Answer' (really more of a 'Comment') is to try for some clarity.

I assume that "measured within .01ms" means that, in effect, measurements are rounded to the nearest .01ms. If so, this information is not directly relevant to the problem. Later you say that $S = 1.0403 \ne .01,$ so .01 can't refer directly to the measurement error.

The information that $\bar X = 3.675,\, S = 1.0403$ is relevant to finding confidence intervals. Your 95% CI is correct. I did not check the 90% and 99% CIs. Because of the 95% CI it is clear that the population mean latency $\mu$ is nowhere near 10ms. To compute the CI, I assume you used $\bar X \pm 1.96 S/\sqrt{n},$ where $S/\sqrt{n} \approx 0.000425$ is the standard error of the statistic $\bar X.$ This expresses the error of $\bar X$ as an estimate of the population mean $\mu.$

If by 'found' you mean that you or someone else computed $\bar X$ and $S$ from data, then you should not use the word 'parameter' to refer to $\bar X$ or $S$. These are statistics, which are (sample) estimates of (population) parameters.

When you are asked for the 'probability that the latency will exceed' 10ms, that must mean the probability that a future individual measurement from this same process exceeds 10ms. The standard deviation $\sigma$ for a single measurement is estimated by $S.$ Also, the population mean $\mu$ is estimated by $\bar X.$ Thus, as in the comment, you seek $$P(X > 10) = P\left(Z > \frac{10 -\mu}{\sigma}\right) \approx P\left(Z > \frac{10-3.675}{1.0403} = 6.08\right ) \approx 0,$$

where $Z$ is a standard normal random variable.

So provided that the estimates 3.675 and 1.0403 are reasonably accurate, it would be very rare indeed for an individual observed latency to be above 10.


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