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Question 1182495: Santos and Casey began arguing about who did better on their tests, but they couldn't decide who did better given they took different tests.
Santos took a test in Art History and earned a 77.3, and Casey took a test in Math and earned a 69.
Use the fact that all student's test grades in the Art History class had a mean of 61.4 and a standard deviation of 9.5 to answer the following questions.
a. Z-score for Santo's test grade
z = _______
b. X-score for Casey's test grade.
z= _______
c. Which person did relatively better?
Santos
Casey
They did equally well
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
For the entire art history class, we have
mu = population mean score = 61.4
sigma = population standard deviation of the scores = 9.5
The population being just this art class only.
Santos got a 77.3 on the test, so this is the raw score x.
Convert to a z score
z = (x-mu)/sigma
z = (77.3-61.4)/9.5
z = 1.67368421052631
z = 1.67
This is the approximate z score for Santos
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Part (b)
We run into a problem because Casey took a math test, and not an art history test.
Let's assume that both classes had the same population parameters (mu and sigma)
If so, then,
z = (x-mu)/sigma
z = (69-61.4)/9.5
z = 0.8
this represents Casey's z score but again only if the assumption above holds true. If Casey's math class had different mu and/or sigma values, then we won't get the same z score. It seems strange how your teacher didn't mention this.
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Part (c)
Again, we have to rely everything on the assumption made at the top of part (b). That assumption being mu and sigma are the same for art history and math.
If that assumption is true, then we see that Santos did better compared to Casey because his z score is higher (ie z = 1.67 is larger than z = 0.80).
If the assumption is false, then we cannot answer part (b) nor (c) because we simply don't have enough information. I would ask your teacher for clarification on this.
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