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

__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:

**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.

**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

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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