Glucose levels in patients free of diabetes are assumed to f

Glucose levels in patients free of diabetes are assumed to follow a normal distribution with a mean of 111 and a standard deviation of 10. What is the probability that a patient has a glucose level between 90 and 110?

Solution

Normal Distribution
Mean ( u ) =111
Standard Deviation ( sd )=10
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 90) = (90-111)/10
= -21/10 = -2.1
= P ( Z <-2.1) From Standard Normal Table
= 0.01786
P(X < 110) = (110-111)/10
= -1/10 = -0.1
= P ( Z <-0.1) From Standard Normal Table
= 0.46017
P(90 < X < 110) = 0.46017-0.01786 = 0.4423