Here we have an unknown population mean $\displaystyle \mu $, and a known standard deviation $\displaystyle \sigma $ based on a simple random sample of size n, which we are also given.

$\displaystyle \bar{X} = 75 $

$\displaystyle n = 50 $

$\displaystyle \sigma = 10 $

Since our sample is large enough (usually n > 30 will suffice), we consider the sample mean $\displaystyle \bar{X} $ normally distributed, with the same expectation for the population mean $\displaystyle \mu $, but with standard error $\displaystyle \frac{\sigma}{\sqrt{n}} $. We standardize and get a random variable

$\displaystyle

Z = \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} =

\frac{75-\mu}{1.4142} $

that depends on the population mean parameter $\displaystyle \mu$ (which is to be estimated), but with a normal distribution independent of $\displaystyle \mu$. Thus, we can find the range endpoints -z and z, independent of the population mean parameter, which is the range that Z falls somewhere in between with 95% probability. So we set the following equation up:

$\displaystyle P(-z \leq Z \leq z) = 1 - \alpha = 0.95 $

$\displaystyle \implies P(Z \leq z) = 1 - \frac{\alpha}{2} =0.975 $

$\displaystyle \implies z = 1.96 $ (which you could have looked up easily enough (or had it memorized)

So now we have

$\displaystyle 0.95 = 1 - \alpha = P(-z \leq Z \leq z) =

P(-1.96 \leq \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \leq 1.96)

$

$\displaystyle

= P\bigg{(}75 - 1.96(1.4142)

\leq \mu \leq \bar{x} + 1.96 (1.4142 )\bigg{)}

$

$\displaystyle

= P(75 - 2.7719 \leq \mu \leq 75 + 2.7719)

$

$\displaystyle CI = (72.2281 ; 77.7719) $