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 1098409: 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?
I do not know how to do this. Can someone please explain to me step by step how to solve a problem like this? Thank you
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z score for the 98th percentile is +2.055. Go to the z-table where .9800 is and look to the left for the ones and tenths value, then go to the top for the hundredths value
z=(x-mean)/sd
2.055(16)=x-100
32.88+100=x
x=132.88
=====================
z(0.95)=+1.645
+1.645(16)=x-100
x=126.32
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