Posterior for Discrete prior from normal sample with known variance 
Given a sample $(1,2,5,2.5,3,7)$, and known $\sigma^2 = 4$ with possible values for prior mean of $(0,0.5,1,1.5)$ all equally likely, find the posterior distribution and the posterior probability that $\mu\leq 1$

So in R I have:
> sample = c(1,2,5,2.5,3,7)
> m = c(0,0.5,1,1.5)
> prior = (rep(1/4,4))
> lilklihood = dnorm(mean(sample), mean = m, sd = 2/sqrt(length(sample)))
> posterior = prior*liklihood/sum(prior*liklihood)
> posterior
[1] 0.002021756 0.021735920 0.160607934 0.815634390


So my posterior distribution just is this vector correct? The thing I find is that all my values are actually less than 1, so is posterior probability that $\mu\leq 1$ just 1?

The resulting posterior is the posterior probability associated with four possible values for the mean; therefore;
$$
\begin{aligned}
&P(\mu = 0 \vert \textrm{data}) &&= 0.002021756,\\
&P(\mu = 0.5 \vert \textrm{data}) &&= 0.021735920,\\
&P(\mu = 1 \vert \textrm{data}) &&= 0.160607934,\\
&P(\mu = 1.5 \vert \textrm{data}) &&= 0.815634390,\\
\end{aligned}
$$
therefore $P(\mu \leq 1 \vert \textrm{data})$ is given by
$$
\begin{aligned}
P(\mu \leq 1 \vert \textrm{data}) &= P(\mu = 0 \vert \textrm{data}) + P(\mu = 0.5 \vert \textrm{data}) + P(\mu = 1 \vert \textrm{data})\\
&= 0.18436561\\
&\approx 18\%.
\end{aligned}
$$
