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n(1-p)=100(.85)=85, this works too\\  large population\\  so $\hat{p}=N(0.15,\sqrt{0.15*0.85}{100})$  this reduces to a z score problem, $z=\frac{.10-.15}{0.0357}=1.40$. so you can look this up on thie the  z table and it p=0.0808.\\ \subsection{Are sample proportions enough}  knowing that there is some variability associated with out sampling procedure, why not provide an interval?\\  Its usually a sample proportion($\hat{p}$), not the p( true population proportion) this is known.\\  \subsection{standard deviation vs Standard error}  since we often dont know p, we cant find the true standard deviation, so we use an estimate called the $\textbf{standard error}$\\  \[SE(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]\\