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Solution
Exam A
mean = 500
standard deviation = 20
[tex]\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ Z=\frac{480-500}{20} \\ Z=-\frac{20}{20}=-1 \end{gathered}[/tex]since scores on each exam are normally distributed
then z score = -1
Exam B
mean = 500
standard deviation = 50
[tex]\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ -1=\frac{x-500}{50} \\ -50=x-500 \\ -50+500=x \\ x=450 \end{gathered}[/tex]Therefore Christopher score on Exam B = 450