Confidence interval on part of normal distribution

Question

Given an sample of data set $X$ which is normal-distributed to $\mathscr N(\mu=240,\sigma)$, I want to find the 95% confidence interval of $\max(\mu-200,0)$. As the $\max(\mu-200,0)$ cut the normal-distribution curve at the point of $200$, the sample of $\max(\mu-200,0)$ is not more normal distributed. Therefore what is a reasonable $95\%$ confidence interval?

Answer 1

Your distribution contains a $\delta$-function at $x=0$ of probability mass equal to the probability mass of your normal distribution less than the value $x=200$. (This will involve an error function.) The remaining probability mass will be due to the portion of your Gaussian to the right of $x = 200$. As such, your $95\%$ confidence interval will extend from $x=0$ up to the value $x$ for which $5\%$ of the total probability mass (or $10\%$ of the truncated Gaussian) is above $x$. Clearly, this is expressed as an error function.