Find Probability that latency exceeds 10 ms given sample mean and variance 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!
 A: 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. 
