# Examples of Z-score Calculations

One type of problem that is typical in an introductory statistics course is to find the z-score for some value of a normally distributed variable. After providing the rationale for this, we will see several examples of performing this type of calculation.

## Reason for Z-scores

There are an infinite number of normal distributions. There is a single standard normal distribution. The goal of calculating a z - score is to relate a particular normal distribution to the standard normal distribution. The standard normal distribution has been well-studied, and there are tables that provide areas underneath the curve, which we can then use for applications.

Due to this universal use of the standard normal distribution, it becomes a worthwhile endeavor to standardize a normal variable. All that this z-score means is the number of standard deviations that we are away from the mean of our distribution.

## Formula

The formula that we will use is as follows: z = (x - μ)/ σ

The description of each part of the formula is:

• x is the value of our variable
• μ is the value of our population mean.
• σ is the value of the population standard deviation.
• z is the z-score.

## Examples

Now we will consider several examples that illustrate the use of the z-score formula. Suppose that we know about a population of a particular breed of cats having weights that are normally distributed. Furthermore, suppose we know that the mean of the distribution is 10 pounds and the standard deviation is 2 pounds. Consider the following questions:

1. What is the z-score for 13 pounds?
2. What is the z-score for 6 pounds?
3. How many pounds corresponds to a z-score of 1.25?

For the first question, we simply plug x = 13 into our z-score formula. The result is:

(13 – 10)/2 = 1.5

This means that 13 is one and a half standard deviations above the mean.

The second question is similar. Simply plug x = 6 into our formula. The result for this is:

(6 – 10)/2 = -2

The interpretation of this is that 6 is two standard deviations below the mean.

For the last question, we now know our z -score. For this problem we plug z = 1.25 into the formula and use algebra to solve for x:

1.25 = (x – 10)/2

Multiply both sides by 2:

2.5 = (x – 10)