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In a recent survey, the total sleep time per night among college students was approximately Normally distributed with mean μ

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a. Standard deviation = Sigma/sqrt(n) = 1.24/sqrt(150) = 0.101

b. 95% of the time we'll expect mean to between a lower value of Mean - 2*Stdev = 6.72-2*1.24 = 4.24 and an upper value of Mean +2*Stdev = 9.2 hours

c. P(average is below 6.9) = P(Xbar<6.9) = P(Z< (6.9-6.72)/.101) = P(Z<1.782) = 0.9626

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