Percentiles: Assume test scores are approximately....

[camcascod]

Assume certain standardized test scores are approximately normally distributed with a mean of 500 and a standard deviation of 20. Find the following percentiles.

The 90th and the 40th

I just need an equation that will get me to the correct answer. I know you have to find the z-score, but that is where I am getting confused. Any help that you can give will be greatly appreciated.

 

[galactus]

Look up 0.90 and 0.40 in the body of the table, then find the corresponding z-score.

Use it to solve for x in:

\(\displaystyle \L\\z=\frac{x-{\mu}}{\sigma}\)

You have \(\displaystyle {\mu}=500 \;\ and \;\ {\sigma}=20\)

For the first one, that means 90% of the grades are on or below that grade. So, if you scored that or higher, you did pretty good.

 

[camcascod]

Answers

I came up with 524 by using the equation: 1.2=x-500/20 for the 90th and I came up with 495 for the 40th percentile by using the equation -.25=x-500/20. Does this sound correct to you?

 

[galactus]

the 90th percentile has a z-score of 1.28.

-0.25 is close enough for the other.

 

[camcascod]

Thank you