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Question 794:



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  1. The find a percentile rank we first find the z-score associated with a score of 80 = (80-100)/15 = -1.333. Now we lookup the area that is 1.33 standard deviations BELOW the mean and we get .0912 making a score in the 9.1 percentile.
  2. For between area we find the areas for both scores (just as in part a, and subtract the smaller area from the larger area).  For a score of 120, we get a z-score of 1.333, which is in the 90.9th percentile The percent of scores in between 80 and 120 is then 90.9-9.1 = .8175 or 81.7%
  3. The standard score is just a z-score for 90 = (90-100)/15 = -.6667.
  4. We find the z-score of 120 (from part b) = 1.3333 and the area above this in the normal curve = .0912 or 9.12%
  5. We find the z-score associated with the 87th percentile using excel's formula =NORMSINV(.87) and we get a z-score of 1.1264. So we just need to know what values is 1.1264 standard deviations above the mean. Multiplying this times the sd we get 1.1264*15= 16.8959. Now we just add that to the mean of 100 and we get a cutoff score of 116.8959.

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