5 Suppose X Nmu 3 sigma2 9 Find PX 3 6 Solution Normal

-5- Suppose X ~ N(mu = 3, sigma^2 = 9). Find P{|X - 3| > 6}.

Solution

Normal Distribution
Mean ( u ) =3
Standard Deviation ( sd )=3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

|X-3|>6
=> x-3>6, -(x-3)>6
=> x>6+3 , -x+3>6
=> x>9, -x>3
=> x>9, x<-3

To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < -3) = (-3-3)/3
= -6/3= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(X > 9) = (9-3)/3
= 6/3 = 2
= P ( Z >2) From Standard Normal Table
= 0.0228
P( X < -3 OR X > 9) = 0.0228+0.0228 = 0.0455