Suggesting why something isn't likely to be normally distributed? I encountered this problem, summarised as: 
For a random variable X representing volume of water usage :
population mean = μ 
population SD σ = 65 
A sample of 80 has mean x̄ = 118 
Numerically justify why X is unlikely to be a normally distributed?
Here's the screen clipping of the actual question (in my summary I used the random variable X whereas in the original it's V).

The answer said it's because 
x̄ − nσ = 118 – 65n <0 and volume usage cannot be negative. 
I'm not sure where x̄ − nσ = 118 – 65n <0 came from. 
A screenshot of the given solution:

Many thanks
 A: The explanation given is really poor.  Usually, indeed, n does stand for sample size.  Here, though, it does not.
The point is that (as noted) water usage cannot be negative.
If water usage is normally distributed with mean 118 and SD 65 then the minimum is only 118/65 = 1.81 sd below the mean. Where might N come from? 
In the sample of 80, we expect there to be some people to be more than 1.81 sd below the mean, and that's impossible.  And the population distribution can't possibly be normal because it has a lower limit. 
A: I am not quite sure I understand the reasoning in the answer.
The sample mean won't be normally distributed but it might easily be very close.
Certainly the original observations can't be really close to normal, since the mean is only about 1.8 standard deviations above 0 and the variable can't go below zero. 
However the reasoning given in the answer is not about the distribution of the original observations, it's explicitly about the distribution of $\bar{X}$ (in spite of the question asking about the variable, not the mean).
The answer seems to be relying on a rule of thumb for gauging whether a semi-bounded random variable would be likely to be near to normally distributed (a rule which you haven't passed on to us, but which has presumably been given to you in come form). 
Presumably the rule says that if the sample mean isn't at least $n$ standard deviations from bound (zero in this case), then sample means won't be close to normally distributed.
It sounds like the rule is not especially useful, however -- in sample sizes like $n=80$, the normality of the variate itself is rarely of particular concern.  I'd be very curious to read what justification was given for it and why they think it matters.
Here's an example of a population distribution for a positive random variable that has mean 118 and standard deviation 65:

In this case it's mildly skew. The thing is we don't expect it to be exactly normal, the issue is whether it's close enough for whatever purpose is at hand. 
Now here's what the sample mean for a sample of size 80 from that distribution looks like:

The distribution of the mean is in black; the normal distribution with the same mean and variance is in green. Note that this is simply the first example I tried; it wasn't especially chosen to make the mean as normal looking as possible, so no doubt one could easily find a way to make the two visually indistinguishable.
So sample means may be quite close to normal and that may be more relevant, depending on what normality was thought to be needed for in the first place.
In practice the question at issue is also not "is the sample mean normally distributed?"* it's whether the non-normality we have is bad enough that we can't reasonably use whatever (unspecified) procedure we wanted to which assumes normality.
* we don't need a rule of thumb for that .. it won't be 
This particular example inference on the mean could behave perfectly reasonably; whether or not some other distribution than the one I was playing with would work depends on (a) the distribution of the original variable as well as (b) the number of standard deviations the mean is from the bound and (c) the sample size. The rule of thumb considers the last two combined in a particular way, but it doesn't seem to be useful for assessing whether normality would be reasonable.
A: The question is about a model for $V$ the volume of water used by an individual guest. 
The problem is not to, 'estimate the population mean and assign a confidence interval from a sample of $n$ individuals'.
The $n$ in the solution formula only makes sense if thought of as a quantile of the standard normal distribution. In this case with $n \approx 2$ we have $ 118 - 2 \cdot 65 = -12$.  
This would mean impossible values of $V$ being generated for more than one in forty guests, on average. Making the normal a very poor model.
