Answer:
0.148 or 14.80%
Step-by-step explanation:
Mean overhead reach (μ) = 200 cm
Standard deviation (σ) = 8.9 cm
The z-score for any reach, X, is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For a X = 209.30 cm:
[tex]z=\frac{209.3-200}{8.9}\\ z=1.045[/tex]
A z-score of 1.045 corresponds to the 85.20th percentile.
Therefore, the probability that an individual distance is greater than 209.30 cm is:
[tex]P = 1-0.8520\\P=0.148=14.80\%[/tex]
The probability is 0.148 or 14.80%
