I just realized I asked a question with a very similar title a while back. This is not a duplicate. It is another conjecture though.
Conjecture: $\lim\limits_{x\to\infty}\mu \{ f(2), f(3), f(5), ... , f(x) \}=\frac{\pi}{2}-1$ where $x\in\mathbb{P}$, $f(x)$ equals the amount of $1$s divided by the total number of digits in the binary representation of $x$.
$\mu$ is the arithmetic mean and $\mathbb{P}$ represents the set of all primes.
Example of $f(x)$ because of my poor explanation (I tried my best):
$f(5)\rightarrow 101\rightarrow \frac{\text{number of $1$s (or the digit sum)}}{\text{total number of digits}}\rightarrow\frac{2}{3}$
I have no idea why this conjecture would be true but it appears that it may converge, at least for the first $2000$ primes. In the picture, the green line is $y=\frac{\pi}{2}-1$ and the black points are $\mu \{ f(2), f(3), f(5), ... , f(x) \}$.
My question is whether or not my conjecture is true?
Am I missing something obvious here? Also, I find it very interesting that it looks similar to a quasiperiodic function.
This conjecture seems unlikely to me. I suspect the apparent agreement is an artifact of not trying large enough numbers. I predict that your conjecture is wrong.
If $x>2$ is prime, then its binary representation starts with a 1 (because all binary representations do) and ends with a 1 (because all primes are odd). Heuristically, we could expect that every other bit is equally likely to be 0 or 1.
Under this heuristic, a $b$-bit prime can be expected to have about $(b+2)/2$ bits set to 1, on average. This corresponds to a fraction of $(b+2)/2b$ bits set to 1.
Under this heuristic, if you look at all of the primes that are up to $n$ bits long, then you'd expect the value of your expression ($\mu\{\cdots\}$) to be roughly $(n+2)/2n$, or a little bit more. Notice that the first 1900 primes are 15 bits long or less. So if you evaluate your expression on such a value of $x$, this heuristic would predict your expression would have the value about $(n+2)/2n$ where $n=15$, i.e., $17/30 \approx 0.567$, or a bit more. Notice that $\pi/2-1$ is about $0.57$.
So your observed data is consistent with my heuristic.
Also, my heuristic predicts that if you pick a much larger value of $x$, then the value will drop significantly, and that the correct limiting value is $0.5$, not $\pi/2-1$. So I predict that your conjecture is wrong.
To test this, you could write code to evaluate this expression for much larger values of $x$, and see how well it fits my heuristic vs your conjecture.