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Question:

How many times do we have to throw a normal coin so that, we are at least $99\%$ sure, that the percentage of heads will be between $49.5\%$ and $50.5\%$?

Solution:

$$\begin{align} 99\% \ \text{CI} &= p \pm 2.575 \cdot \sqrt{\frac{p(1-p)}{n}} \\ (0.495, 0.505) &= 0.5 \pm 2.575 \cdot \sqrt{\frac{0.25}{n}} \\ 0.005 &= 2.575 \cdot \frac{0.5}{\sqrt{n}} \\ n &= \left(2.575 \cdot \frac{0.5}{0.005}\right)^2 \\ &= 66306.25 \end{align}$$

then $66307$ tosses would be required.

Could you please explain the solution in a simple way? Or if it's possible to give an alternative solution?

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    Your solution method is fine, so I don't know what you mean by a "simple way".2012-06-03
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    What is it you don't understand? This seems like a standard computation using the normal distribution2012-06-03
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    Where does 2.575 come from???2012-06-03
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    @user32810 Probably from a table of values for the standard normal2012-06-03
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    Is this table somewhere to be found???2012-06-03
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    http://www.math.ucdavis.edu/~gravner/MAT135A/materials/standardnormaltable.pdf2012-06-03

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