# 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

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

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.

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