A population has mean $\mu=32$ and standard deviation $\sigma=2$. Find $\mu_{\bar{x}}$ and $\sigma_{\bar{x}}$ for samples of size $n=25$. Round your answers to one decimal place if needed.
\[
\begin{array}{l}
\mu_{\bar{x}}=\square \quad \square \quad 0 \quad \square \\
\sigma_{\bar{x}}=\square \quad \square
\end{array}
\]

Step 1 ：The mean of the sample means, denoted as \( \mu_{\bar{x}} \), is equal to the population mean, \( \mu \). So, \( \mu_{\bar{x}} = \mu = 32 \).

Step 2 ：The standard deviation of the sample means, denoted as \( \sigma_{\bar{x}} \), is equal to the population standard deviation, \( \sigma \), divided by the square root of the sample size, \( n \). So, \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).

Step 3 ：Let's calculate \( \sigma_{\bar{x}} \) using the given values of \( \sigma = 2 \) and \( n = 25 \).

Step 4 ：After executing this calculation, we get the value of \( \sigma_{\bar{x}} = 0.4 \).

Step 5 ：Final Answer: \[ \begin{array}{l} \mu_{\bar{x}}=\boxed{32} \ \sigma_{\bar{x}}=\boxed{0.4} \end{array} \]