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Given a normally distributed population with mean $\mu$ and say I constructed a confidence interval of $\mu \pm a$ with a confidence level of 95%. Is the following statement correct?
There is a 0.95 probability/chance that the population mean is captured within this confidence interval?
In the sense of the Bayesian interpretation of probability, that is actually correct. But it is easy to be misled by that if you don't also understand how frequentists look at it.
In the sense of the frequentist interpretation of probability, it is not correct. Because from the frequentist point of view, you construct one confidence interval, and now the mean is either in that interval or it isn't. Where is the repeatable experiment? The repeatable experiment is the construction of the confidence interval itself. That is, from the frequentist point of view, we "have 95% confidence that the population mean is in the confidence interval" because if we were to take many samples and construct many confidence intervals, 95% of the samples would produce 95% confidence intervals which overlap with $\mu$.
You will get different answers for your question depending on philosophical understanding. I like to say that the proposed statement is at best inaccurate because the population mean "exists" but is unknown, therefore there is no randomness to account for (and thus no probability).
I prefer to say the following:
"The confidence interval was produced with a method that captures the mean $\mu$ with $95~\%$ probability."
Note that the probability is associated with the method (say, the random sampling and the formulae) rather than with its actual, numerical output.
