An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 212 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.)

Answer: (0.545, 0.647)Step-by-step explanation:Let p be the population proportion of all dies that pass the probe.Given : An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 212 of these passed the probe.i.e.Sample size : n= 356Sample proportion of all dies that pass the probe: [tex]\hat{p}=\dfrac{212}{356}\approx0.596[/tex]Critical value for 95% confidence interval ( Using z-value table) : [tex]z=1.96[/tex]Now, the 95% (two-sided) confidence interval for the proportion of all dies that pass the probe will be :[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]i.e. [tex]0.596\pm (1.96)\sqrt{\dfrac{0.596(1-0.596)}{356}}[/tex]i.e. [tex]0.596\pm (1.96)(0.026)[/tex] [tex]\approx0.596\pm 0.051=(0.596-0.051,\ 0.596+0.051)\\\\=(0.545,\ 0.647)[/tex]Hence, the 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.= (0.545, 0.647)