Question 923874
mean of 50 and a standard deviation of 15.   {{{z = blue(x - 50)/blue(15)}}}    
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a)What is the lowest rating you should give to an “exemplary” employee if you follow the Univ. of Texas HR guidelines? 
z = invNorm(.90) = 1.282
15(1.282) + 50 = 69.23
70, lowest rating you should give to an “exemplary” employee
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 b) What is the lowest rating you should give to an “competent” employee if you follow the Univ. of Texas HR guidelines?
z = invNorm(.30) = -.524
15(-.524) + 50 = 42.14, 
43, the lowest rating you should give to an “competent” employee