Suppose that a particular entrance examination into Public Service is designed so that the marks will be normally distributed with a mean of 300 and a standard deviation of 60. If only the top 15% of candidates are selected, what mark would a student be expected to obtain for admission?
I did .5 - .15 to find the percentage that would be below top 15% in the upper half.
went to my normal distribution table, found 3508 (closest listed to 3500) = 1.04 Z value.
Plugged 1.04 into z = (x - mean) / standard D
Z x S.D. + Mean = X
1.04 x 60 + 300 = 362.4
x = 362.4
Correct? Thanks in advance.