1 Suppose Z is the standardized normal random variable and Z

1. Suppose Z is the standardized normal random variable and Z~ N (0, 1²). Please use z-table and show Step by Step to get the following probabilities.

a. P(Z < 1.66)=

b. P(Z > 1.23) =

c. If P(Z > a) = 0.9772, a =

d. P(Z>-1.44) =

e. P(-1.44< Z <1.66)=

f. P(1.23< Z<1.76) =

g. P(Z<1.36) + P(Z>1.36) =  

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=12
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 1.66) = (1.66-0)/12
= 1.66/12= 0.1383
= P ( Z <0.1383) From Standard Normal Table
= 0.555                  
b)
P(X > 1.23) = (1.23-0)/12
= 1.23/12 = 0.1025
= P ( Z >0.103) From Standard Normal Table
= 0.4592                  
c)
P ( Z < x ) = 0.9772
Value of z to the cumulative probability of 0.9772 from normal table is 1.999
P( x-u/s.d < x - 0/12 ) = 0.9772
That is, ( x - 0/12 ) = 2
--> x = 2 * 12 + 0 = 23.988                  
d)
P(X > -1.44) = (-1.44-0)/12
= -1.44/12 = -0.12
= P ( Z >-0.12) From Standard Normal Table
= 0.5478