# Michael's question at Yahoo! Answers concerning normally distributed data

#### MarkFL

Staff member
Here is the question:

Help With A Statistics Problem! 10 Points!?

SAT math scores are scaled so that they are approximately normal, with a mean of 506 and a standard deviation of 124.
A college wants to send letter to students scoring in the top 29% on the math portion of the exam. What SAT math score should they use as the dividing line between those who get letters and those who do not?

What score would they use as the dividing line if they choose to send letters to only the top 15% instead?
Here is a link to the question:

Help With A Statistics Problem! 10 Points!? - Yahoo! Answers

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

#### MarkFL

Staff member
Re: Michael's question at Yahoo! Answers concerning normally distrubted data

Hello Michael,

For the first part of the question, to find the score wich divides the top 29% from the remainder, we need to consult our chart for areas under the standard normal curve. We are looking for the z-score with an area of 0.21, since 50% will be to the left of the mean, 0.21 will be between the mean and the z-score we want, and the remaining 0.29 will be to the right of this z-score.

Consulting the chart, we find this z-score is closest to 0.55. If we use numeric integration, we find it is closer to 0.553384719556.

I used my TI-89 Titanium and the command:

solve((1/√(2π))∫(e^(-t^2/2),t,0,z)=0.21,z)

Now we wish to convert this z-score to a raw data score, via the relation:

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

Solve for $x$ to get:

$\displaystyle x=z\sigma+\mu$

We are given:

$\displaystyle \sigma=124,\,\mu=506$

and so:

$\displaystyle x\approx0.553384719556\cdot124+506=574.619705224944\approx575$

So, scores which are at least 575 would be in the top 29%.

Using the same method to determine the bottom score for the top 15%, we may use the command:

solve((1/√(2π))∫(e^(-t^2/2),t,0,z)=0.35,z)

to get:

$z\approx1.03643338949$

and so:

$\displaystyle x\approx1.03643338949\cdot124+506=634.5177402967599\approx635$

So, scores which are at least 635 would be in the top 15%.