Wiki Page, Math 13 - Perez Sandbox : Module 12 - Introduction to Normal Probability Distributions (3 of 5)

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2026-01-17T21:15:14-08:00

The Empirical Rule gives us a method for identifying likely and unlikely values. But it is limited to probability questions involving X-values that fall exactly 1, 2, and 3 standard deviations away from the mean.

We will now consider probability questions that cannot be answered using the Empirical Rule.

EXAMPLE Male foot length

Let's go back to our example of male foot length and ask a question that cannot be answered using the Empirical Rule.

What is the probability that a man's foot length is more than 13 inches?

Because 13 inches is not exactly 1, 2, or 3 standard deviations away from the mean, we can give only a very rough estimate of the probability at this point.

To find the probability, we will determine the number of standard deviations 13 is above the mean and then use technology to determine the probability.

We will start by practicing how to find the number of standard deviations an X-value is from the mean. After we practice this, we will learn to use technology to find the probability.

How many standard deviations above the mean is a male foot length of 13 inches?
13 inches is 2 inches above the mean.

How many standard deviations is this?
Since a standard deviation is 1.5 inches, 2 inches is standard deviations above the mean.

We can combine these two steps into one equation:

standard deviations.

In the language of statistics, we found the z-score for a male foot length of 13 inches to be . Or, to put it another way, we have standardized the value of 13 into standard deviation units. A foot length of 13 inches is 1.33 standard deviations above the mean.

In general, the standardized value z tells how many standard deviations below or above the mean the original value is. The formula for the z-score is

Using our familiar notation, we can rewrite this formula as

For values of x above the mean ( ), the z-score will be positive; for values of x below the mean( ), the z-score will be negative.

EXAMPLE Standardized value

Let's practice finding another standardized value (z-scores).

What is the standardized value for a male foot length of 8.5 inches? What does this number tell us?

. A foot length of 8.5 inches is 1.67 standard deviations below the mean.

A z-score allows us to compare X-values from different normal probability distributions. Here are several examples.

EXAMPLE Comparing standardized values for foot lengths

Women's foot lengths also have a normal probability distribution, but the mean is 9.5 inches and standard deviation is 1.2 inches.

Sam's foot length is 13.25 inches, and Candace's foot length is 11.6 inches. Which of the two has a longer foot relative to his or her gender group?

We cannot just compare the foot lengths. Why? Because these foot lengths come from different distributions. Candace's foot length is barely above the mean for men, but her foot is much larger than the mean for women.

To make the comparison, we standardize the values using z-scores.

Sam: (Sam's foot length is 1.5 standard deviations above the mean foot length for men).

Candace: (Candace's foot length is 1.75 standard deviations above the mean foot length for women).

Even though Sam's foot is longer than Candace's, Candace's foot is longer relative to other women than Sam's foot is relative to other men.

EXAMPLE Comparing standardized values for test scores

The test scores in Maria's class have a normal probability distribution with a mean of 70 and a standard deviation of 5. The test scores in Lamar's class have a normal probability distribution with a mean of 85 and a standard deviation of 10.

Maria scored a 76 and Lamar scored an 80. Why did better on the test relative to their classmates?

Answer: Because Maria and Lamar are from different classes with different probability distributions, we need to standardize their test scores in order to make a fair comparison.

Maria: z = (76 - 70)/5 = 6/5 = 1.20. Maria scored 1.2 standard deviations above the mean for her class.

Lamar: z = (80 - 85)/10 = –5/10 = –0.5. Lamar scored 0.5 standard deviations below the mean for his class.

Relative to their respective classes, Maria performed better on the test, even though her score was lower than Lamar's.

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