So in the last post, we talked about the normal distribution, and at the very end, discussed that if you knew the mean and standard deviation of a population for a particular variable, than you can compute the probabilities associated with a particular value of that variable within that population. The problem is, to do so, you have to use a really long equation that involves math and stuff, and if you're reading this, chances are you're not a big fan of math. I know I'm not. What we need, then, is a simpler way to get those probabilities. And it turns out there is just such a way: a **standardized normal distribution**. What's the standard normal distribution, you ask? Well, it's quite simple: it's a normal distribution with a mean of 0 and a standard deviation of 1. And because we know the properties of the normal distribution, we can convert any particular normal distribution into a standardized normal distribution with a simple equation:

Z = (X - μ)/σ

That is, the z-score for any particular value of the variable X is X minus the mean (μ) for X, divided by the standard deviation (σ) of X. If you do this for all of the scores, you'll get the standardized normal distribution. So if you choose a value of X equal to the mean, then you will subtract the mean from itself, which yeilds 0. Thus, the mean in the standard normal distribution is z = 0. If you choose a score that is one standard deviation above the mean and subtract the mean from it, then what you'll have left is the standard deviation. When you then divide the standard deviation from that (thus, from itself), you'll have z = 1. If you pick a score that's 1.5 standard deviations from above mean, subtract it from the mean, and then divide by the standard deviation, you'll get z = 1.5. And so on. You can do the same with negative scores. If you have a score that is one standard deviation below the mean, and subtract that from the mean, you'll get -σ. Divide σ from that, and you'll get -1.

The standardized normal distribution is associated with standard probabilities, so you don't have to use the equation, you can just look them up! That's what makes it so great. For example, we already know that the mean is associated with a probability of .5 (that is, there's a .5 chance that you'll get a score less than the mean). But now we can look up the probability associated with 1 standard deviation (z = 1). If you want to play around with these, you can check out the calculator here. For our purposes, though, it will be good to memorize a few z-score probabilities. Here they are:

- The probability of getting a score between z = -1 and z = 1 is about
**0.68**. That is, there's a 68% chance that if you take a random instance of X from the population, its value will be between -1 and 1 standard distribution from the mean. If you want cumulative probabilities, the chances of getting a score up to z = -1 is .16, the probability of getting a score up to z = 1 is .84. The chances of getting a score outside of the -1 to 1 range is 1 - 0.68, or 0.32. - The probability of getting a score between z = -2 and z = 2 is about
**0.954**, or there's a 95.4% chance of getting a score between -2 and 2 standard deviations from the mean. - The range associated with a 95% probability (which will be important later on) is z = -1.96 to z = 1.96.

There's one standardized normal distribution that you're probably all familiar with: the distribution of IQ scores. IQ scores are standardized so that they will have a mean of 100, and depending on the test, a set standard deviation (for the Weschler, it's 15). Here's a graph of the standardized normal distribution for the Weschler (I made it in Excel, so it's not perfect):

Because it's standardized, we know that 68% of IQ scores will be between 85 and 115. This is also where percentiles come from. If you have an IQ of 130, then 98% of IQ scores are at or below your IQ, and you are therefore in the 98th percentilre (technically, 97.77, but I round therefore I am).

So that's the standard normal distribution and Z-scores. Next up will be samples and sampling distributions.

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Thank you very much for this series! I completed my psychology degree 8 years ago and have been working in fields where I don't apply statistics or the classics of psychology. Lately I have started to re-read theories and methods in psychology and have found your site very useful for freshing up my rusty knowledge. Now you are starting this comprehensive statistics series... fantastic! Thanks!

Off topic, but Tag

I am not a statistician, but I am just about to sink my teeth into some stats in a bunch of papers I am reading, and your excellent series is timely and well written. Thank you.

Thankz you, who ever you are...