= Solution
Each observation has <normal distribution> <probability density function>
$$
f(x_i\mid\mu)=\frac1{\sqrt{2\pi}}\exp\!\left[-\frac12(x_i-\mu)^2\right].
$$
The observations are <independent and identically distributed>, so their joint <probability density function>, viewed as a function of the unknown mean, is the product:
$$
\boxed{L(\mu)=\prod_{i=1}^nf(x_i\mid\mu)
=(2\pi)^{-n/2}\exp\!\left[-\frac12\sum_{i=1}^n(x_i-\mu)^2\right]}.
$$
Independence justifies multiplication; the common mean and known unit <variance> justify using this same density factor for every observation.
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