Why doesn't the standard deviation represent a normal distribution? Why doesn't the standard deviation of a sufficiently large sample represent a normal distribution that we can make inferences from?
Let me list my thought process, so hopefully someone can highlight if I'm on the right track, or have some misunderstanding (which I probably do).


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*As you take a sample of sufficient size, the distributions of its values will begin to approximate the parameters of its population (population mean, standard deviation). This is assuming that each value is independent and has an equal chance of being selected.


As an example: If I take the blood pressure of 1000 people, the mean and standard deviation will approximate the true population. Let's pretend the mean for our sample was 120 and the standard deviation is 10.
Say an individual is then measured, who's blood pressure is 80. Why then do I need to use the standard error to estimate how far they deviate from the mean? 
I assume there is some inherent property of taking the ratio of the variance/sample (std. dev / sqrt n) that nicely shifts the probability distributions so that one standard error contains 68% of values, and two contain 95% etc. Is this true? Have I missed something obvious? It seems bizarre to me that to make an inference we need to rely on the standard deviation of the sampling distributions, when we only have one sample which nicely approximates the population we are looking for.  
 A: 
As you take a sample of sufficient size, the distributions of its values will begin to approximate the parameters of its population (population mean, standard deviation). 

What I think you mean is that sample quantities tend to approach population quantities. This is often true for a great many statistics, including raw moments (like means), central moments (like variances), quantiles (like medians), and transformations of them, like standard deviation. Further, the empirical distribution function (ECDF) approaches the population distribution (CDF). But the distribution doesn't approach the parameters - distribution and parameters being different kinds of objects.

As an example: If I take the blood pressure of 1000 people, the mean and standard deviation will approximate the true population. 

... the true population mean and standard deviation respectively. They don't approximate the true population as a whole.
Assuming the required quantities exist (if we're dealing with a Cauchy distribution for example, the sample mean doesn't tend to come close to anything; it bounces about like a randomly drawn single value and keeps doing that in the same way as $n$ increases). Take a look at the weak and the strong law of large numbers.

Say an individual is then measured, who's blood pressure is 80. Why then do I need to use the standard error to estimate how far they deviate from the mean?

You don't. You appear to be conflating standard deviation and standard error (and conflating sample quantities with population quantities). The first is for the distribution of individual measurements.
If the population was normally distributed, the standard deviation tells you about the proportions of individual measurements within so many standard deviations of the population mean. 
The 'standard error' is the standard deviation of the sampling distribution of a statistic. If you were talking about sample means for samples of size $n$, then the standard error of the sample mean would be $\sigma/\sqrt{n}$. 

I assume there is some inherent property of taking the ratio of the variance/sample (std. dev / sqrt n) that nicely shifts the probability distributions so that one standard error contains 68% of values, and two contain 95% etc. Is this true? 

If the sample means were normally distributed (or very close to it) - this is not necessarily the case, even at $n=1000$ - then if you took many samples each of size $n$, then about 68% of sample means would be within one standard error ($\sigma/\sqrt{n}$) of the population mean, 95% within two, and so on.
As $n$ grows (as you take means of larger and larger samples), that standard error will go to 0. The bounds within which 68% of sample means lie (under repeated sampling at a particular $n$) grow nearer to the population mean (the "$/\sqrt{n}$" describes that effect).
