SOLUTION: An analysis of the final test scores for a selection test revealed that they approximate a normal distribution with a mean of 70 and a standard deviation 10.The examiner wants to a

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Question 1043314: An analysis of the final test scores for a selection test revealed that they approximate a normal distribution with a mean of 70 and a standard deviation 10.The examiner wants to award the grade “A” to upper 10%of the candidates and fail of the candidates who score the last 10% of the marks
(i) What is the minimum mark to obtain an “A” grade?
(ii) What is the minimum mark a candidate to obtain in order to get pass marks?.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The z-value for the 90th percentile is 1.28
z=(x-mu)/sd)
1.28=(x-mu)/sd
12.8=x-mu, since sd is 10 and we multiply both sides by 10.
x=mu+12.8=82.8
It is the opposite for the bottom 10%, where the z=-1.28
x-mu=-12.8
x-mu-12.8=57.2
The minimum mark to get an A is 92.8 if tenths are allowed.
The minimum mark to pass is 57.2