Question 150223
The average hourly wage of employees of a certain company is $9.83. Assume the variable is normally distributed. If the standard deviation is $4.58, find the probability that a randomly selected employees earns less than $5.43. 
---------------------
Find the z-score of 5.43
z(5.43) = (5.43-9.83)/4.58 = -0.960699
P(z < -0.960699) = 0.1684 or 16.84%
=======================
Cheers,
Stan H.