How to calculate the entropy of random normally distributed samples?
It's (fitted) parameters are mean = 35 units, std.dev. = 8 units and I read the values as 8 bit integers.


The choice of parameters hint at a normal distribution. Is this the case? You might want to obtain a histogram and perhaps substitute mean and standard deviation with modal mean and range. This will also help with an estimation of entropy.


  I'm trying to determine a theoretical rate of entropy generation per reading, so for example 3.4 bits /reading or the like.


A first step would be to estimate the Shannon Entropy. In this case, your alphabet is the range of 256 symbols that are possible to be formed with an 8-bit integer and you would be estimating the entropy of a source that generates these symbols with probabilities dictated by the histogram of a long enough sequence of symbols.

If these 256 symbols appeared with equal probabilities in your sequence then you would need all 8-bits to represent them. However, if a subset of symbols appeared with higher probabilities than others then, on average, it would be possible to represent this sequence with less bits per symbol. To determine Shannon's Entropy, obtain the histogram of a long enough sequence, divide each entry by the sum of all entries so that it is now a probability density function (PDF), i.e. it has a total sum of 1.0, and then calculate:

$$H(X) = -\sum_{k=0}^{N} x_i \log_2(x_i)$$

Where $X$ is your PDF (derived above) and $N$ is the length of the histogram.


  PS. As entropy is actually generated by the observer rather than by the process itself, I'm particularly interested in the significance of the bit depth of the individual readings.


Entropy is a characteristic of the source. This particular way of deriving entropy applies to a particular kind of source called ergodic and this is why I am also emphasising the "long enough sequence" (For more information please see this and specifically section 5.)

I am not sure what you mean by "significance of the bit depth". The idea is this: Suppose that you have an alphabet with 256 symbols. Obviously, you need 8-bits to represent them. But, if you were to discover that your source is actually generating numbers in the range $0..32$ for 98% of the time and a single 256 value for the rest 2% of the total length of a given sequence, then, for 98% of the time you would be wasting 3 bits, because $2^{5} = 32$ and you would need all 8 bits for only 2% of the time. Imagine these bits on a sequence:

00000011  00000100  00001110  00001001  00001100  10001101  00011001  00001110

If you could somehow squeeze those 3 bits out of your stream, you could save time of transmission (therefore fit more data through a channel) or storage space.

This is why figures such as 2.5 bits make sense. You can't define half a bit BUT over two symbols (i.e. 3 bits + 3 bits = 6 bits) you would be saving 1 bit (because 2.5 bits + 2.5 bits = 5bits).

For more information on encodings that allow you to effect exactly that "squeezing", please see variable length coding.

Hope this helps.

EDIT:
If you are absolutely sure that the source is normally distributed, then substitute $X$ for an expression of the normal PDF and the sum for an integral over the sufficiently probable figures