6 Determine the following probabilities for the standard nor

6. Determine the following probabilities for the standard normal random variable Z. (a) P(-1

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -1) = (-1-0)/1
= -1/1 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 1) = (1-0)/1
= 1/1 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(-1 < X < 1) = 0.84134-0.15866 = 0.6827                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -2) = (-2-0)/1
= -2/1 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 2) = (2-0)/1
= 2/1 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(-2 < X < 2) = 0.97725-0.02275 = 0.9545                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -3) = (-3-0)/1
= -3/1 = -3
= P ( Z <-3) From Standard Normal Table
= 0.00135
P(X < 3) = (3-0)/1
= 3/1 = 3
= P ( Z <3) From Standard Normal Table
= 0.99865
P(-3 < X < 3) = 0.99865-0.00135 = 0.9973                  
d)
P(X > 3) = (3-0)/1
= 3/1 = 3
= P ( Z >3) From Standard Normal Table
= 0.0013