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Statistics Tutorials - 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.

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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