Confidence Interval for $\mu$ Let $X_i$ be Uniformly distributed (i.i.d.) between [$\mu-1,\mu+1$]. For i element of the Natural numbers. A sample of $X_1=0.2$ and $X_2=0.7$ has been drawn.
$\hat\mu=\frac{X_1+X_2}{2}$
Calculate a 95% CI for $\mu$ in the form of $[\mu-k,\mu+k]$ for k>0.
Use the following $X_1+X_2$ follows a triangular distribution with mean $2\mu$ and a support of $[2\mu-2,2\mu+2]$
My ideas: can i use that $\mu=0$
So that F($\hat\mu-\mu)=F(\hat\mu)$ and also for the support of the triangular distribution. I have never heard the term support is it where the distribution is non zero?
I would have stated that (B being one of the values for the interval)
$F(\hat\mu<B)=0.975$
$F(\frac{X_1+X_2}{2}<B)=F(X_1+X_2<2B)$ can i do that?
i am so new to all of this and my maths background is not very profound, so my questions may be a little stupid.
Anyway many thanks
 A: Thanks for including the 'Self Study' tag and showing your thoughts so far. Here is a method that organizes your idea towards a solution: 
For the triangular distribution of $T = X_1 + X_2,$ which has
support $(2\mu - 2, 2\mu + 2),$ find the number $\delta$ such that
$$P(2\mu - 2 + \delta  \le T \le 2\mu + 2 - \delta) = .95.$$
Draw a picture. Elementary geometry and algebra should suffice.
Then $\bar X = T/2,$ so that 
$$P(\mu - 1 + \delta/2 < \bar X \le \mu + 1 - \delta/2) = .95.$$
Finally, solve the inequality of the event so that you have
$$P(\bar X -1 + \delta/2 \le \mu \le \bar X + 1 - \delta/2) = .95.$$
This is called 'pivoting'.
Finally, a 95% CI for $\mu$ is if the form $(\bar X - 1 +\delta/2, \bar X + 1 - \delta/2).$

Notes: (1) Just to check, I used my $\delta,$ del in the code, to simulate (in R) using $\mu = 5$ to see what proportion of a million such confidence intervals $(L,U)$ actually covers $\mu.$ Answer is consistent
with 95%.
set.seed(324)
mu = 5;  del = sqrt(.2)       # How to evaluate 'del'?
x1 = runif(10^6, mu-1, mu+1)
x2 = runif(10^6, mu-1, mu+1)
a = (x1 + x2)/2  # mean
L = a - 1 + del/2
U = a + 1 - del/2
mean(L < mu & U > mu)
## 0.950169

(2) If $T$ is the sum of, say, $n = 10$ or more observations, then
the distribution of $T$ is nearly normal (instead of triangular) with mean and SD that are fairly easy to find. Then you can use an appropriate CI based on a good normal approximation.
