Question

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?

Answer 1

Answer:

$P(90<X<110)=P(\frac{90-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{110-\mu}{\sigma})=P(\frac{90-100}{15}<Z<\frac{110-100}{15})=P(-0.67<z<0.67)$

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

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(90<X<110)=P(\frac{90-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{110-\mu}{\sigma})=P(\frac{90-100}{15}<Z<\frac{110-100}{15})=P(-0.67<z<0.67)$

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.  

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