Describing a coin toss using Central Limit Theorem

Question

I did solve this one, however I am not sure if I have solved it correctly, so I am asking for a second opinion here.

This is the problem:

A fair coin is flipped n times. Let X be the number of times that the outcome is head. What is the smallest value of n for which the given inequality is satisfied?

$0.99 ≤ P(0.49 ≤ \frac{X}{n}≤ 0.51)$

Ok so, I tried to consider the random variable of the result of a coin toss as a uniform distribution in which 1 is heads and 0 is tails. So we get a mean of 0.5 and a standard deviation of 0.5.

Now, lets say y is the average of all the outcomes of the values obtained from tossing a coin n times. Then y would be the sum of heads (considering tails is 0) divided by n. So we can say y follows a normal distribution with mean 0.5 and standard deviation of $\frac{(0.5)}{(\sqrt{n})}$. So we can directly substitute y for $\frac{X}{n}$ in the inequality above.

If I apply the transformation $Z = \frac{(y - 0.5)}{(0.5/\sqrt{n})}$ and I solve the inequality above, I get $16641 ≤ n$. Rounding that I get that n should be at least $16641$ for the inequality to hold. I am not sure if that is correct though. Could someone offer some help?

Thank you in advance.

Answer 1

Comments interspersed below.

Ok so, I tried to consider the random variable of the result of a coin toss as a uniform distribution in which 1 is heads and 0 is tails. So we get a mean of 0.5 and a standard deviation of 0.64.

As @LeeDavidChungLin mentioned in a comment, your calculation of $\sigma$ is incorrect here. (If you'd like to show your work for it, we can comment in further detail.)

Now, lets say y is the average of all the outcomes of the values obtained from tossing a coin n times. Then y would be the sum of heads (considering tails is 0) divided by n. So we can say y follows a normal distribution with mean 0.5 and standard deviation of $\frac{(0.64)}{(\sqrt{n})}$. So we can directly substitute y for $\frac{X}{n}$ in the inequality above.

OK other than the mistaken $0.64$ referenced above.

If I apply the transformation $Z = \frac{(y - 0.5)}{(0.64/\sqrt{n})}$ and I solve the inequality above, I get $13.77 ≤ n$. Rounding that I get that n should be at least $14$ for the inequality to hold. Yet I don't really feel like this is the right answer because it is such a small number.

I can't follow the part where you say, "and I solve the inequality above, I get $13.77 \leq n$," but I'm pretty sure there's a second mistake there. Can you show your work for that part in greater detail? In particular, how did you use the $0.99 = \dots$ part of the original equation?