Probability of $|H-T|$ in 10,000 coin tosses If a fair coin is thrown $10,000$ times. Using the binomial convergence to normal,find $P|H-T|\le 80$
My intuition say that mean is 0.But I am not able to proceed further. 
 A: Let denote the random variable for the number of heads as $H$ and the random variable for the number of tails as $T$. Then the sum of the outcomes of $H$ and $T$ is $10,000$:$H+T=10,000\Rightarrow T=10,000-H$ 
And it is given the inequality $|H-T|\leq 80$. We have to get rid of the absolute value. For this purpose we make a case differentiation.
1) Let $H\geq T$. The inequality becomes $H-T\leq 80\Rightarrow H\leq T+80$. 
Insterting the expression for T
$H\leq 10,000-H+80\Rightarrow 2H\leq 10,080\Rightarrow H\leq 5,040$
2) Now let $H< T$. The inequality becomes $T-H\leq 80\Rightarrow H\geq T-80$. 
Insterting the expression for T
$H\geq 10,000-H-80\Rightarrow 2H\geq 9,920\Rightarrow H\geq 4,960$
In total the interval is $4,960\leq H\leq 5,040$ in $10,000$ coin tosses. And $H$ is distributed as $H\sim Bin(10,000;0.5)$
Now you can  use the De Moivre-Laplace theorem to approximate:
$P(4,960\leq H\leq 5,040)\approx \Phi\left(\frac{5,040-\mu}{\sigma} \right)-\Phi\left(\frac{4,960-\mu}{\sigma} \right)$
$P(4,960\leq H\leq 5,040)\approx \Phi\left(\frac{5,040-5,000}{\sqrt{10,000\cdot 0.5\cdot 0.5}} \right)-\Phi\left(\frac{4,960-5,000}{\sqrt{10,000\cdot 0.5\cdot 0.5}} \right)$
=$\Phi(0.8)-\Phi(-0.8)$
$\Phi(z)$ is the cdf of the standard normal distribution, where $Z=\frac{H-\mu}{\sigma}$. It is symmetric around the mean. Thus $\Phi(-z)=1-\Phi(z)$. We get
$=\Phi(0.8)-\left[ 1-\Phi(0.8) \right]$
$=2\cdot \Phi(0.8)-1$
The value of $\Phi(0.8)$ can be looked up in this table or you can use this calculator.
I leave it to you to evaluate the value of $\Phi(0.8)$.
