Find a Z value such that 66 of the standard normal curve lie

Find:

a) Z (value) such that 66% of the standard normal curve lies to the left of z. (Round your answer to two decimal places.)

b) Z (value) that 94.5% of the standard normal curve lies to the right of z.(Round your answer to two decimal places.)

Solution

a)
P ( Z < x ) = 0.66
Value of z to the cumulative probability of 0.66 from normal table is 0.412
P( x-u/s.d < x - 0/1 ) = 0.66
That is, ( x - 0/1 ) = 0.41
--> x = 0.41 * 1 + 0 = 0.412 ~ 0.41
              
b)
P ( Z > x ) = 0.945
Value of z to the cumulative probability of 0.945 from normal table is -1.6
P( x-u/ (s.d) > x - 0/1) = 0.945
That is, ( x - 0/1) = -1.6
--> x = -1.6 * 1+0 = -1.598 ~ -1.56