Solution:-
a) The z-scores for 10th and 90th percentile is -1.282 and 1.282.
b) The mean and standard deviation of the performance score is 253 and 80.343.
p-value for the 10th percentile = 0.10
z-score for the p-value = -1.282
By applying normal distribution:-
...................................(1)
p-value for the 90th percentile = 0.90
z-score for the p-value = 1.282
By applying normal distribution:-
.....................................(2)
By solving (1) and (2) we get:-
c) The performance score manager must exceed to get an A is 404.125.
p-value for the top 3% = 1 - 0.03 = 0.97
z-score for the p-value = 1.881
By applying normal distribution:-
x = 404.125
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