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Your input: find the margin of error for the sample size $n=64$, standard deviation $\sigma=7$, and confidence level $95.0 \%$ using normal distribution.

First, find the critical value: $z_{\frac{\alpha}{2}}=1.95996398454005$

Next, find the standard error of the mean: $SE=\frac{\sigma}{\sqrt{n}}=\frac{7}{8}$

Finally, the margin of error is $ME=z_{\frac{\alpha}{2}} \cdot SE=1.95996398454005 \cdot \left(\frac{7}{8}\right) \approx 1.71496848647255$

Answer: $ME=1.71496848647255$