# Chebyshev's Theorem

• Feb 11th 2009, 10:15 PM
Yan
Use the Chebyshev's theorem and 1-(p(1-p)/(nc^2)) to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6

Mr F edit: Merged post from a new thread:

Use the Chebyshev's theorem and Binomial distribution to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6.
• Feb 15th 2009, 12:24 PM
Moo
Hello,
Quote:

Originally Posted by Yan
Use the Chebyshev's theorem and Binomial distribution to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6.

Let X be the number of heads that you get. This follows a binomial distribution of parameters (n,p), where p=0.5, since it's a balanced coin.

The proportion H of heads is given by $H=\frac{X}{900}$

You want to prove that $P\{0.4

The mean of a binomial distribution is np.
So $\mu(X)=900 \cdot 0.5=450$

$\boxed{0.4 $\leftrightarrow [360

So $P\{0.4 $=1-P\{|X-\mu(X)|>90\}$

But we know by Chebyshev's inequality :
$P\{|X-\mu(X)|>90\} \leq \frac{V(X)}{90^2}$

The variance of a binomial distribution is $np(1-p)$.
So $V(X)=900 \cdot 0.5^2=225$

Hence $P\{|X-\mu(X)|>90\} \leq \frac{225}{90^2}=\frac{1}{36}$

------> $P\{0.490\} \geq 1-\frac{1}{36}=\frac{35}{36}$