Need helping determining 20% on normal distribution curve

Question

A manager wants to identify the 20% of employees with the greatest number of missed days at work. The mean number of missed days in the past year is 5.0. The standard deviation for missed days of work in the past year is 1.5. The distribution of scores for this variable is normal. What is the fewest number of missed days at work in the past year an employee could have that would still place them among the highest 20% of employees.

I have drawn the picture on the curve to try to get a better visual of what the question is asking. I understand that if I can obtain 20% of the highest numbers above the mean, then I can find the answer by choosing the smallest number of the 20%, relatively speaking. However, I am unsure of how to find the 20%. Please help! Thank you!

Answer 1

Prob( z < k) = 0.8

k=0.84162 <--- REVERSE TABLE LOOKUP

0.84162 = (x-5)/1/5

X = 6.26243

Answer 2

Obtain a z-score table: http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf

20% has a z-score of 0.842 (use interpolation)

0.0842 = (X - 5) / 1.5

Can you solve for X and answer?