Say you select samples randomly from the same population. The sampling mean x^ is the random variable (a value obtained from the random process of selection)defined as the mean of the values of a given sample. The CLT then says that (among other things) if
you take a large-enough number of random samples, all of the same size N, and for
each sample s with values $x_s1,x_s2,....,x_sN $ , you calculate:
$x_s$:=$\frac{x_s1+....+xs_N}{N}$
That the collection of all these values of x^:=sampling mean has a normal distribution with mean equal to the population mean, and has a standard deviation equal to the population standard deviation divided by $ n^{1/2}$.
Then the values of $x_s$ will be (are) normally-distributed, with mean $\mu_s$=
$\mu_{pop}$ , where $\mu_{pop}$ is the true population mean, i.e., the value you would get if you were to sample every single member of your population (but you can tell that doing this is often impractical and/or too costly) ,and then divide by the size of the population.
Say, now (assuming we don't know the true value of the population mean, otherwise no point in collecting sample data), you collect a sample s' of size N from your population, and you get a sample mean $x_s'$. Now, unless you know in advance the value of $\mu$ , there is no way of knowing whether it equals $x_s'$; the best you can do is to use the CLT to determine an interval centered at $x_s'$ that contains the value of $\mu$, with a certain probability. For this, you use the CLT: specifically, the value of $x_s'$ will be a certain number k of deviations from the mean. Now, using the fact that x^ is normally-distributed, you use the probability of obtaining a value that is k deviations from the mean, and this gives you the interval. Let me exapand on this a little later.
Not making assumptions about the distribution of salaries, I'd say the question is impossible to answer. You do know that your sample will have the same expectation value as the population mean, and that the variance of the mean of the sample can be calculated, taking advantage of the smallness of $n/N$, as
$$\sqrt{\frac{\sigma^2}{n}(1-\frac{n-1}{N-1})} \approx \sqrt{\frac{\sigma^2}{n}} = \$15,132$$
If we take the distribution to be normal, then answering your question is a walk in the park from here. But that's not stated, so I don't think we can do better than Chebyshev's inequality. $\$20,000$ is $k=1.322$ of the value we got for the standard deviation of the sample mean. So the probability of that the mean of a sample is beyond $\$20,000$ cannot be more than $1/k^2=0.572$. So the sought probability is at least $42.8\%$.
Best Answer
Suppose I show you a fair coin. You take this coin and flip it 10 times, and record the number of heads and the number of tails you observed. Even though it is "fair" in the sense that each coin toss has equal probability of landing heads versus landing tails, is it guaranteed that you will observe exactly 5 heads and 5 tails? Of course not. Indeed, if you were to flip the coin an odd number of times, it isn't even possible to get an equal number of heads and tails, despite the coin being fair.
So, this illustrates that a sample mean is not the same as a population mean. The sample mean is a statistic, which is subject to sampling variation--in other words, it is the result of the observations we made, and may change each time we take a sample because there is randomness involved in the process of sampling.
Now, regarding your specific question, we are given that the distribution of salaries has mean $\mu$ and standard deviation $\sigma = 82000$. Then by the Central Limit Theorem, the sample mean $\bar x$ of the salaries of $n = 80$ randomly selected players is approximately normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n} = 9167.88$. The probability that $\bar x$ is within $15000$ of the population mean is then $$\Pr[|\bar x - \mu| \le 15000] = \Pr\left[\left| \frac{\bar x - \mu}{\sigma/\sqrt{n}} \right| \le \frac{15000}{9167.88} \right] = \Pr[|Z| \le 1.63615],$$ where $Z$ is a standard normal random variable (mean = 0, standard deviation = 1). Now we can look this probability up in a $z$-table, which gives us $0.898191$, or about an $89.8\%$ chance that the sample mean is within $15000$ of the population mean.