Problem
ermine the area under the standard normal curve that lies to the right of the $z$-score of 0.81 . \begin{tabular}{c|cccccccccc} $\mathbf{z}$ & $\mathbf{0 . 0 0}$ & $\mathbf{0 . 0 1}$ & $\mathbf{0 . 0 2}$ & $\mathbf{0 . 0 3}$ & $\mathbf{0 . 0 4}$ & $\mathbf{0 . 0 5}$ & $\mathbf{0 . 0 6}$ & $\mathbf{0 . 0 7}$ & $\mathbf{0 . 0 8}$ & $\mathbf{0 . 0 9}$ \\ \hline $\mathbf{0 . 7}$ & 0.7580 & 0.7611 & 0.7642 & 0.7673 & 0.7704 & 0.7734 & 0.7764 & 0.7794 & 0.7823 & 0.7852 \\ $\mathbf{0 . 8}$ & 0.7881 & 0.7910 & 0.7939 & 0.7967 & 0.7995 & 0.8023 & 0.8051 & 0.8078 & 0.8106 & 0.8133 \\ $\mathbf{0 . 9}$ & 0.8159 & 0.8186 & 0.8212 & 0.8238 & 0.8264 & 0.8289 & 0.8315 & 0.8340 & 0.8365 & 0.8389 \end{tabular}
Solution
Step 1 ：The problem is asking for the area under the standard normal curve that lies to the right of the z-score of 0.81. The z-score table provided gives the cumulative probability of the standard normal distribution up to a given z-score.
Step 2 ：To find the area to the right of a z-score, we need to subtract the cumulative probability at that z-score from 1. Looking at the table, the cumulative probability at z=0.81 is 0.791.
Step 3 ：So, we need to calculate \(1 - 0.791\).
Step 4 ：The area under the standard normal curve that lies to the right of the z-score of 0.81 is \(\boxed{0.209}\).
