Likelihood of a given sample mean? (p-values versus confidence intervals)

Question

My issue is with comparing p-values to confidence intervals. The different values being churned out are too great, and I don't understand why. So, an example to illustrate my problem:

A sample of size $n=100$ has mean $\bar{x}=1000$. The population has standard deviation $\sigma=5$. Is the population mean $\mu=990$ likely?

1. So, $990$ has a $z$-score of $-2$, so the probability of the true mean $\mu$ lying between $1000\pm10$ is $1-0.2275\times2=0.545$.

2. On the other hand, for a confidence interval to contain $990$ we require $z^{\ast}\geq20$, as we need $10\geq z^{\ast}\frac{\sigma}{\sqrt{n}}\Rightarrow z^{\ast}\geq20$, which is silly!

Okay, so method (1) is flawed, as (I believe) there is a $0.545$ probability of a given sample point being in this range, not the mean of a given sample. But I don't understand why the difference is so huge - why the $z$-scores are so so different.

Answer 1

Keeping this in the form of guidance and hints:

Note that the z-score of the sample mean isn't 2.

When you calculate the correct standard deviation of a sample mean, you should be able to get the correct z-value.