Probability of senior citizens in a one million residence In a city of over $1000000$ residents, $14\%$ of the residents are senior citizens. In a simple random sample of $1200$ residents, there is about a $95\%$ chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the $5\%$ interval options ]
$N=1000000$ residents;
$p=14\%=0.14$ Senior Citizens
$n=1200$ residents simple random sample
$p=95\%=0.95$ chance that the $\%$ Senior Citizens is in the which interval?
$\quad\big[(9\%-19\%)\,$ or $\,(10\%-18\%)\,$ or $\,(11\%-17\%)\,$ or $\,(12\%-16\%)\,$ or $\,(13\%-15\%)\big]\,?$
 A: This is an application of a 95% confidence interval for a binomial proportion.
We calculate the confidence interval as such:
$$ p \pm z_{\frac{\alpha}{2}} \cdot SE $$
where the standard error is defined as: $$SE = \sqrt{\frac{p\cdot(1-p)}{n}}$$
It is quite straight-forward to "plug in" the data:
$p = 0.14$  (The probability of "success")
$n = 1200$  (The size of the sample)
$z = 1.96$  ("This value is the standard normal percentile having right-tail probability equal to $\frac{\alpha}{2}$." [1]) We get this value from the normal distribution, as we want a 95% Confidence Interval, so $\alpha=0.05$. $\frac{\alpha}{2}=0.025$ The corresponding z-score is 1.96.
The computation is easy to carry out:
$SE = \sqrt{\frac{0.14\cdot(1-0.14)}{1200}}$
$SE = \sqrt{\frac{0.14\cdot(1-0.14)}{1200}}$
$\therefore SE = 0.01002$
$\therefore Interval(p) = 0.14 \pm 1.96*0.01002$
$\therefore Interval(p) = (0.1203608,0.1596392)$
Rounded:
$Interval(p) = (0.12,0.16)$
So that will be the correct interval. 
Let me know if you need any clarification.
[1] Cited from Alan Agresti's Introduction to Categorical Data Analysis, 2nd Ed.
A: Let $X$ be the number of seniors in the sample. We are presumably sampling without replacement, but $1000000$ is very large compared to the sample size, so we can assume that the distribution of $X$ is binomial, $n=1200$, $p=0.14$. So $X$ has mean $(1200)(0.14)$ and standard deviation $\sqrt{1200(0.14)(0.86)}$. 
Let $Y=\frac{X}{1200}$, the sample proportion of seniors. 
Then $Y$ has mean $0.14$, and standard deviation $\frac{\sqrt{(0.14)(0.86)}}{\sqrt{1200}}\approx 0.0100$. 
Note that the distribution of $Y$ is very close to normal. So with probability $0.95$, $Y$ is within $(1.96)(0.0100)$ of the mean $0.14$. The nearest in the given list is $12\%$ to $16\%$. 
