The time spent in days waiting for a heart transplant in two

The time spent (in days) waiting for a heart transplant in two states for patients with type A+ blood can be approximated by a normal distribution. Assume u-132, standard deviation=18.3. (a) What is the shortest time spent waiting for a heart that will still place a patient in the top 15% of waiting times? (Round to two decimal places as needed) (b) What is the longest time spent waiting for a heart that would still place a patient in the bottom 1% of waiting time? (Round to two decimal places as needed)

Solution

Normal Distribution
Mean ( u ) =132
Standard Deviation ( sd )=18.3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P ( Z > x ) = 0.15
Value of z to the cumulative probability of 0.15 from normal table is 1.04
P( x-u/ (s.d) > x - 132/18.3) = 0.15
That is, ( x - 132/18.3) = 1.04
--> x = 1.04 * 18.3+132 = 150.9588 ~ 150.96  
b)
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.326
P( x-u/s.d < x - 132/18.3 ) = 0.01
That is, ( x - 132/18.3 ) = -2.33
--> x = -2.33 * 18.3 + 132 = 89.4342 ~ 89.43