Statistics Tutorials

How to Compute Z-Scores

Assume that $X$ has a normal distribution, with mean $\mu$ and standard deviation $\sigma$ . This is typically written as

X \sim N( \mu, \sigma^2 )

Then, the Z-score associated to $X$ is defined as

Z = \displaystyle{\frac{X - \mu}{\sigma}}

Z-scores are an excellent way to normalize scores, in the sense that it provides the ability of reducing the calculation of normal probabilities to probabilities associated to the standard normal distribution

You have infinite possible normal distributions by choosing different means and standard deviations, so think for a second how convenient it is to be able to normalize them and reduce them to one well known distribution, for which we have for example probability tables readily available.

The magic of Z-scores is that they allow to compare "pears" with "apples", as you could have normally distributed variables of completely different scale, like the height of eight-graders and the weight of male lions, and then by converting them into z-scores you can now compare to see their standing with respect to each of their populations.

Example: Consider the random variable

X

, which as a normal distribution, with mean

\mu = 34

and standard deviation

\sigma = 4

. Compute the z-score of

X = 41

Answer :

Using the definition of z-score, we use the following formula:

Z = \displaystyle{\frac{X - \mu}{\sigma} = \frac{41 - 34}{4} }= \frac{7}{4} = 1.75

What does the z-score represent?

The z-score gives measures how far the random variable $X$ is from its mean $\mu$ . This measure is not arbitrary, it indicates how many standard deviations the value of $X$ is away from $\mu$ . In other words, a z-score of 1.75 indicates that the value of $X$ is 1.75 standard deviations away from its mean. Since the z-score is positive, that means that the value of $X$ is 1.75 standard deviations to the right of its mean, to be more precise.

Applications of Z-scores

Application Example: Peter took his finance exam last week, and he got 89/100. The mean for his class was 77, with a standard deviation of 15. Jenna took her math test last week too, and she got 84/100. The mean for her class was 75, with a standard deviation of 5. Their were arguing on who did better, who do you think did better relative to their class?

Answer : We need to use z-scores. For Peter we have

Z = \displaystyle{\frac{X - \mu}{\sigma}} = \displaystyle{\frac{89 - 77}{15}} = \frac{12}{15} = 0.8

On the other hand, for Jenna:

Z = \displaystyle{\frac{X - \mu}{\sigma} = \frac{84 - 75}{5}} = \frac{9}{5} = 1.8

The z-score associated with Jenna's score test is higher than the z-score test associated with Peter's score test, which means that Jenna did better than Peter, relative to her class.