Finding the probability of X_Bar with sample variance included? The question I am asked is $P(\bar{X} > 3 + 0.4984S)$, where I am additionally provided $n = 25, \mu = 3.0, \sigma^2_\text{pop} = 3.0$. $\bar{X}$ is the sample mean and $S$ is the sample variance. The population is normal. 
I understand the concept of finding such probabilities; what I mean is if you ask me to graph a question like this, I can draw it out for you with every detail. I just don't understand how to get an answer to this. I am not given sample variance so I can't solve x_bar > some number. 
I have also tried manipulating the equation, from which I receive $P((\bar{X} - 3)/(0.4984S) > 1)$. Then I said that $0.4984S = \sigma / \sqrt{n} = 2 / 5$. But that is also incorrect, according to my professor. 
Please give me some tips numerically or conceptually as to how I should approach solving these kinds of problem. 
Thanks!

$$P(\bar{X} > 3 + 0.4984S) = 1 - P(\bar{X} \le 3 + 0.4984S)$$
where
$$P(\bar{X} \le 3 + 0.4984S) = P(\frac{\bar{X} - 3}{S} \le 0.4984)$$
$$= P(\sqrt{25}\frac{\bar{X} - 3}{S} \le 0.4984\sqrt{25})$$
3 is the true mean so we have
$$= P(Z \le 0.4984\sqrt{25}) = P(Z \le 2.492)$$
Then look up 2.492 on the Z-table or computer
