How to use standard normal table in question 5 Could you ple

Solution

Normal Distribution
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
b)
P(X < 0) = (0-0)/1
= 0/1= 0
= P ( Z <0) From Standard Normal Table
= 0.5                  

c)
P(X > 1) = (1-0)/1
= 1/1 = 1
= P ( Z >1) From Standard Normal Table
= 0.1587                  
P(X < = 1) = (1 - P(X > 1)
= 1 - 0.1587 = 0.8413                  

f)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.5) = (0.5-0)/1
= 0.5/1 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(X < 1) = (1-0)/1
= 1/1 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(0.5 < X < 1) = 0.84134-0.69146 = 0.1499                  

h)
P ( Z > x ) = 0.8
Value of z to the cumulative probability of 0.8 from normal table is -0.84
P( x-u/ (s.d) > x - 0/1) = 0.8
That is, ( x - 0/1) = -0.84
--> x = -0.84 * 1+0 = -0.842