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 $\,(13\%-15\%)\big]\,?$

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 $16\%$.