I am given the following question in one of the lectures I was looking through.
Suppose a random sample of size $n = 400$ is to be selected from a population of size $N = 2000$. A quantitative characteristic of $x$ is of interest. Suppose that the population has mean $\mu= 53$ and standard deviation $\sigma= 10$. Let $\bar{x}$ denote the sample average. Use the normal distribution to calculate the probability that $\bar{x}$ falls between $52.1$ and $53.9$, that is, calculate $P(52.1 < \bar{x} < 53.9)$
(a) $0.75$
(b) $0.87$
(c) $0.93$
(d) $0.95$
My steps were as follows:
$$Z = \frac{(52.1-53)}{\frac{10}{\sqrt{400}}} = -1.8$$
and
$$Z = \frac{(53.9-53)}{\frac{10}{\sqrt{400}}} = 1.8$$
I went to a $z$ score table and got the values of $.0359$ and $.9641$ respectively. I want the area between these values which is $$0.9641 - 0.0359 = 0.9282$$ so I figured the answer is (c) $0.93$, but the correct answer is (d) $0.95$.
Where am I going wrong in my calculations?
You must use the finite population correction factor, because: $$\frac nN=\frac{400}{2000}=0.2>0.05.$$ So, the standard deviation of the sampling distribution of the sample means (or standard error) is: $$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\cdot \sqrt{\frac{N-n}{N-1}}$$ so you must have: $$P(52.1 < \bar{x} < 53.9)=\\ P\left(\frac{52.1-53}{\frac{10}{\sqrt{400}}\cdot \sqrt{\frac{2000-400}{2000-1}}}<z<\frac{53.9-53}{\frac{10}{\sqrt{400}}\cdot \sqrt{\frac{2000-400}{2000-1}}}\right)\approx\\ P(-2.01<z<2.01)\approx 0.95.$$