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\begin{array}{l}\sigma=20 \\n=10 \\\bar{x}=185\end{array}The nute and alternative hypothesis are:-\begin{array}{l}H_{0}: \mu=175 \\H_{1}: \mu>175 \text { (Right tailed test) }\end{array}Here for 2, Type 1 Probability when critical region is \bar{x}>175.As we know:-z=\frac{\bar{x}-\mu}{S / \sqrt{n}}\begin{aligned}P(\bar{x}>185) & =1-P\left(Z<\frac{185-175}{20 / \sqrt{10}}\right) \\& =1-P(Z<1.58) \\P(\bar{x}>185) & =1-0.9429 \\P(\bar{x}>185)= & =.0571 \approx 0.06 \\P & =0.06 \text { Ang. }\end{aligned} ...