Using the Standard Normal Table found in your textbook find

Using the Standard Normal Table found in your textbook, find the z-scores such that: The area under the standard normal curve to its left is 0.5 2 = The area under the standard normal curve to its left is 0.9826 2 = The area under the standard normal curve to its right is 0.1423 2 = The area under the standard normal curve to its right is 0.9394

Solution

a)
P ( Z < x ) = 0.5
Value of z to the cumulative probability of 0.5 from normal table is 0
P( x-u/s.d < x - 0/1 ) = 0.5
That is, ( x - 0/1 ) = 0
--> x = 0 * 1 + 0 = 0                  
b)
P ( Z < x ) = 0.9826
Value of z to the cumulative probability of 0.9826 from normal table is 2.111
P( x-u/s.d < x - 0/1 ) = 0.9826
That is, ( x - 0/1 ) = 2.11
--> x = 2.11 * 1 + 0 = 2.111                  
c)
P ( Z > x ) = 0.1423
Value of z to the cumulative probability of 0.1423 from normal table is 1.07
P( x-u/ (s.d) > x - 0/1) = 0.1423
That is, ( x - 0/1) = 1.07
--> x = 1.07 * 1+0 = 1.07                  

d)
P ( Z > x ) = 0.9394
Value of z to the cumulative probability of 0.9394 from normal table is -1.55
P( x-u/ (s.d) > x - 0/1) = 0.9394
That is, ( x - 0/1) = -1.55
--> x = -1.55 * 1+0 = -1.55