A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed? 
A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

I need a clarification on how to use binomial distribution formula in this problem. What i got was: 
$$x = 55, \quad \mu = 50, \quad SD = 5$$
$$ z = (55-50)/5 \implies z = 0.84 \implies 1- z = 0.16$$
$$p(x) = 0.16$$
Is this the right answer? 
 A: The exact answer is obtained from the Binomial distribution. Various pieces of software, including Wolfram Alpha, will do the computation for you. This approach is painful to carry out by hand. 
Perhaps you are expected to use the normal approximation to the binomial. If $X$ is the number of heads, then $X$ has approximately normal distribution, mean $50$, standard deviation $\sqrt{100(1/2)(1/2)}=5$.
We want to find the probability that $X\ge 56$. We find, approximately, the probability of the complement, the probability that $X\le 55$. This, approximately, is the probability that a normal with mean $50$ and standard deviation $5$ is $\le 55$.
But perhaps you are expected to use the continuity correction. Then
$$\Pr(X\le 55)\approx\Pr\left(Z\le \frac{55.5-50}{5}\right)\tag{1}$$ 
where $Z$ is standard normal. If you are not expected to use the continuity correction, use $55$ instead of $55.5$.
In either case, the answer to the original question is, approximately, $1$ minus the number obtained in (1).
A: As lab bhattacharjee mentioned, the probability of exactly $k$ heads is  
$\binom{n}{k} p^k (1-p)^{n-k}$
Let us write $X$ to denote the number of heads, and $Q$ to denote the desired probability, $P(X > 55)$. Note that by symmetry we have $P(X > 55) = P(X < 45)$. Also,
$P(X > 55) + P(X < 45) + P(45 \leq X \leq 55) = 1$
so $2Q + P(45 \leq X \leq 55) = 1$. Therefore the problem reduces to evaluating $Z = P(45 \leq X \leq 55)$.
We can use another symmetry argument to evaluate $Z$ by observing that $P(45 \leq X < 50) = P(50 < X \leq 55)$. This means that we will only need to calculate $\binom{n}{k} p^k (1-p)^{n-k}$ for $50 \leq k \leq 55$, substantially reducing the workload.
A: HINT:
From this,
the probability of getting exactly $k$ successes in $n$ trials  is $$\binom nk p^k(1-p)^{n-k}$$
Here $\displaystyle p=\dfrac12\implies p^k(1-p)^{n-k}=\left(\frac12\right)^n, n=100$
and $\displaystyle\sum_{k>55}^{100}\binom nk=\frac{(1+1)^{100}}2-\sum_{k=50}^{55}\binom nk$
A: Considering $n = 100$ is very large we are going to approximate this as normal distribution
$$\mu = np = 100*0.5 =50$$
$$\sigma =\sqrt{npq} = \sqrt{100*0.5*0.5} = 5$$
Using the mean and standard deviation we can calculate a Z-score:
$$z = \frac{x - \mu}{\sigma} = \frac{55 - 50}{5} = 1$$
$$P(z > 1) \approx 0.16$$
