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OpenStudy (anonymous):
STATISTICS: The mean time was 103.57s, and the standard deviation is 4.52s. If the normal model is appropriate, what percent of times will be less than 99.05s?
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OpenStudy (kropot72):
\[z=\frac{X-\mu}{\sigma}=\frac{99.05-103.57}{4.52}=-1\] Now you need to refer to your z-table
OpenStudy (kropot72):
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