There are a set of twins, Peter and Mary. They are both very competitive. They are both taking statistics and have to take a test. The professor has 2 different types of test formats, A and B. Mary took type A and Peter took type B. Format A exam value was from 0-100. The average was 50 points and the Standard Deviation was 20 points. For Test format B, the exam value was 0-100, the average was 50 pts, and the standard deviation was 30 pts. Who did better on the test?

To give the full answer to this question we need to know the scores Peter and Mary got. Since that information is not available, I will, for purposes of showing what to do, assume that Peter took exam A and got a score of 75, and Mary took Exam B with a score of 82.

If we assume that the scores on the exams are normally distributed then we can look at the percentile each person fell into for their Exam.

For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.

You can translate into standard normal units by:
Z = ( X - μ ) / σ

Moving from the standard normal back to the original distribution using:
X = μ + Z * σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

An applet for finding the values:
http://www-stat.stanford.edu/~naras/jsm/FindProbability.html

calculator:
http://stattrek.com/Tables/normal.aspx

how to read the tables:
http://rlbroderson.tripod.com/statistics/norm_prob_dist_ed9.html

In this question we have, for Exam A
X ~ Normal( μ = 50 , σ² = 400 )
X ~ Normal( μ = 50 , σ = 20 )

And the percentile Peter’s score is in:

Find P( X < 75 )
P( ( X - μ ) / σ < ( 75 - 50 ) / 20 )
= P( Z < 1.25 )
= 0.8943502

So Peter did better than 89.44% of the the others taking this exam.

For Mary we have:
X ~ Normal( μ = 50 , σ² = 900 )
X ~ Normal( μ = 50 , σ = 30 )

Find P( X < 82 )
P( ( X - μ ) / σ < ( 82 - 50 ) / 30 )
= P( Z < 1.066667 )
= 0.8569388

So Mary did better than 85.69% of the others taking Exam B.

Even though Mary had a higher numeric score than Peter, she is in a lower percentile for the exam she took. Peter did better on the exam he took than Mary did on the exam she took.