The scores on a test have a mean of 100 and a standard deviation of 15. If the manager wishes to only select the top 25% of test takers, what is the cutoff value? Assume the test results are normally distributed.
Wanna make sure I'm doing it right, cause it doesn't feel like it. The closest to 25% in simple Z scores I could get was 0.67, at .2486.
So plugging that into the equation of x gives me
(0.67*15) + 100 = x
110.05 = x
And since you can't get .05 of a point, the cutoff value would be 110.
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I'm just worried that I should have taken 0.67-0.5 instead, since it's the latter half of the normal distribution chart, but that would make the answer closer to the mean, which wouldn't make sense if I'm looking for the top 25%.
The z-score for a cumulative probability of 0.75 = 0.674
Rearranging the formula to find a z-score, we get:
\[\large X=\mu+z \times \sigma\]
Plugging in the given values gives
\[\large X=100+0.674\times15=you\ can\ calculate\]