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The question is related to computing the likelihood function of a simple problem for MCMC. The full problem can be found here on page 2.
The question in the title is written more succinctly in the screenshot below: How do we compute $N(\mu|x,\sigma)$ given $\mu$, $\sigma$, $x$?
Here is the full excerpt in a description of the Metropolis-Hastings algorithm (p.144), when contemplating a move from [current] $\mu=110$ to [proposed] $\mu=108$:
Compare the height of the posterior at the value of the new proposal against the height of the posterior at the most recent sample. Since the target distribution is normal with mean 100 (the value of the single observation) and standard deviation 15, this means comparing $N(100|108, 15)$ against $N(100|110, 15)$. Here, $N(μ|x, σ )$ indicates the normal distribution for the posterior: the probability of value $μ$ given the data $x$ and standard deviation $σ$.
Hence, we are in a situation where the targeted posterior is $$\pi(\mu|x,\sigma)\propto \varphi(x;\mu,\sigma)\propto\exp\{-(x-\mu)^2/2\sigma^2\}$$ as the prior density on $\mu$ is constant. Therefore, $N(μ|x, σ )$ denotes the normal density with mean $x$ and standard deviation $\sigma$. And the Metropolis-Hastings acceptance probability is $$\min\{1,\exp[-(\underbrace{100}_x-\underbrace{108}_{\mu^\text{new}})^2/2\sigma^2+(\underbrace{100}_x-\underbrace{110}_{\mu^\text{old}})^2/2\sigma^2]\}$$ Let me add that the description is very poor, in particular because the author has inverted the roles of $\mu$ and $x$ in the sentence
this means comparing $N(100|108, 15)$ against $N(100|110, 15)$.
which should be
this means comparing $N(108|100, 15)$ against $N(110|100, 15)$.
Of course, mathematically, $N(100|108, 15)=N(108|100, 15)$, so the comparison is correct. But it unnecessarily confuses the reader.
