Approximating a probability from a sample of size $200$ I'm having some issues with this problem.

The median age of residents of US people is $35.6$ years. If a survey of $200$ residents is taken, approximate the probability that at least $110$ will be under $35.6$ years of age. 

So $n=200$ and $p = 110/200 = .55$
$$Z= \frac{X - 200 \cdot 0.55}{\sqrt{200 \cdot 0.55 \cdot (1-0.55)}}$$
 A: What's the issue? You're almost done.
Plug X = 35.6 into the formula you gave, calculate $Z$, and then find the area to the left of this $Z$-value in the standard normal distribution.
We get $Z = -10.5747$, and then  this on-line calculator  tells us that the associated probability is $0.00003$.
A: The median of a distribution is $m$ such that $F(m)=.5$ that is $P(x\leq m)=.5$. Here, you are given that the median is 35.6. This suggests that any randomly drawn individual will have a .5 probability of being below that age. 
The probability that exactly $n$ people out of $200$ will be below the age of 35.6 has a binomial $B(200,.5)$ distribution. To directly calculate the probability 110 or more will be below the median age would be cumbersome using the binomial distribution. However here, since $N$ (the sample size) is quite large, this distribution is well approximated by the normal distribution with mean $Np=200*.5=100$ and standard deviation $\sqrt{Np(1-p)}=\sqrt{200*.5*.5}=\sqrt{50}$. 
That is to say, the number of people who are below the median age from a sample of 200 is approximately normal with mean 100 and standard deviation 50. Let X be the number of people below the median age. The value $Z=\frac{X-100}{\sqrt{50}}$ is distributed standard normal (mean $0$ and standard deviation $1$). 
Plugging in $p=.5$ (probability an observation is below the median) $X=110$ gives $Z=1.41421$. Using a cumulative standard normal table tells us that the area above $1.41421$ is $.079$.
