NormalDistribution: Problem Question:Cans of regular Coke are labeled as containing $12 \mbox{ oz}$. 
 Statistics students weighted the content of 6 randomly chosen cans, and found the mean weight to be $12.11$. 
 Assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of $12.00 \mbox{ oz}$ and a standard deviation of $0.13 \mbox{ oz}$. Find the probability that a sample of 6 cans will have a mean amount of at least $12.11 \mbox{ oz}$.
I think the answer should be : $(z-μ)$/$σ$ = $(12.11-12)/0.13$ = $0.846$,which $z=.7995$ What I did worng? Can someone help me? Thank!

You need to find $z$ that corresponds to the sample mean from this exercise. So first a comment on your notation - it should be:
$$z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$$
Of course $\mu_{\bar{x}}=\mu$. But $\sigma_{\bar{x}}\neq\sigma$, as you have computed with. Look up the standard deviation of the sample mean. (Since this is homework I won't give everything away.)
The $z$ that you have computed corresponds to choosing a single can that would weigh $12.11$. While that has a certain likelihood associated to it, it would be more rare to find the mean of 6 cans to be that far from $12.00$.
