Question #26126

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Find the IQ score separating the top 10% from the others.

Expert's answer

We have normal distribution with mean 100 and standard deviation & 15:

m = 100

sigma = 15

And we know:

P ( x > mean + z*sigma ) = 0.1

or& percentage of values expected to lie in and outside the symmetric interval& mean +- z*sigma = 80 %

From the tables: z = 1.281552 ~ 1.28

Finally,

z*sigma = 1.28*15 = 19.2

Therefore, 10% of all adults have IQ score :

100+19.2=119.2 or more

Answer: 119.2

m = 100

sigma = 15

And we know:

P ( x > mean + z*sigma ) = 0.1

or& percentage of values expected to lie in and outside the symmetric interval& mean +- z*sigma = 80 %

From the tables: z = 1.281552 ~ 1.28

Finally,

z*sigma = 1.28*15 = 19.2

Therefore, 10% of all adults have IQ score :

100+19.2=119.2 or more

Answer: 119.2

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