Answer :
The coin is flipped at least 52 times in order to obtain a 99% confidence interval of width of at most . 18 for the probability of flipping a head.
Hence, Option F is correct answer.
In a sample with a number n of people surveyed with a probability of a success of π , and a confidence level of (1-α), we have the following confidence interval of proportions.
[tex]\pi[/tex] ± z[tex]\sqrt{\frac{\pi \p(1-\pi )}{n} }[/tex]
z is the z-score that has a pvalue of [tex]1-\frac{\alpha }{2}[/tex]
Since, the coin is fair, so [tex]\pi =0.5[/tex].
The margin of error is:
[tex]M=z\sqrt{(\frac{\pi (1-\pi )}{n} )}[/tex]
99% confidence level:
So [tex]\alpha =0.01[/tex] ,z is the value of Z that has a p value of 1-[tex]\frac{0.01}{2}[/tex]=0.995,
so Z= 2.575
How many times would we have to flip the coin ?
We have to flip the coin in order to obtain a 99% confidence interval of width of at most 18 for the probability of flipping a head at least n times.
n is found when M=0.18 .
So
M=[tex]z\sqrt{\frac{\pi (1-\pi )}{n} }[/tex]
[tex]0.18=2.575\sqrt{\frac{0.5*0.5}{n} }[/tex]
[tex]0.18\sqrt{n} =2.575*0.5[/tex]
[tex]\sqrt{n}=\frac{2.575*0.5}{0.18}[/tex]
[tex]\sqrt{n} ^{2} =(\frac{2.575*0.5}{0.18} )^{2}[/tex]
n = 51.16
Thus, we have to flip the coin at least 52 times.
Learn more about the Probability here: https://brainly.com/question/24756209
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