SOLUTION: IQ scores on the Stanford-Binet Intelligence Test are normally distributed with a mean μ = 100 and standard deviation σ = 16. (a) In order to qualify for Mensa, an organi

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Question 1098246: IQ scores on the Stanford-Binet Intelligence Test are normally distributed with a mean μ = 100 and standard deviation σ = 16. (a) In order to qualify for Mensa, an organization of people with high IQ scores, one must have an IQ at the 98th percentile. What IQ score is required to qualify for Mensa? (b) Statisticians typically consider an event "unusual" if it has less than a 5% chance of occurring. What would be the cut-off IQ for an "unusually" high IQ?
If someone could help me with this problem, i would appreciate it
Found 2 solutions by Boreal, rothauserc:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
z score for the 98th percentile is +2.055
z=(x-mean)/sd
2.055(16)=x-100
32.88+100=x
x=132.88
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z(0.95)=+1.645
+1.645(16)=x-100
x=126.32

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
a) The z-score associated with the 98 percentile is 2.054 (use table of z-scores)
:
2.054 = (X - 100) / 16
:
X - 100 = 2.054 * 16 = 32.864
:
X = 132.864
:
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IQ required to join Mensa is 132.864
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b) (0.05/2) + 0.95 = 0.975
:
z-score associated with 0.975 is 1.96
:
1.96 = (X - 100) / 16
:
X - 100 = 1.96 * 16 = 31.36
:
X = 131.36
:
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An unusually high IQ is 131.36
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