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.
Here are some hints to help you structure your thinking:
- If $\sigma_{X_i} = 30$ what is the standard deviation of $\frac{1}{75}\sum_{1}^{75} X_i$? (Let's call this $\sigma_{\bar X}$)
- What is the $Z$ score of the difference $D=76-80$?
- What is the probability that a standard normal random variable will take on values less than $Z$?
To put OP out of his misery...
We are told that the router prices are normally distributed with a standard deviation of 30 and a mean of 80. What this problem is asking you to do is determine how likely it is that Maria's sample of 75 routers came from a normally distributed population with mean 80, standard deviation 30.
The first thing to do is calculate the standard deviation of the sample average of 75 iid observations from this population. As you correctly calculated, that would be $\frac{30}{\sqrt{75}} \approx 3.46$. So, 95% of the time, a sample average of 75 iid observations from a normally distributed population of mean 80, standard deviation 30 will fall i the interval $80 \pm 2\times 3.46$.
Maria's sample average is 76. The question is asking you for the probability of getting less than or equal to 76 for the sample average if it came from a population with mean 80 and standard deviation 30.
How many standard deviations below the mean is the sample average of $76$?
We already calculated that the standard deviation of a 75 observation mean to be 3.46, so 76 is $\frac{76-80}{3.46}\approx -1.15$ standard deviations below the expected sample average.
How likely is this? We simply need to look up the above value (-1.15) in a standard normal table (you should have been taught how to do this already) or you can use excels NORMSDIST(-1.15) to get a lower tailed probability. This will give you 0.12 or $12\%$
Best Answer
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 $42.8\%$.