Question about how to use Z-score to help us determine which test the student did better Here is the question:
A student scores 56 on a geography test and 267 on a mathematics test. The geography test has a mean of 80 and a standard deviation of 20. The mathematics test has a mean of 300 and a standard deviation of 22.
If the data for both tests are normally distributed, on which test did the student score better?
You can find the solution here: https://study.com/academy/answer/a-student-scores-56-on-a-geography-test-and-267-on-a-mathematics-test-the-geography-test-has-a-mean-of-80-and-a-standard-deviation-of-20-the-mathematics-test-has-a-mean-of-300-and-a-standard-deviati.html
My problem:
Q1: I attempted the problem without looking at the answer. I do not understand why the solution use z-score here and not use the numbers: 56 and 261. Why can I just use the numbers 56 and 267? Since 267>56, the student did better in math.
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Q2: I do not understand how the z scores help determine which test the student did better. I realize I do not understand z scores.
Thank you so much.
 A: Doing better doesn't mean getting the higher score. Different exams and different students taking them make a direct comparison of scores written on the exam papers impossible. (These are sometimes called 'raw' scores.)
What you might mean by 'doing better' is to ask on which exam the student scored better relative to others who took each test. Using Z scores helps to do this.
For geography the student has 56 on an exam with class
mean $\mu = 80$ and SD $\sigma = 20.$ The $Z$ score is the number of SDs above the mean. (Calculations and graphs in R.)
$$Z_g = \frac{56 - 80}{20} = -1.2.$$
Not just below average, but 1.2 standard deviations below average.
For math,
$$Z_m = \frac{267 - 300}{22} = -1.5.$$
Not just below average but one and a half standard deviations below average. (Such z-scores are sometimes called 'standard scores'.)
Here are graphs of the two normal distributions and
the student's standing in each, marked by a vertical line.
par(mfrow=c(1,2))
 curve(dnorm(x, 80, 20), 20, 140, col="blue", ylab="PDF", 
       main="Geography")
  abline(h=0, col="green2")
  abline(v= 56, col="red")

 curve(dnorm(x, 300, 22), 200, 400, col="blue", ylab="PDF", 
       main="Mathematics")
  abline(h=0, col="green2")
  abline(v= 267, col="red")
par(mfrow=c(1,1))


For the $Z$-scores, here are plots of the standard normal
distribution with standard scores shown as vertical lines.

par(mfrow=c(1,2))
 curve(dnorm(x), -3.5, 3.5, ylab="PDF", xlab="z", col="blue", 
       main="Geog Z-Scores")
  abline(h=0, col="green2"); abline(v=0, col="green2")
  abline(v=-1.2, col="red")
 curve(dnorm(x), -3.5, 3.5, ylab="PDF", xlab="z", col="blue", 
      main="Math Z-Scores")
  abline(h=0, col="green2"); abline(v=0, col="green2")
  abline(v=-1.5, col="red")
par(mfrow=c(1,1))

Finally, you can use printed normal tables or software to
see that 11.5% of the geography class scored lower than our student and that 6.7% of the mathematics class scored lower than our student.
In each of the four plots, the area under the
density curve to the left of the vertical red line
represents the percentage of the class scoring below our student. (All density curves enclose a
total area of $1 = 100\%).$
Our student did somewhat worse on the math exam than on the geography exam. For example, if both classes fail the lowest 10% of their students, then our student will pass geography (barely), but not mathematics. (I hope this student is doing better in some classes, maybe art, creative writing, philosophy, theater, a foreign language, etc.)
pnorm(-1.2)
[1] 0.1150697
pnorm(-1.5)
[1] 0.0668072

