M's question at Yahoo Answers regarding normally distributed data

[MarkFL]

Here is the question:

Help with Z Score Problem?

Help with this one problem! Please show work if possible

IQ scores (as measured by the Stanford-Binet intelligence test) are normally distributed with a mean of 85 and a standard deviation of 19. Find the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 125. (Round your answer to the nearest hundred thousand.)

Here is a link to the question:

Help with Z Score Problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello m,

We know that in a normal distribution, half of the data is to the left of the mean $\mu$. So, we want to find the area under the normal curve to the right of the mean, and to the left of the datum value 125. To do this, we need to convert this datum to a $z$-score using:

\(\displaystyle z=\frac{x-\mu}{\sigma}=\frac{125-85}{19}\approx2.11\)

Now, consulting a table, we find the area associated with this $z$-score is:

$0.4826$

Hence, the area $A_L$ under the curve to the left of $x=125$ is:

\(\displaystyle A_L=0.5+0.4826=0.9826\)

Since the total area under the curve is $1$, we know the area $A_R$ to the right of $x=125$ is:

\(\displaystyle A_R=1-A_L=0.0174\)

Multiplying this number with the given population, we may state the number $N$ in that population with an IQ greater than 125 is:

\(\displaystyle N=0.0174\,\times\,2.8\,\times\,10^8\approx4,900,000\)

To m and any other guests viewing this topic, I invite and encourage you to post other pre-calculus statistics questions here in our http://www.mathhelpboards.com/f23/ forum.

Best Regards,

Mark.