SOLUTION: given a normal distribution with a mean of 6000 and standard diviation 1500 what score is at the 30th percentile?
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Question 412458: given a normal distribution with a mean of 6000 and standard diviation 1500 what score is at the 30th percentile? Answer by ewatrrr(24785) (Show Source):
Hi
given a normal distribution with a mean of 6000 and standard diviation 1500
what score is at the 30th percentile?
*Note:
-.5244 = (x -6000)/1500 |NORMSINV(0.3) = -.5244
-.5244*1500 + 6000 = x = 5213.4