Question 1107340: : In the example in problem 18, assume that the mean is 300 and the standard deviation is 25. Using a normal curve table, what scores would be the top and bottom scores to find a)the middle 50% of architects b)the middle 90% of architects and c)the middle 99% of architects?
q 18 Suppose that the scores of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have Z scores a)above .10 b)below .10 c)above.20 d)below .20 e)above 1.10 f)below 1.10 g)above -.10 and h)below -.10
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The middle 50% need probability of .2500 on both sides.
This is z=+/-0.67
z*sigma=+/-16.75
scores are (283.25, 316.75)
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z for the middle 90% is +/-1.645
z*sigma=+/-41.1
scores are (258.9, 341.1)
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middle 99% z is +/- 2.576
z*sigma=+/-25*2.576=+/-64.4
scores are (235.6, 364.4)
>0.1=0.4602
<0.1=0.5398
>0.20=0.4207
<0.20=0.5793
>1,10=0.1357
<1.10=0.8643
>-0.10=0.5398
<-0.10=0.4602
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