Answer
$0.9893$
Work Step by Step
$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$
Hence here the z-score is: $\frac{80-64}{7}\approx 2.3$
Then, using the table: $P(x\leq 80)\approx P(z\leq 2.3)=0.9893$
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