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Solution
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: $$$ME=1.71496848647255$$$
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