# How to Compute Z-scores for a Normal Distribution

TR Smith is a product designer and former teacher who uses math in her work every day.

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.

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• H.J. 2 years ago

Hi, what is the z-score for a distribution with a mean of 34.5 and a standard deviation of 12.7? I want to find the probability that a measurement falls between 40.6 and 47.9. Thanks.

• Author

TR Smith 2 years ago from Eastern Europe

Hi HJ,

Assuming the distribution in question is Gaussian (aka normal or a bell curve) then you compute the z-scores

z-upper = (47.9-34.5)/12.7 = 1.0551

z-lower = (40.6-34.5)/12.7 = 0.4803

With a z-score table you find P(z ≤ 1.0551) and P(z ≤ 0.4803) and subtract these numbers to find the probability that observations will fall between 40.6 and 47.9. Remember this is only accurate for distributions that are Gaussian, it won't work for others.