# How to Compute Z-scores for a Normal Distribution

The standardized normal distribution has a mean of 0 and a standard deviation of 1. However, most populations in which a certain measurable characteristic is normally distributed don't have a mean of 0 and a standard deviation of 1 for that characteristic. In order to compute the probability that a sample observation will lie between two measurements, it is necessary to convert the numbers to z-scores, basically a process which converts an arbitrary normal distribution into the standard normal distribution.

To compute z-scores, all you need to know are the mean and standard deviation of the set you are studying. Once you have the z-scores, you can look up the probabilities in a **z-score table** for the standard normal curve. Several examples are given below to help illustrate the procedure.

## Example 1

You are studying the length of men's feet in a certain country and you know that the length is normally distributed with a mean of 10.3 inches and a standard deviation of 1.8 inches. You want to know the probability that a man's foot is between 9 inches and 12 inches.

First, we convert the lower bound of 9 inches into a z-score by computing

z1 = (9 - 10.3)/1.8 = -0.72.

Next, we convert the upper bound of 12 inches into a z-score by computing

z2 = (12 - 10.3)/1.8 = 0.94

In both cases the z-score is simply found by evaluating the expression

(x - mean)/(standard deviation)

where x is the measurement of interest. Using a z-score table and looking up the values of z1 = -0.72 and z2 = 0.94, we find that

N(z1) = 0.2296

N(z2) = 0.8212

The probability that an observation lies between 9 and 12 inches is the difference of N(z2) and N(z1), so

probability = 0.8212 - 0.2296 = **0.5916**

So more than half of the men are expected to have a foot length between 9 inches and 12 inches.

## Example 2

A factory manufactures mechanical parts that have a mean diameter of 20 cm with a standard deviation of 0.031 cm. If the diameter is normally distributed, what is the probability that a part will have a diameter of 20.07 cm or higher?

Since the measurement of interest is 20.07, the mean is 20, and the standard deviation is 0.031, the z-score we need is

z = (20.07 - 20)/0.031 = 2.26

Looking up the value in a z-score table gives us N(z) = 0.9881. But this is the probability that an observation will have a z-score *less than* 2.26. In order to find the probability that an observation will have a *greater* z-score, we must compute the complement, 1 - 0.9881, which equals **0.0119.** So the probability that a part will have a diameter greater than 20.07 cm is 0.0119.