I have been recently studying Bayesian statistics and more precisely the problem of finding posterior distributions. I am able to understand the my textbook's problems, but I realize that I understand the issue on cookbook level. This is why I would like to ask a question, which I found and hope to get better insight.
The problem is: Suppose we are interested in finding the mean value of the velocity of a large number of particles. We draw and measure a sample of size $N$ and get a sample mean value of 70m/s. We assume that the population mean is normally distributed with an unknown mean $\mu$ and std. deviation 10 m/s. The prior distribution of the mean $\mu$ is a normal distribution with mean 100m/s and st. deviation 20m/s.
The question is how to get the posterior as a function of the sample size $N$
The prior distributions should be
$$ P(\mu)=\frac{1}{\sqrt{2 \cdot 20^2 \pi}}\exp{\left(-\frac{(\mu-100)^2}{2 \cdot 20^2}\right)} $$ The next step is to define the likelihood.Since we are interested in a function of sample size $N$ my guess is to use the expression for the sample standard deviation $$ \sigma_{\overline{x}}=\frac{\sigma}{\sqrt{N}} $$
The sample mean is $\mu_{\overline{x}}=70$, this should yield for the likelihood
$$ P(\mu_{\overline{x}}=70|\mu)=\frac{1}{\sqrt{2 \cdot \frac{10^2}{N} \pi}}\exp{\left(-\frac{(\mu-70)^2}{2 \cdot \frac{10^2}{N}}\right)} $$
This should give us for the posterior expression of the sort $$ P(\mu|\mu_{\overline{x}}=70) = P(\mu_{\overline{x}}=70|\mu) \cdot P(\mu) $$ Or in other words the posterior gives the probability to have population mean $\mu$ if we had $\mu_{\overline{x}}=70$. I fear that my argument is faulty, I am not sure. Would appreciate your help.