if x is normally distributed with mean 12 and standard devia

if x is normally distributed with mean 12 and standard deviation 2 then p(x £ 9) is

a. P(z 3/2)

b. P(z 3/4)

c. P(z 9/10)

d. P(z 2/3)

Solution

Normal Distribution
Mean ( u ) =12
Standard Deviation ( sd )=9
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > -1.5) = (-1.5-12)/9
= -13.5/9 = -1.5
= P ( Z >-1.5) From Standard Normal Table
= 0.9332                  
P(X < = -1.5) = (1 - P(X > -1.5)
= 1 - 0.9332 = 0.0668                  
b)
P(X > -0.75) = (-0.75-12)/9
= -12.75/9 = -1.4167
= P ( Z >-1.417) From Standard Normal Table
= 0.9217                  
P(X < = -0.75) = (1 - P(X > -0.75)
= 1 - 0.9217 = 0.0783                  
c)                  
P(X < 0.9) = (0.9-12)/9
= -11.1/9= -1.2333
= P ( Z <-1.2333) From Standard Normal Table
= 0.1087                  
d)
P(X < 0.667) = (0.667-12)/9
= -11.333/9= -1.2592
= P ( Z <-1.2592) From Standard Normal Table
= 0.104