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.
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