# Chebyshev's Theorem

• Feb 11th 2009, 09:15 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, 11:24 AM
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 $\displaystyle H=\frac{X}{900}$

You want to prove that $\displaystyle P\{0.4<H<0.6\} \geq \frac{35}{36}$

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

$\displaystyle \boxed{0.4<H<0.6} \leftrightarrow [900 \cdot 0.4<X<900 \cdot 0.6]$ $\displaystyle \leftrightarrow [360<X<540] \leftrightarrow \boxed{-90<X-\mu(X)<90}$

So $\displaystyle P\{0.4<H<0.6\}=P\{-90<X-\mu(X)<90\}=P\{|X-\mu(X)|<90\}$ $\displaystyle =1-P\{|X-\mu(X)|>90\}$

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

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

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

------> $\displaystyle P\{0.4<H<0.6\}=1-P\{|X-\mu(X)|>90\} \geq 1-\frac{1}{36}=\frac{35}{36}$