So suppose we flip a coin $n$ times, if we define $X= \text{ the number of tails in those $n$ flips}$, which is the smallest value of $n$ that satisfies that $P[49 \leq X \leq 51]\geq 0.95$.

I got the hint to use a normal approximation i.e:

$$P[49 \leq X \leq 51] = P \left[ \frac{49-\frac{n}{2}}{\sqrt{\frac{n}{4}}}\leq \frac{X-\frac{n}{2}}{\sqrt{\frac{n}{4}}}\leq \frac{51-\frac{n}{2}}{\sqrt{\frac{n}{4}}} \right]$$

And then you have to find $n$ such that

$$\int^{\left.\left(51-\frac{n}{2}\right)\right/\sqrt{\frac{n}{4}}}_{\left.\left(49-\frac{n}{2}\right)\right/\sqrt{\frac{n}{4}}} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx \geq 0.95$$

I'm stuck in that part could you help me? Thanks a lot

  • $\begingroup$ Oh you're right, silly error when transcribing, will edit $\endgroup$
    – user371816
    Oct 31, 2017 at 16:09
  • $\begingroup$ I think what you want is the number of tails to be between 49% and 51% of n rather than 49 and 51, right? $\endgroup$
    – A.G.
    Oct 31, 2017 at 16:14
  • $\begingroup$ You are right, misread the question. How does the question change? $\endgroup$
    – user371816
    Oct 31, 2017 at 16:18
  • $\begingroup$ just replace 49 and 51 by .49n and .51n. $\endgroup$
    – A.G.
    Oct 31, 2017 at 16:20

1 Answer 1


What you want is $.49 n\leq X \leq .51 n$. It is "well known" that 95 % of a normal distribution happens for $\pm$ 1.96 standard deviations. Therefore you want $$ {.51\,n - {n\over2} \over \sqrt{n\over 4}}\geq 1.96\rightarrow n\geq 9604. $$

  • $\begingroup$ So, I'm a bit lost in the second part of your comment, would you mind expanding on the 1.96 standard deviations? $\endgroup$
    – user371816
    Oct 31, 2017 at 16:16
  • $\begingroup$ @user371816 en.wikipedia.org/wiki/1.96 $\endgroup$
    – A.G.
    Oct 31, 2017 at 16:18
  • 2
    $\begingroup$ @user371816 : $$ \int_{-1.96}^{1.96} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx \approx 0.95. $$ Using R, I see that the number rounded here to $1.96$ is $\approx 1.959964,$ so $1.96$ is more accurate than you would expect with something rounded to the nearest $0.01. \qquad$ $\endgroup$ Oct 31, 2017 at 16:21
  • $\begingroup$ Just a further question, I get where the 1.96 comes from, but why would we want ${.51\,n - {n\over2} \over \sqrt{n\over 4}}\geq 1.96$ what does that mean? $\endgroup$
    – user371816
    Oct 31, 2017 at 16:29
  • 1
    $\begingroup$ @user371816 : That was a guess about what question you should have asked rather than about what question you did ask. The point is that the $\operatorname{Binomial}(n,1/2)$ distribution is approximated by the normal distribution with the same expected value, $n/2,$ and the same standard deviation, $\sqrt{n/4\,}.$ $\qquad$ $\endgroup$ Oct 31, 2017 at 17:28

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