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Human intelligence is often measured with an Intelligence Quotient (IQ) test. IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. If one person were selected at random to take an IQ test, what is the probability that he or she would score between 90 and 110?

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Answer:


P(90<X<110)=P((90-\mu)/(\sigma)<(X-\mu)/(\sigma)<(110-\mu)/(\sigma))=P((90-100)/(15)<Z<(110-100)/(15))=P(-0.67<z<0.67)


P(-0.67<z<0.67)=P(z<0.67)-P(z<-0.67)


P(-0.67<z<0.67)=P(z<0.67)-P(z<0.67)=0.749-0.251=0.498

Explanation:

Let X the random variable that represent the IQ scores, and for this case we know the distribution for X is given by:


X \sim N(100,15)

Where
\mu=100 and
\sigma=15

We want to find this probability


P(90<X<110)

And we can use the z score formula given by:


z=(x-\mu)/(\sigma)

If we apply this formula to our probability we got this:


P(90<X<110)=P((90-\mu)/(\sigma)<(X-\mu)/(\sigma)<(110-\mu)/(\sigma))=P((90-100)/(15)<Z<(110-100)/(15))=P(-0.67<z<0.67)

And we can find this probability with this difference:


P(-0.67<z<0.67)=P(z<0.67)-P(z<-0.67)

And in order to find these probabilities we can use the normal standard table or excel and we got.


P(-0.67<z<0.67)=P(z<0.67)-P(z<0.67)=0.749-0.251=0.498

User Dsalaj
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