Suppose I have a $95 \%$ confidence interval for $\mu$ with population variance $\sigma^{2}$ known. It will be of the form $$[\bar{X}-1.96 \sigma/\sqrt{n}, \bar{X}+1.96 \sigma/\sqrt{n}]$$
The statistical interpretation of this is that $95\%$ of the confidence intervals will contain the true mean? I know individually, there is a $50\%$ chance for a confidence interval to contain the mean. But collectively, my above sentence is valid?
