Calculate the credible interval for posterior distribution ~ N(69.07,0.53^2) I'm not really sure as to how to calculate the credible interval for this posterior distribution I'm given
~ N(69.07, 0.53^2)
And I need to find the probability of the interval, of length 1, which has the highest probability. 
I know this interval is about to average ie 69.07+-0.5 but I don't know how to calculate the probability of this interval
 A: Your posterior distribution is $\mathsf{Norm}(\mu = 69.07, \sigma=0.53).$
Finding the interval $(L_{ci}, U_{ci})$ of length $U_{ci}-L_{ci} = 1$ that contains
the greatest probability under the posterior PDF requires some sort of
computation.
As you say, intuitively this interval will be about $69.07 \pm 0.5$ because
that is were the density function is highest.
Not knowing the kinds of computations with which you are familiar, I will
do a 'grid search' in R statistical software to illustrate the idea. This
is a brute-force method that looks at all 'reasonable' intervals of length $1$
(30,001 intervals with
endpoints rounded to four places), finds the probability of each, and
picks the one with the largest probability. It confirms that the interval
$(68.57, 69.57)$ is indeed the correct one. It has about probability
about 65.5%. [One can prove that this is the shortest 65.5% posterior credible interval.]
Here is some R code:
 L = seq(67,70, by = 0.0001);  U = L+1
 pr = pnorm(L+1, 69.07, .53) - pnorm(L, 69.07, .53)
 L.ci = L[pr == max(pr)]  # lower end of interval (length 1) with largest probability
 U.ci = L.ci+1;  L.ci;  U.ci;  max(pr)
 ## 68.57
 ## 69.57
 ## 0.6545217

 curve(dnorm(x, 69.07, .53), 67, 71, lwd=2,  ylab="PDF", main="Posterior Distribution")
 abline(h = 0, col="green2"); abline(v=c(L.ci,U.ci), col="blue")
 abline(h = dnorm(L.ci, 69.07, .53), col="red")

The figure below (made with the last three lines of code) shows the
interval boundaries, and illustrates that the the interval 'uses' the
largest values of the posterior density.

Note: It is common to use 'probability symmetric' intervals--especially
with symmetrical (or nearly-symmetric) posterior distributions. Below is
R code to find a 95% credible interval $(68.03, 70.11)$ that cuts probability 2.5% from each tail of the
posterior distribution.
 qnorm(c(.025,.975), 69.07, .53)
 ## 68.03122 70.10878

