the mean and standard deviation of the sample mean x bar?

depends on your sample, it may not be very similar to population

the distribution of the sample mean x bar as n increases?

if n is sufficiently large then,

we can assume x bar~N(mean, (sigma^2)/n)

the distribution of the sample mean if x bar is itself normal?[/quote]

N(mean, (sigma^2)/n)

that is because small samples has increased variability