Answer
$91.92\%$.
Work Step by Step
$z=\frac{\text{data item-mean}}{\text{standard deviation}}$.
Hence here: $z=\frac{100-121}{15}=-1.4$
Using the table, the percentage of data items in a normal distribution that lie above it is $100\%$ minus the percentile that lies below it (we obtain this from the table): $100\%-8.08\%=91.92\%$.
