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The time required to finish a test in normally distributed with a mean of 80 minutes and a standard deviation of 15 minutes. What is the probability that a student chosen at random will finish the test in more than 110 minutes?
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34% 16% 82% 2%
\[P(X>110)=P\left(\frac{X-80}{15}>\frac{110-80}{15}\right)=P(Z>2)\]
If you're not familiar with \(z\) tables and the computation of these sorts of probabilities, the other way to determine the value of \(P(Z>2)\) is to ask - what proportion of a population lies 2 standard deviations away from the mean?|dw:1406080407791:dw|
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