A random sample of size n = 24 measurements is drawn from a normal population. The sample has a mean of 103.6 and a standard deviation of 12.5. If the true population is 100, find the probability of taking a random sample of 24 measurements and getting a mean of at least 103.6.
What I know:
$Z = \frac{\overline{X}- \mu }{ \frac{ \sigma }{ \sqrt{n} }} $
$\overline{X}$ is the sample mean of 24 random measurements.
The mean of $\overline{X}$ is the population mean which is 103.6.
The standard deviation $s$ of $\overline{X}$ is given by $s = { \frac{ \sigma }{ \sqrt{n} }} = { \frac{12.5}{\sqrt{24}}} = 2.5516$
Therefore,
$P(\overline{X} \geq 103.6) $
$= P(Z \geq { \frac{ 103.6 - \mu}{ s }}) $
$= P(Z \geq { \frac{ 103.6 - 103.6}{ 2.5516 }})$
$ = P(Z \geq 0)$
Then by using the z-score table would that then be that the $P(Z \geq 0) = .5000$ ?
Start by fixing a couple of mistakes--mostly after "Therefore":
(a) The standard deviation (SD) of $\bar X$ is not the population SD $\sigma$, but $\sigma/\sqrt{n}$ as you show at the very start.
(b) The sample SD 12.5 estimates $\sigma.$ With $n$ as large as 24, your book may want you to go ahead and use $\sigma \approx 12.5.$
(c) The sample mean $\bar X = 103.6$ is not the population mean $\mu.$ (In your problem you are told to assume that the true value is $\mu = 100.)$
Read the chapter that precedes this problem in your textbook to see if you can get the right answer once you fix these mistakes.