5.5: The Empirical Rule and Standard Normal (Z) Distribution

Imagine you just received your exam score back from a national standardized test. You scored a 720, but you're not sure if that's good, average, or below expectations. The raw score alone doesn't tell you much. What really matters is how your score compares to everyone else's, and whether it's above or below average. Was your score typical, or was it unusually high?

This is where z-scores and the standard normal distribution come in. They let us translate raw scores into a common frame of reference, showing exactly how far a score is from the average (mean), measured in standard deviations. In this section, we’ll learn how the Empirical Rule and z-scores help us reframe raw data in terms we can easily interpret, compare, and explain.

Definition: Z-score

A z-score tells you how far a data point is from the mean in standard deviation units.

\[ z = \frac{x - \mu}{\sigma} \]

Where:

• \( x \) = your data value
• \( \mu \) = the dataset mean
• \( \sigma \) = the standard deviation

A positive z-score means the value is above the mean, and a negative z-score means it is below the mean.

Computing a z-score normalizes a value in a distribution. In the standard normal distribion (\(\mathcal{N}(0,1))\), a value is equal to its z-score. (Check this!)

Example: Converting to a Z-score

Suppose we find out that the test you scored a 720 on has a mean of 700 and a standard deviation of 40.

We calculate the z-score:

\( z = \dfrac{720 - 700}{40} = \dfrac{20}{40} = 0.5 \)

This means you scored one half of a standard deviation above the average.

While z-scores are most commonly associated with the standard normal distribution, they can still be useful for comparing individual data points across different data sets, even when those distributions are not normal.

This works because a z-score doesn't rely on the shape of the distribution, it simply tells us how many standard deviations away a value is from the mean of its own distribution.

Note: Z-scores are useful to compare between different normal distributions!

You can use z-scores to compare values from different distributions, like comparing a math test score to a science test score, even if those datasets have different units, means, and standard deviations.

Why? Because z-scores standardize the values. They translate everything into "how far from the mean, in standard deviations", which is a unitless scale.

Example: Comparing Across Tests

Suppose you score an 82 on a math test with a class mean of \(\mu = 75\) and \( \sigma = 5 \), and you score a 87 on your history test, where the class mean is 85 and \( \sigma = 4 \).

Good job on your tests! Note that your history score is higher than the math score, but the history class average is also higher than the math class average! Rather than comparing the scores directly, we can compare where they fall in the distribution, which tells us if you're doing better in each class relative to your peers, although one should not necessarily compare themselves to others!

Let’s compute the z-scores:

• Math: \( z = \frac{82 - 75}{5} = 1.4 \)
• History: \( z = \frac{87 - 85}{4} = 0.5 \)

We can see from the z-scores that your math performance, relative to the class, was better than your history performance, regardless of the fact that your history score was higher.

We have calculate standard deviations, and described values relative to them using z-scores, but we have a tool to build a better intuitive idea of what being "1 standard deviation above the mean" actually implies.

The Empirical Rule (68–95–99.7 Rule)

The Empirical Rule tells us how data is distributed in a normal distribution when we know the mean and standard deviation. It estimates the proportion of data within 1, 2, and 3 standard deviations of the mean.

Here’s how it breaks down:

• About 68% of the data falls within 1 standard deviation of the mean (\( \mu \pm \sigma \))
• About 95% falls within 2 standard deviations (\( \mu \pm 2\sigma \))
• About 99.7% falls within 3 standard deviations (\( \mu \pm 3\sigma \))

Shown are the various components of the distribution up to 3 standard deviations. All values are approximate.

The Empirical Rule gives us a quick way to estimate the spread of the data when we know it’s approximately normal. It helps us identify what range is “typical” or “unusual.”

Using a Z-Table (Standard Normal Table)

The Z-table allows us to find the proportion of values that fall below (to the left of) a given z-score in a standard normal distribution. This area represents a cumulative probability.

Excerpt from Z-table (Left-Tail Cumulative Area)

z-scores and probabilities

Z	0.00	0.01	0.02	0.03	0.04
0.0	0.5000	0.5040	0.5080	0.5120	0.5160
0.1	0.5398	0.5438	0.5478	0.5517	0.5557
0.2	0.5793	0.5832	0.5871	0.5910	0.5948

For example, a z-score of 0.20 corresponds to an area of approximately 0.5793, or 57.93% of values falling below that score.

Worked Example: Exam Scores and Z-scores

Suppose exam scores for a national standardized math test are approximately normally distributed with a mean of 500 and a standard deviation of 100.

Javier scores a 630. You want to find out how well Javier did compared to everyone else, specifically, what proportion of test takers scored below him.

Step 1: Find the Z-score

Use the z-score formula:

\( z = \frac{x - \mu}{\sigma} = \frac{630 - 500}{100} = \frac{130}{100} = 1.30 \)

This tells us that Javier scored 1.3 standard deviations above the mean.

Step 2: Use the Standard Normal Table

Next, use the z-table to find the proportion of scores to the left of \( z = 1.30 \).

Look up 1.3 on the leftmost column and use the 0.00 column (since we want exactly 1.30):

• \( P(Z < 1.30) = 0.9032 \)

This means that approximately 90.32% of test takers scored below Javier.

Final Answer

Javier’s score of 630 places him in the 90th percentile of all test takers.

Think About It:

If a student scores 2 standard deviations below the mean on a national exam, where would that person fall in the population? Use the Z-table to help you estimate.

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