Given that MSMC students sleep an average of 44522 hours per

Given that MSMC students sleep an average of 4.4522 hours per night with a standard deviation equal to .77511.hours. z=7-6.6522/.77511=.45 What percent of the students sleep between 7 and 9 hours? z=9-6.65232/77511 = 3.028 What percent sleep less than 6 hours? z=6-6.6/.77511=-0.67 What percent of students sleep between 8 and 11 hours?

Solution

Mean ( u ) =6.6522
Standard Deviation ( sd )=0.77511
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
b)
P(X < 6) = (6-6.6522)/0.77511
= -0.6522/0.77511= -0.8414
= P ( Z <-0.8414) From Standard Normal Table
= 0.2001                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8) = (8-6.6522)/0.77511
= 1.3478/0.77511 = 1.7388
= P ( Z <1.7388) From Standard Normal Table
= 0.95897
P(X < 11) = (11-6.6522)/0.77511
= 4.3478/0.77511 = 5.6093
= P ( Z <5.6093) From Standard Normal Table
= 1
P(8 < X < 11) = 1-0.95897 = 0.041