The distribution of the IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 15. According to the standard deviation rule, only 0.15% of people have an IQ over what score?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

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

[tex]X \sim N(100,15)[/tex]

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.0015[/tex] (a)

[tex]P(X<a)=0.9985[/tex] (b)

Since we want the 0.15% of the people in the right tail since says above.

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9985 of the area on the left and 0.0015 of the area on the right it's z=2.97. On this case P(Z<2.97)=0.9985 and P(z>2.97)=0.0015

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99985[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.9985[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=2.97=\frac{a-100}{15}[/tex]

And if we solve for a we got

[tex]a=100 +2.97*15=144.55[/tex]

So the value of height that separates the bottom 99.85% of data from the top 0.5% is 144.55.