Once again, utilize the Central Limit Theorem. Notice that the symmetrical confidence interval
\begin{equation*}
P(\overline{x} - E \lt \mu \lt \overline{x} + E) = 1-\alpha.
\end{equation*}
is equivalent to
\begin{equation*}
P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt \frac{\overline{x} - \mu}{\sigma / \sqrt{n}} \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha
\end{equation*}
in which the middle term can be approximated using a standard normal variable and therefore this statement is approximately
\begin{equation*}
P \left ( \frac{-E}{\sigma / \sqrt{n}} \lt Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \alpha.
\end{equation*}
Using the symmetry of the standard normal distribution about Z=0 gives
\begin{equation*}
\Phi (z_{\alpha/2} ) = \Phi \left ( \frac{E}{\sigma / \sqrt{n}} \right ) = P \left ( Z \lt \frac{E}{\sigma / \sqrt{n}} \right ) = 1 - \frac{\alpha}{2}
\end{equation*}
and so to determine E again requires the inverse of the standard normal distribution function. Using an appropriate \(z_{\alpha /2}\) (as determine in a manner described in the previous section) gives a confidence interval for the mean
\begin{equation*}
\overline{x} - z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} \lt \mu \lt \overline{x} + z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}
\end{equation*}
with confidence level \(1-\alpha\) and margin of error
\begin{equation*}
E = z_{\alpha /2} \frac{\sigma}{\sqrt{n}}.
\end{equation*}