Question

Hi, can you help me answer this question, please, thank you!

Answer 1

It is given that standard deviation = 2

Margin error =0.5 mm

At 99% confidence interval the Z is;

`\begin{gathered} \text{ }\alpha\text{ = 1-99\%} \\ \alpha\text{ = 1 - 0.99} \\ \alpha=0.01 \\ (\alpha)/(2)=0.005 \\ Z_{(\alpha)/(2)}=2.58 \end{gathered}`

Sample size express as;

`\begin{gathered} n=(Z_{(\alpha)/(2)}*(\sigma)/(E))^2 \\ n=(2.58*(2)/(0.5))^2 \\ n=106.50 \\ n=107 \end{gathered}`

Answer : A) 107

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