1. 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.

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

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