SOLUTION: 4. Suppose that a school district wants to start a program for gifted students. The participants in the program are to be chosen on the basis of IQ scores (normal with  =
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Question 1028318: 4. Suppose that a school district wants to start a program for gifted students. The participants in the program are to be chosen on the basis of IQ scores (normal with = 100, = 15). If the school district wants only the top 2% of students to participate in the program, what should be the IQ score that students must exceed to be accepted?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
from what i understand:
mean is 100
standard deviation is 15.
you want the iq score that represents the top 2%.
look up in z-score table for area under the distribution curve of .98.
that would be the area to the left of the z-score in most z-score tables that are full tables, meaning they show z-scores of roughly -3.5 to + 3.5 or thereabouts.
if you have a half table that only shows z-scores from 0 to + 3.5 or thereabouts, then you need to translate differently.
i'll assume you have a full table.
if you have a half table, then let me know and i'll change the procedure to accommodate that.
assuming the full table, .....
an area to the left of the z-score of .98 means the area to the right of that z-score is 1 - .98 = .02
that equates to the student being in the top 2%.
the table i used has the closest area above .98 to be .9803 and the closest area below .98 to be .9798.
.9798 give you a z-score of 2.05
.9803 give you a z-score of 2.06
.98 will be somewhere between there, probably somewhere around 2.053 or 2.054.
i confirmed, using my calculator, that an area of .98 to the left of the indicated z-score will lead to a z-score of 2.0537.
now to translate that z-score to a raw score.
your mean is 100
your standard deviation is 15.
formula for z-score is z = (x-m)/s
x is the raw score you are looking for.
m is the mean of 100
s is the standard deviation of 15
z is the z-score which you know to be 2.0537
formula becomes 2.0537 = (x - 100) / 15
solve for x to get x = 15 * 2.0537 + 100 = 130.8055.
an iq of 130.8055 or above would qualify as being in the top 2%.
this answer might be more exact than you would get using your table.
more than likely your table led you to a z-score of 2.05 or 2.06.
a z-score of 2.05 would lead to a raw score of 15 * 2.05 + 100 = 130.75
a z-score of 2.06 would lead to a raw score of 15 * 2.06 + 100 = 130.9
a z-score of 2.06 means that students with that z-score would be in the top 1 - .9803 = .0197 or 1.97% of their class.
a z-score of 2.05 means that students with that z-score would be in the top 1 - .9798 = .0202 or 2.02% of their class.
either one would be close, but to ensure that you were in the top 2% or less of the class, the z-score of 2.06 would be preferable.
not knowing exactly what your instructors are looking for, this is the best that i can do.
if you need further clarification, or can provide more details about what you need, then send me an email and i'll try to improve on the answer accordingly.
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