IQs: percentage with IQs below 80, above 130, 95 to 115

skyguy

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Sep 11, 2006
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The IQs of football players at a particular college (you pick the school) are normally distributed with mean equal to 100 and standard deviation equal to 20. Determine the percentage of the students with IQs:

a) below 80
b) above 130
c) between 95 and 115
 
Use the formula:

\(\displaystyle \L\\z=\frac{x-{\mu}}{\sigma}\)

I'll do the second one and you do the other two.

\(\displaystyle z=\frac{130-100}{20}=\frac{3}{2}=1.5\)

Look up 1.5 in the z-table. It corresponds to 0.9332.

But they want the percentage greater than 130, so subtract from 1.

1-0.9332=0.0668 or 6.68% have an IQ above 130
 
These use the normal distribution. Convert each raw score into a z score:

z = (x - mu) / s
x = raw score
mu = mean
s = std dev

then use the z table (standard normal distrib table) to find the probabilities.

(a) z = (80-100)/20 = -1.0
P(below 80) = P(z < -1.0)

(b) z = (130-100)/20 = 1.5
P(above 130) = P(z > 1.5)

(c)
z1 = (95-100)/20 = -0.25
z2 = (115-100)/20 = 0.75
P(between 95 and 115) = P(z is between -0.25 and 0.75)

Your textbook will explain exactly how to find the probabilities for the z scores.
 
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