Assume X is normally distributed with a mean of 10 and a sta

Assume X is normally distributed with a mean of 10 and a standard deviation of 5.  Determine the following:

P(X 15)

P(X 12)

P(9 X 20)

P(5 X 15)

Solution

Normal Distribution
Mean ( u ) =10
Standard Deviation ( sd )=5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 15) = (15-10)/5
= 5/5 = 1
= P ( Z >1) From Standard Normal Table
= 0.1587                  
          
P(X < = 15) = (1 - P(X > 15)
= 1 - 0.1587 = 0.8413                  
          
b)
P(X < 12) = (12-10)/5
= 2/5= 0.4
= P ( Z <0.4) From Standard Normal Table
= 0.6554                  
P(X > = 12) = (1 - P(X < 12)
= 1 - 0.6554 = 0.3446                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 9) = (9-10)/5
= -1/5 = -0.2
= P ( Z <-0.2) From Standard Normal Table
= 0.42074
P(X < 20) = (20-10)/5
= 10/5 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(9 < X < 20) = 0.97725-0.42074 = 0.5565                  

d)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 5) = (5-10)/5
= -5/5 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 15) = (15-10)/5
= 5/5 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(5 < X < 15) = 0.84134-0.15866 = 0.6827