A z score can be used to compare values from two normal distributions with different means and known population standard deviations.

Suzie Student has, by mistake, been put in both her high school's algebra I class and the AP Calculus BC class. She is waiting for her counselor to drop her from the class, but in the meantime, she has a test on the same day in both classes. When she gets her grades back, she finds that she got a 97% on the algebra test, but only a 79% on her calculus test. Obviously she did better on the algebra test, right?

Well, not exactly. Sure, she's going to get more points for her algebra test, but so is everyone else in that class. Algebra I is a much easier class than calculus (supposedly), so more people got higher scores on the algebra test. We need a way to compare how she did on each test to the rest of the class.

# The Z Score to the Rescue!

x - μ
z = -------
σ

This is the formula used to standardize data to a z score. This essentially is finding the number of standard deviations *x* is from the mean.

If

*z* is positive,

*x* is

above the mean. If it's negative, it's

below the mean. So if a test score is 2σ above the mean, z=+2.

It turns out that the N(μ, σ) distribution (assuming scores come from a normal distribution) is **N(87, 18.18)** for the algebra test and **N(75, 7.02)** for the calculus test. To find the z scores for the two distributions:

97 - 87
z_{A} = --------- = +.550
18.18
79 - 75
z_{C} = --------- = +.570
7.02

So her algebra test is +.550 standard deviations above the mean, and her calculus test is +.570 standard deviations above the mean. Compared with the rest of the classes, Suzie actually did better on her calculus test than on her algebra test. Now she just needs to figure out how to explain that to Mom.

Yates, Daniel, David Moore, and George McCabe. __The Practice of Statistics__.