Given a standardized normal distribution with a mean of 0 an

Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), what is the probability that Z is less than -0.21 or greater than the mean?

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
                  
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < -0.21) = (-0.21-0)/1
= -0.21/1= -0.21
= P ( Z <-0.21) From Standard Normal Table
= 0.4168
P(X > 0) = (0-0)/1
= 0/1 = 0
= P ( Z >0) From Standard Normal Table
= 0.5
P( X < -0.21 OR X > 0) = 0.4168+0.5 = 0.916834