1Previous information on SEL students indicate that X the am

1.Previous information on SEL students indicate that X, the amount of sleep in hours in a 24-hour period, is normally distributed with a mean, m, of 6.75 and a standard deviation, s, of 1.25. Assume this describes the sleep habits of this semester\'s SEL students as well.

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a) Find the probability that someone sleeps less than 8 hours.

b) Find the probability that someone sleeps more than 5 hours

c) Find the probability that someone sleeps between 7 and 10 hours

Solution

Normal Distribution
Mean ( u ) =6.75
Standard Deviation ( sd )=1.25
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 8) = (8-6.75)/1.25
= 1.25/1.25= 1
= P ( Z <1) From Standard Normal Table
= 0.8413                  
b)
P(X > 5) = (5-6.75)/1.25
= -1.75/1.25 = -1.4
= P ( Z >-1.4) From Standard Normal Table
= 0.9192                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 7) = (7-6.75)/1.25
= 0.25/1.25 = 0.2
= P ( Z <0.2) From Standard Normal Table
= 0.57926
P(X < 10) = (10-6.75)/1.25
= 3.25/1.25 = 2.6
= P ( Z <2.6) From Standard Normal Table
= 0.99534
P(7 < X < 10) = 0.99534-0.57926 = 0.4161