Good question - this is from a statistical hypothesis test (here, a one-tailed z-test, with p-value).

Assume the frequencies, F, are similar to the distribution NormalDistribution[50.0012, 0.0584802]. Then we have two hypotheses:

H0: This was caused by a statistical misfortune (the null hypothesis)

H1: This was caused by a process problem (the alternative hypothesis)

We will test this at the 99.99% significance level (so for us to be certain H1 was the case, the probability this would occur if the process was in statistical control must be below ? = 0.0001). The probability this was an accident, assuming the process was in control, was p = Prob[F < 49.5] = 5.1191*10^-18 < 0.0001 = ?. So we have sufficient evidence, at the 99.99% significance level, to reject H_0. Therefore, we can be almost (but never completely; such is the way of statistics) certain that this could not be a statistical misfortune. So the process was not in control at the time of the blackouts.

With regards to the variable names, I have just amended that. Thank you for pointing that out!

Although I agree that a normal distribution is not perfect here, observe how far the 49.5 Hz reading lies from the mean:

In[1]: (Mean[Last/@frequencies] - 49.5)/StandardDeviation[Last/@frequencies]

Out[1]: 8.57123

Regardless of choice of distribution, 8.57 standard deviations from the mean is a remarkable outlier. Which distribution would you suggest as opposed to the normal (I considered Student's T, but this was not successful)?

I had already tried this, but was not pleased with the results.

In[1]: dist = FindDistrubution[Last/@frequencies]

Out[1]: StudentTDistribution[50.0054, 0.0739269, 8.64764]

In[2]: Show[Histogram[Last /@ frequencies, Automatic, "ProbabilityDensity"], 
 Plot[PDF[dist, x], {x, 49.7, 50.3}, 
  PlotStyle -> Thick]]

which gave:

which doesn't seem like a particularly good fit. Others from the best solutions do not seem particularly accurate either.