How to find confidence interval of expected sample results from true population proportion? Let's say I know the true population proportion of a particular attribute, say 15% of people are left-handed. Imagine I perform a random sample of 1000 people around the world. How do I compute a 95% confidence interval of the number of people I expect to be left-handed from this sample? 
 A: Exact binomial computation. If $X \sim \mathsf{Binom}(n=1000, p=0.15),$ then
$P(128 \le X \le 172) = 0.954,$ according to exact
computations using R statistical software.
[In R qbinom the inverse of a binomial CDF, and dbinom is a binomial PDF.]
qbinom(c(.025,.975), 1000, .15)
[1] 128 172
sum(dbinom(128:172, 1000, .15))
[1] 0.9538581

If this is from a problem in an elementary statistics
or probability class that is not using software, then
you may be expected to get a good approximate value
by using the fact that the distribution of $X$
is approximately $\mathsf{Norm}(\mu = np, \sigma = \sqrt{np(1-p)}),$ where $n = 1000, p = 0.15.$
You could use either software or printed tables of
the standard normal CDF for that computation.
One method using R is shown below. The area under
the density curve of $\mathsf{Norm}(150, 11.29)$
between 127.5 and 172.5 is about 0.954. So the normal
approximation method gives essentially the same integer
boundaries.
n = 1000;  p = .15
mu = n*p;  sg = sqrt(n*p*(1-p))
mu;  sg
[1] 150
[1] 11.29159
qnorm(c(.025,.975), mu, sg)
[1] 127.8689 172.1311
diff(pnorm(c(127.5,172.5), mu, sg))
[1] 0.9536984

The following figure shows the binomial distribution (blue histogram) and the approximating normal density curve (black curve).

