Respuesta :
Answer:
153 times
Step-by-step explanation:
We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14
Width = 0.14
ME = [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 = 0.5
z 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
