Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are 36 white and 64 black balls in the bin. After we have randomly picked some ball we put it back to the bin. How many times we must pick balls from the bin to be sure, that probability of frequency of picking white ball vary from 0.36 at least 0.12 is equal to 0.1.


$P\left ( \left | \frac{X}{n} - 0.36 \right | \geq 0.12 \right )=0.1$

Answer: $n \geq 40 $. How did they get this result?

share|cite|improve this question
Here $0,36 = 0.36$? – Patrick Li Jan 13 '13 at 23:02
Yes, I fixed this. – pavlucco Jan 13 '13 at 23:14
up vote 3 down vote accepted

There are two ways to approach this problem, one is exact, the other using a normal approximation, as alluded to in the title of your question.

If we denote by $X_n$ the number of white balls in the first $n$ draws, then $$ \frac{X_n-.36n}{\sqrt{.36\times.64\times n}} $$ is approximately standard-normally distributed. Consequently the probability that the relative frequency of white balls $X_n/n$ differs from the expected probability $.36$ by at least $.12$ is equal to $$ P\left(\left|\frac{X_n}{n}-.36\right|\geq .12\right)=P\left(\left|\frac{X_n-.36 n}{\sqrt{.36\times.64\times n}}\right|\geq \frac{.12n}{\sqrt{.36\times.64\times n}} \right)\approx P\left(|N|\geq \frac{.12n}{\sqrt{.36\times.64\times n}}\right), $$ where $N\sim\mathcal{N}(0,1)$ is a standard normal random variable. In order to find $n$ such that the last probability is equal to $.1$ you can use the cumulative distribution function $\Phi(x)=P(N\leq x)$ of the normal distribution and observe that $$ P(|N|\geq x)=1-P(|N|<x)=1-\Phi(x)+\Phi(-x)=0.1 \Rightarrow x=1.64485. $$ We thus find $n$ by solving $$ \frac{.12n}{\sqrt{.36\times.64\times n}}=1.64485 \Rightarrow n=43.2887. $$

Hence, for any $n\leq 43$ (large enough for the normal approximation to be valid), the probability that the relative frequency of white balls $X_n/n$ differs from the expected probability $.36$ by at least $.12$ is at least $.1$

share|cite|improve this answer
How did you get 1.64485 in: $$ P(|N|\geq x)=1-P(|N|<x)=1-\Phi(x)+\Phi(-x)=0.1 \Rightarrow x=1.64485. $$ Every time I try to calculate this I recieve diffrent result. – pavlucco Jan 14 '13 at 0:06
@pavlucco: Using the symmetry of the normal distribution you can deduce that $\Phi(-x)=1-\Phi(x)$, so that you only need to solve $2\Phi(-x)=.1$ or $\Phi(-x)=.05$. This can be looked up in tables and is also implemented in many calculators. Alternatively, you may enter Solve[CDF[NormalDistribution[0,1],x]==0.05,x] at . – Eckhard Jan 14 '13 at 8:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.