Posted by **Anonymous** on Tuesday, February 23, 2010 at 11:39am.

College students average 7.8 hours of sleep per night with a standard deviation of 45 minutes. If the amount of sleep is normally distributed, what proportion of college students sleep for more than 9 hours?

- statistics -
**dongo**, Tuesday, February 23, 2010 at 11:51am
You know the value of expectation, 7.8h, you know the deviation, 0.75h and you know that the amount of students, sleeping more LESS than (or equal) 9 hours is:

/bigPhi((9h-7.8h)/0.75h)=

/bigPhi(1.6)

Hence the required probability to find such a student (sleeping MORE than 9 hours) is:

P=1-/bigPhi(1.6)

- statistics -
**PsyDAG**, Wednesday, February 24, 2010 at 12:57pm
Z = (x-μ)/SD

Z = (9-7.8)/.75 = 1.6

In table in the back of your stat text called something like "areas under normal distribution," look up Z score for smaller area to get proportion.

I hope this helps a little more.

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