Q:

Probability with average of measures

I am trying to solve Problem 2 from this problem set. 

Let $\mathbf{Y}$ be the avreage of $5$ independent measurements.
  For a single measurement one have $\sigma^2 = 0.060^2$ and $\mu = 6.8$.
$\textbf{b})$ What is the probability that $\mathbf{Y}$ deviates more than $0.06$ from $\mu$ ?

I tried to use the probability that one measurement deviated more than $0.06$ from $\mu$. And then the total probability should be $1 - P(\text{all are below } 0.06)$.
But I see that this is wrong. 
I also tried calculating 
$$
    1 - P\left( Z < \frac{5\cdot 6.8- (5\cdot 6.8 + 0.06)}{\sigma/\sqrt{5}} \right)= 1 - P(Z<\sqrt{5\ \!})
$$
Which also turns out to be false. The correct answer should be 
P = $0.026$, but I can not quite get there. I am not experienced with dealing with 
the average of independent measurements, so any literature, hints or tips is very welcome. 

A:

The questions says "results which are assumed to be independent and normally distributed", which is important to note.
In effect you are being asked to find the distribution of $Y$ (which is a mean, not a sum as in your expression).  

What is the expected value of $Y$? (It is not $5\cdot 6.8$.) 
What is the standard deviation of $Y$? (As you seem to say, it is $\sigma/\sqrt5$.)

Now, what is the probability of $Y$ deviates from $\mu$ by more than $0.06$?  Remember that $Y$ might be greater than or less than $\mu$, and the deviation here probably means the absolute value of the difference.

