Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head? (note that the z-score was rounded to three decimal places in the calculation)a) 217b) 153c) 212d) 209e) 150f) None of the above

Answer:153 timesStep-by-step explanation:We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14Width = 0.14ME = [tex]\frac{width}{2}[/tex]ME = [tex]\frac{0.14}{2}[/tex]ME = [tex]0.07[/tex][tex]ME\geq z \times \sqrt{\frac{\widecap{p}(1-\widecap{p})}{n}}[/tex]use p = 0.5z at 95.8% is 1.727(using calculator)[tex]0.07 \geq 1.727 \times \sqrt{\frac{0.5(1-0.5)}{n}}[/tex][tex]\frac{0.07}{1.727}\geq sqrt{\frac{0.5(1-0.5)}{n}}[/tex][tex](\frac{0.07}{1.727})^2 \geq \frac{0.5(1-0.5)}{n}[/tex][tex]n \geq \frac{0.5(1-0.5)}{(\frac{0.07}{1.727})^2}[/tex][tex]n \geq 152.169[/tex]So, Option B is true Hence  we have to flip 153 times the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head