Answer:
97.73%
Step-by-step explanation:
Mean (μ) = 170
Standard deviation (σ) = 5
Assuming a normal distribution, for any observed value, X, the z-score is determined by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X = 180:
[tex]z=\frac{180-170}{5}\\ z=2[/tex]
A z-score of 2 corresponds to the 97.73-th percentile of a normal distribution.
Therefore, the percentage of data that is less than 180 is 97.73%.