Yes

What is that mean?Or is it simply the square root of the variance?

Further on I am asked to find the table of the proportion of means within one and two stadard deviations of E(X), and I use the following code:

mean(abs(xbar10-mu)<sqrt(var/n))

You should be either using Chebyshev's inequality or treating the sample means as if they are normally distributed with the same mean as the population and standard deviations as above.

Now without seeing the actual wording of the question I am suspicious about using the population variance in this work rather than the unbiased estimate of the population variance from your samples, in which case you end up with a t-distribution rather than normal distribution at the end.which would mean the sqrt(var/n) is the standard deviation, correct?

CB