Originally Posted by
lisczz Do you mean $\displaystyle -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z $?
We can split that problem into two inequalities:
1) $\displaystyle -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}}$ and 2) $\displaystyle \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z $
For the first one, we can first multiply both sides of the inequality by $\displaystyle \frac{B}{\sqrt{n}}$,
and we arrive at $\displaystyle \frac{-Z(B)}{\sqrt{n}} = \bar{x} - B$.
Then, add $\displaystyle B$ to both sides:
$\displaystyle \frac{-Z(B)}{\sqrt{n}} + B = \bar{x} $
We can factor out $\displaystyle B$:
$\displaystyle B(\frac{-Z}{\sqrt{n}}+1) < \bar{x}$
and then divide both sides by $\displaystyle \frac{-Z}{\sqrt{n}}+1$, so the final inequality is:
$\displaystyle B < \frac{\bar{x}}{\frac{-z}{\sqrt{n}}+1}$.
It's the same process for the second inequality. Once you have both inequalities, you can mesh them together to form one big inequality with $\displaystyle B$ in the center.