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Sagot :
It is given that standard deviation = 2
Margin error =0.5 mm
At 99% confidence interval the Z is;
[tex]\begin{gathered} \text{ }\alpha\text{ = 1-99\%} \\ \alpha\text{ = 1 - 0.99} \\ \alpha=0.01 \\ \frac{\alpha}{2}=0.005 \\ Z_{\frac{\alpha}{2}}=2.58 \end{gathered}[/tex]Sample size express as;
[tex]\begin{gathered} n=(Z_{\frac{\alpha}{2}}\times\frac{\sigma}{E})^2 \\ n=(2.58\times\frac{2}{0.5})^2 \\ n=106.50 \\ n=107 \end{gathered}[/tex]Answer : A) 107
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