The output of a certain integrated-circuit production line is checked daily by inspecting a sample of $100$ units. Over a long period of time, the process has maintained a yield of $80$ percent, that is, a proportion defective of $20$ percent, and the variation of the proportion defective from day to day is measured by a standard deviation of $0.04$.
If on a certain day the sample contains $38$ defectives, find the mean of the posterior distribution of $\theta$ as an estimate of that day’s proportion defective. Assume that the prior distribution of $\theta$ is a Beta distribution.
I know $\textbf{posterior} \propto \textbf{likelihood} \times \textbf{prior}$, and it seems like the likelihood function is binomial so $$f(x|\theta)\propto \theta ^x(1-\theta)^{n-x} $$ and the prior is $$h(\theta )\propto \theta^{\alpha - 1} (1- \theta)^{\beta - 1}~,$$ right? It seems like the posterior is supposed to look like a beta distribution, but where do the values $0.04$ and $38$ come into play here?
On this certain day the Binomial as you stated has $n = 100$ and $x = 38$.
The Beta prior as you stated $h(\theta )\propto \theta^{\alpha - 1} (1- \theta)^{\beta - 1}$ has parameters such that
the mean is $\dfrac{ \alpha }{ \alpha + \beta } = 0.2$
the variance is $\dfrac{ \alpha \beta }{ (\alpha + \beta)^2 (\alpha + \beta + 1) } = 0.04$
which determines uniquely the values to be $~\alpha = \frac35$ and $\beta = \frac{12}5$.