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Application of z-Scores

Application of z-Scores

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Author: Dan Laub
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In this lesson, students will learn how to apply z-scores to real-world situations.

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Tutorial

Source: All graphs created by Dan Laub, Image of baby, PD, http://bit.ly/1Sj5ZvI; Image of tire, PD, http://bit.ly/1OHGXXt; Image of z-tables, PD, http://bit.ly/1QV2z2D

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[MUSIC PLAYING] [SCRIBBLING NOISE]

[MUSIC PLAYING]

Hi. Dan Laub here. And in this lesson, we're going to discuss the application of z-scores.

But before we do so, let's discuss the objective for this lesson. By the end of the lesson, you should be able to use a z-score table to determine the probability of values associated with a real-world variable that is normally distributed. So let's get started.

With a normal distribution of a variable, the mean is the center of the distribution while the standard deviation shows the level of variability that exists. When we know the value of a variable that is normally distributed, we can calculate the z-score by using the following formula. The z-score would be equal to the difference between the chosen value and the mean, and we'd take that difference and divide it by the standard deviation.

When we are interested in determining the probability of a specific value occurring, we can use a table to look up areas for a calculated z-score to determine the area under the curve of any normal distribution. And this can tell us a likelihood of a value occurring.

So as an example, let's look at the weight of newborn babies. Let's suppose that the mean weight of newborns is 7 and 1/2 pounds-- or 7.5 pounds-- and the standard deviation, in this case, is 1.25 pounds. Let's suppose we're interested in determining the probability that a newborn weighs less than 6 pounds. How do we do that?

Well, the first thing we do is calculate the z-score. And in this particular case, when we figure out the z-score, we take 6 subtract 7.5 from that to arrive at negative 1.5. When we divide that negative 1.5 by the standard deviation of 1.25, we wind up with a z-score of negative 1.2.

When we're looking at a z-table such as this, the first step is to determine which row we want to look at. That's going to be represented by the spot to the left of the decimal point and this first spot to the right of the decimal point, which, in this case, is negative 1.2. And so we highlight that row down here.

The second step is to look at the digit that is two spots to the right of the decimal point, which, in this case, is a 0. And so we use this column right here. And as we work our way down and see the point at which both of the row and the column intersect one another, we see a probability here of 0.1151. And what that tells us is that we have a probability associated with the baby weighing less than 6 pounds of 0.1151.

Now, what if we're interested in looking, for instance, at the probability that a newborn might weight more than 10 pounds. Well, in this case, we could calculate the z-score much the same way we did with the previous calculation. And in this case, 10 minus 7.5 gives us 2.5. And when we divide that by 1.25, we get a z-score of 2.

The same process is involved in looking at the z-table. We look at 2.0 and which row that entails, which is the one highlighted down here. And since there are no numbers other than 0 to the right of the decimal point, we look at the very first column, which is the 0.00 column. In this case, we wind up with a probability of 0.9772.

Now, what's interesting in this case is, remember, we are determining or we're trying to determine the probability that a newborn weighs more than 10 pounds. The value given to us here is telling us the area underneath the curve to the left of plus 2 standard deviations, as you see on the graph right here. And so what we're actually needing to do here is subtract that value from 1 to get the area that's on the right-hand side of that line that you see on the normal distribution. So in this case, the probability that a newborn would be greater than 10 pounds would be 1 minus 0.9772 or the probability, in this case, is equal to 0.0228.

As a second example, let's look at something a little bit different. Let's look at, say, the average lifespan of car tires. And so suppose that the population mean, in this case, is going to be 60,000 miles, and the standard deviation has been determined to be 8,000 miles. What if we're interested in determining the probability that a tire lasts less than 45,500 miles? Well, in this case, we calculate the z-score. 45,500 minus 60,000, and divide that difference by 8,000, which gives us a z-score of negative 1.81.

When we look that up on the z-table, remember, we first start out with the negative 1.8. And we see that row right here. And since the second decimal point is a 0.01, we look at the second column over here-- the one labeled 0.01. And we find the intersection. It turns out the probability in this case is equal to 0.0351.

What does this tell us? Well, it tells us the probability that a tire would last less than 45,500 miles, which, in this case, would be 0.0351. What if we were interested in determining the probability that a tire would have a lifespan greater than 80,300 miles?

Well, in this case, we'd calculate the z-score much like we've done in the previous examples. And in this case, 80,300 minus 60,000 gives us 20,300. When we divide that by 8,000, we wind up with a z-score of 2.54.

Using the z-table, we find the row that represents 2.5 and we find the column that represents 0.04. And it turns out, in this case, the probability provided to us is 0.9945. Now, that's not the answer we're looking for here. That would be the area underneath the normal distribution curve-- the standard normal distribution curve up to 2.54 standard deviations.

We're interested in the area to the right of that, and so we would need to subtract that value from 1. And when we do, we get a probability 0.0055. In other words, it is very unlikely in this particular case that a tire would have a lifespan greater than 80,300 miles.

And so let's go back to our objectives just to make sure we covered what we said we would. We wanted to be able to use a z-score table to determine the probability of values associated with a real-world variable is normally distributed, and we did. In each of these cases, we assumed that we are normally distributed variable-- both the weight of newborn babies as well as the lifespan of a tire.

And that being the case, we went through two different examples and illustrated how we could figure out, using a z-table, the probability of a value being less than a particular number or the probability of a value being greater than a particular number. So again, my name is Dan Laub. And hopefully, you got some value from this lesson.

Notes on "Application of z-scores"

(0:00 - 0:31) Introduction

(0:32 - 1:14) Application of z-scores

(1:15 - 4:05) Example 1

(4:06 - 6:17) Example 2

(6:18 - 6:56) Conclusion

PDF's of z-score tables

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TERMS TO KNOW
  • z-Score

    (value - mean)/(standard deviation)

  • z-Table

    Table for looking up the area starting at z=0 for a positive z-score.