The sample size given in the question is
[tex]n=37[/tex]
The mean weight is
[tex]\bar{x}=50[/tex]
The standard deviation is
[tex]\sigma=8.4[/tex]
The margin of error is calculated using the formula below
[tex]\text{MOE = Z-score(90\% C.I)}\times\frac{\sigma}{\sqrt[]{n}}[/tex]
Using the Z-score table, the Z-score for the 90% confidence interval is
[tex]=1.645[/tex]
By substituting the values in the formula above, we will have
[tex]\begin{gathered} \text{MOE = Z-score(90\% C.I)}\times\frac{\sigma}{\sqrt[]{n}} \\ \text{Margin of error(MOE)} \\ =1.645\times\frac{8.4}{\sqrt[]{37}} \\ =\frac{13.818}{\sqrt[]{37}} \\ =\pm2.272\text{ounces} \end{gathered}[/tex]
Hence,
The final answer is = ±2.272 ounces
