Estimate the $95$% confidence interval of one sample unknown variance and standard deviation

A preliminary sample of $1200$ engineers has been taken, out of which $260$ are females. The sample is used to estimate the proportion of female engineers denoted by $pF$ .

Estimate the $95$% confidence interval for $pF$.

I know how to calculate CI with known variance or standard deviation, but I can't seem to find a way to do it with unknown variance or standard deviation.

There is a quick little approximation for CIs for sequences of bernoulli trials.

$$\dfrac{n_F}{n} \pm z_{1-\alpha/2}\sqrt{\dfrac{n_Mn_F}{n}}$$

Here, $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile from the standard normal distribution, $n_F$ is the number of observed females, $n_M$ is the number of observed males, and $n$ is the sum of both males and females.

So, doing some plugging and chugging yields...

$$\dfrac{260}{1200} \pm \dfrac{1.96}{1200} \sqrt{\dfrac{260\cdot (1200-260)}{1200}}$$

So the CI is approximately $[0.19,0.24]$

It may be worth noting that this estimate is called a Wald Interval and is known to be a bad estimate over all. Agresti-Coulli intervals for binomial proportions are better, but I doubt that is what your instructor is asking.