Calculating and Interpreting Z-Scores
Date: 12/13/2005 at 14:04:49 From: Catherine Subject: Z-Scores Please help me by providing a step-by-step process for calculating the z-score, or standard score. My class textbook is too confusing. There seem to be many ways of calculating z-scores, but no standard method that I can apply. I know how to do standard deviation, but I fall apart when it comes to the z-scores. Can you help?
Date: 12/13/2005 at 17:28:48 From: Doctor Minter Subject: Re: Z-Scores Hi Catherine! A z-score is used in statistics to model any normal distribution as a standard normal distribution. You can consider it "rescaling" the original distribution to fit the properties of a normal distribution. Recall that a normal distribution has a mean of 0 and a standard deviation of 1. If you understand percentages, the concept is very much the same. If you know what proportion of people in a room have brown hair, say three out of four, you know that 75% of the people in the room have brown hair. Does that mean that there are exactly 100 people in the room and that exactly 75 of them have brown hair? That could be the situation, but what if there are only 20 people in the room? Percentages rescale the actual population. That is, the percentage tells you how many people in the room WOULD have brown hair if there WERE 100 people, while still having the same proportion as the original group. A z-score does something not just similar, but EXACTLY like this to a normal distribution to rescale it to a standard normal distribution. As percentages rescale a proportion to a group of 100, z-scoring rescales any normal distribution to a standard normal distribution. Let's do an example of a z-score rescaling: Let's say we have a normal distribution with mean 10 and standard deviation 2. We'd like to know the probability that a random sampling will produce a value greater than or equal to 11.5. The standard normal distribution is a bell curve that peaks at zero when graphed, and has a standard deviation of 1. We need to rescale our distribution do the same. The z-score conversion formula is y - u z = ----- s where y (you might also see it as x, or even another symbol) is our sample value (11.5 in this case), u (the mean, usually denoted by the Greek letter mu, but unfortunately the keyboard confines me to the letters in our alphabet only! In this case, u is 10.), and s is the standard deviation, usually denoted by the Greek letter sigma, which has a value of 2 in this case. Let's find our z-score: 11.5 - 10 z = --------- = 0.75 2 So our z-score is 0.75. What does this tell us? This value says that if we were to obtain a value greater than or equal to 11.5 by sampling our original distribution, the probability of doing so is the same as the probability of obtaining a value greater than or equal to 0.75 by sampling the standard normal distribution. If we look at the table for this particular z-score, we see that the probability of a random sampling of the normal distribution yielding a value greater than 0.75, which is equal to the probability of a random sampling of the original distribution yielding a value greater than 11.5, is equal to 0.2266. That is, 22.66% of random samples from the standard normal distribution will yield values greater than or equal to 0.75, and correspondingly, 22.66% of the random samples of the original distribution will yield values greater than or equal to 11.5. The reason that we do this is because statistics texts have a table of z-scores and their corresponding probabilities. That handy table of values gives us the probability of z (value obtained by a random sampling of the standard normal distribution) being greater than the z-score obtained by the above formula. That probability is exactly equal to the probability that the example problem asked us to find! This process enables us to find the probability that a random sample from ANY normal distribution (there are other distributions also, and this formula does NOT work for them, by the way) is greater than a certain given value. If you want to find the probability that a sample produces a value LESS than a certain value, keep in mind that the standard normal distribution is symmetric about zero, and that its total area is 1. You can use these two properties to adjust the process to yield the desired answer. For example, the probability of a random sampling of our original distribution yielding a value less than 11.5 is equal to the probability that we found earlier subtracted from 1, or 0.7734. I hope that this will help you get started. Please feel free to write again if you need further assistance, or if you have any other questions. Thanks for using Dr. Math! - Doctor Minter, The Math Forum http://mathforum.org/dr.math/
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