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The average time spent sleeping (in hours) for a group of medical residents at a hospital can be approximated by a normal distribution, with mean of 6.1 hours and standard deviation of 1.0 hour. Let x represents a random medical resedent selected at the hospital. Find P(x < 5.3).

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First we calculate the z-score for this situation.

Z=(X- \mu)/(\sigma) = (5.3-6.1)/(1) = -0.8
Finding P(x < 5.3) is the same as finding P(z < -0.8). Using a table of standard normal distribution and z-scores we see that the probability is 0.21186.
User Harshal Kshatriya
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