Like most of the chapters in this book, Chapter 4 has just two ideas in it. One is probability; the other is the standard normal distribution. We'll look at probability first.

Probability, in general terms, is simply the likelihood of something happening. We can compute probability based on the frequency of an outcome and the total number of possible outcomes.

For example, when a baby is born at Memorial Hospital, it may be a boy or a girl. Last month at Memorial, 20 babies were born, 11 of them boys. Let's say Memorial had a contest in which the hospital administrator picked out the name of one baby born last month and awarded the parents a free dinner at Bob's Supper Club. What is the probability (chance) that the name drawn out was that of a boy? Let's call the possible outcome, boy, outcome A and its probability p(A). We can use the frequency of the boy outcome [f(A)=11] and the total number of possible outcomes (N=20) to figure this out. Here's the formula:

Makes sense, doesn't it? If 11/20 of the babies were boys, then there should be an 11/20 chance that the name drawn was a boy's.

Now if you think about this, you see that the proportion of babies that is boys is the same as the probability that any given baby drawn at random will be a boy. The proportion of babies that is girls is the same as the probability that you'll draw a girl at random. So proportions and probabilities are two ways of saying the same thing. When we add the proportion (portion of the whole) that is boys (0.55) to the proportion (portion of the whole) that is girls (0.45), you get the whole, which is 1. Adding up all the probabilities of all of the possible outcomes should always get you to 1.

In addition, you may realize that multiplying the probability (or proportion) by 100 will give you the per cent of babies that is boys (or girls). So these babies are 55% boys and 45% girls. These percentages add up to 100%, which includes all the babies. Makes sense.

And if we want to ask the reverse question, what is the chance the name drawn will NOT be a boy's, how do we compute those odds? Does it make sense that all the possibilities should add up to 20/20 or 1? In other words, if we add the probability of a boy, 0.55 (or 11/20) to the probability of a girl (that is, NOT a boy), 9/20 or 0.45, we should get the probability of the baby being one or the other, 20/20 or 1. In this case, we can compute the probability of outcome A not happening [noted as by subtracting the probability it WILL happen from 1. This is called the converse rule of probability, and is shown mathematically as follows:

This rule holds as long as the possible outcomes are mutually exclusive. That means that the occurrence of one makes the occurrence of the other impossible. For example, when the baby is a boy, that makes it impossible that the baby is a girl. A baby can't be both a boy and a girl, so boy and girl are mutually exclusive outcomes in a particular baby.

Figuring out the probability that an outcome will not happen isn't so difficult when you have only two possibilities--boy and girl, but the formula helps quite a bit when you're dealing with many possibilities. Look at the next example:

In a survey of the over-25 population in Brookville as to highest level of education completed, the following data were reported:

Now, if we draw one individual at random from those included in our survey, what is the probability that person will have reported High school as the highest level of education completed? Call a response of High school, outcome A. Let's think about that: the frequency of people reporting high school as the highest level completed [f(A)] is 46. The total number of possible outcomes is the number of people surveyed, N=1000. So, using our formula:

Now it should make sense to you that the probabilities of all the other outcomes can be computed in precisely the same way. Go ahead and figure these out. Don't look ahead yet; the answers to these follow.

And the probability of drawing a person who marked something else, that is any category other than High school, should equal the sum of the probabilities of each of these events. In other words, the probability of drawing a person who did NOT mark High school as the highest level of education completed should be 0.001 + 0.016 + 0.894 + 0.030 + 0.009 + 0.004 = 0.954. Now that's a lot of work to go to just to find out the probability a person did not mark High school. There's an easier way to get to the same answer. Remember the converse rule? To find the probability of drawing a person who did NOT mark High school as the highest level of education completed, subtract the probability of drawing a person who DID mark High school from 1.

1 - 0.046 = 0.954

Same result, less work. So, to compute the probability of an event NOT happening, subtract the probability it WILL happen from 1. It's accurate and saves time.

You would expect all the probabilities listed above for our survey in Brookville to add up to 1, since 1 denotes all the possible outcomes. Much of the time when you add up probabilities, you won't get exactly 1, but something close, perhaps 0.998 or 1.002. Just as happened when you added up per cents in Chapter 2, rounding can cause small inaccuracies. These will show up in your total. If your sum is off by a great deal, then it helps to look at your figures for a mistake. Small inaccuracies are usually not due to error, but to rounding.

We've already sneaked up to this idea in computing the probability of someone marking something other than High school in the Brookville Education Survey (BES). Let's take another look at this in the following example.

In the BES, what is the probability that a case drawn at random from the survey sample has completed less than high school? Well, there are those who've completed junior high school, those who've completed elementary school, and those who've completed less. The numbers in each category are 30, 9, and 4. These total 43. We can take this frequency and use it to compute probability for the three categories combined:

or we can simply add up their probabilities, which we've already gone to the work of computing.

0.030 + 0.009 + 0.004 = 0.043

Once again same answer, less work.

This procedure, known as the Addition Rule of Probability. also known as the or rule, can be indicated mathematically as follows:

In other words, the probability of observing one or the other of a set of mutually exclusive events outcomes is equal to the sum of the individual probabilities of each event.

This is how we figured out the probability that someone did NOT mark High school, isn't it? We added up the probabilities of all the other outcomes.

Now, let's complicate things a bit. To do so, let's return to a simpler example, the babies at Memorial Hospital. Remember that, of the 20 babies in the drawing, 11 (0.55) were boys and 9 (0.45) were girls. Now, suppose Memorial decided to draw two babies and give dinners to the parents of both. What is the probability that, when the administrator draws out the names of two babies, both will be boys?

Well, the total of ALL the possibilities: drawing two girls; drawing first a boy, then a girl; drawing first a girl, then a boy; and drawing two boys should add up to 1. So we know the old addition method will give us erroneous results, because adding the probability for a boy (0.55) and for another boy (0.55) gives us 1.10, which is more than 1. We need a method that results in a number we can add to all the various other possibilities (girl, girl; girl,boy; boy,girl) to equal 1.

The new rule we use here is called the multiplication rule of probability, the and rule, which tells us that the probability of a combination of two or more mutually exclusive outcomes occurring is equal to the product of their individual probabilities. Here it is in mathematical notation:

Now, let's see if this rule gives us a rational answer to our original question. What is the probability that, when two names are drawn for the free dinner, both will be boys?

0.55 X 0.55 = 0.3025

At least it doesn't come to more than 1. One way to verify that we're on the right track is to figure out the probability of all our other possible outcomes, then see whether these add up to 1. Here it is:

Looks good, doesn't it?

Now we can use this rule to figure out some combinations of events from the BES. What is the probability that if we draw 3 individuals at random from the BES, all of them have marked Professional/graduate school as the highest level of education completed? Using the and rule:

0.016 X 0.016 X 0.016 = 0.000004096

Pretty slim odds, right? Makes sense when you consider that only 16 out of the 1000 are in this category and we're figuring out the likelihood that three draws give us 3 of these 16.

How about the probability that in 3 draws we get two Baccalaureate and 1 High school?

0.894 X 0.894 X 0.046 = 0.0368

And the probability that in 2 draws we get two Baccalaureate?

0.894 X 0.894 = 0.799

By now you're familiar with a frequency distribution. We've looked at frequency distributions for all kinds of studies. These are called empirical distributions because they result from real data gathered in a real study. In addition, we've looked at frequency distributions on this web site for a number of pretend studies. While these don't result from real studies, they could have. That means these are empirical distributions too.

The normal distribution is a theoretical distribution because it couldn't result from real data. It is an ideal only approached in real life. This is because it assumes an infinite number of cases in the study; and no real study can include an infinite number of cases. Even though it is theoretical, the normal distribution is useful because a bunch of characteristics in real life are normally distributed. That means if we understand a normal distribution, we can use it to figure out all kinds of things about stuff we're interested in studying. A normal distribution has certain specific characteristics. Let's look at the graph of a normal distribution.

Note the characteristics of the normal distribution. It has a specific shape; I'll have more to say about this in a few minutes. It contains an infinite number of scores; you know this because the shape of the graph is curved. Remember when you drew frequency polygons in Chapter 2? They all had a sort of jerky line from point to point. Well, the only way to get this smooth curve is to have an infinite number of points between any 2 other points. Also note that the tails of this curve never actually reach the baseline; they approach it ever more closely, but don't actually touch it. That means there is no top and no bottom score in our data--also impossible. That's what makes this distribution theoretical.

Now, let's look at what makes it useful. You will remember I said that many characteristics are normally distributed in the population. This curve has the "normal" shape we're talking about here. It is bilaterally symmetrical; this means the left side is the mirror image of the right side. Note that the mean occurs at the exact center (median) and the highest point (mode) of the distribution. Note the vertical lines drawn through the graph. They mark off points exactly 1 standard deviation apart, beginning at the mean and working outward in both directions. Then note the percentages marked in each of the sections separated with vertical lines. These are what give the curve its precise shape.

On a normal distribution, 34.13% of cases always fall between the mean and one standard deviation above it. Another 34.13% always occur between the mean and one standard deviation below it. The fact that these percentages are identical fits with the fact that the curve is bilaterally symmetrical. Then between one and two standard deviations above the mean we see 13.59 of cases (same percentage below); between two and three standard deviations above the mean occur 2.14% (same percentage below); and beyond three standard deviations away from the mean is 0.13% on either side.

What makes a normal distribution useful to us is that its shape is perfectly predictable. We can count on these percentages occurring in every normal distribution. Now, let's take a look at this new way we're using standard deviations as we look at the normal distribution.

You probably noticed that we're now using standard deviations as a sort of yardstick of distance from the mean. We've been talking about scores that are one or two or three standard deviations above (or below) the mean as marking-off places for our percentages on the normal distribution. When we begin to look at a frequency distribution in terms of how many standard deviations a score is away from the mean, we're talking about a concept called a standard score.

z-scores z-scores are one kind of standard score that do what we've just been talking about. They express a raw score in terms of how many standard deviations it is away from the mean. So a z-score of +2 is the same as a raw score that is 2 standard deviations above the mean, and a z-score of -1 is the same as a raw score that is 1 standard deviation below the mean. Here's an example.

IQ in humans happens to be normally distributed, with a mean of 100 and a standard deviation (s) of 15. This means that an IQ of 100 is exactly at the mean. An IQ of 115 (100 + 1s) is 1 standard deviation above the mean. An IQ of 70 (100-2s) is 2 standard deviations below the mean. Put into z-scores, the IQ of 100 has a z-score of 0 (0 standard deviations away from the mean), 115 has a z-score of +1, and the IQ of 70 has a z-score of -2. Pretty easy, isn't it? It stays easy as long as the IQs we're interested in fall exactly 1 or 2 or 3 standard deviations away from the mean. Gets trickier when we're interested in an IQ of, say 89. We know that 85 is one standard deviation below the mean, and the mean is 100; so we can say fairly easily that the z-score for 89 is between -1 and 0. But exactly what is it?

Here's a formula that should help:

What this formula tells us to do is find the deviation of the score, something you did a lot of in Chapter 3, then divide this deviation by s, the size of 1 standard deviation. That is a logical way to find how many standard deviations a score is away from the mean. Subtracting the mean from the score assures that scores below the mean will get a negative z-score and scores above the mean will get a positive z-score. So to find the z-score of an IQ of 89:

This answer is indeed between -1 and 0. We can verify that this formula gives us good answers by using it to compute z-scores we already know, for IQs of 115 and 70.

Characteristics of z-scores. Standard scores have some interesting characteristics we should look at. To begin with, a distribution of z-scores from a sample has exactly the same shape as the distribution of raw scores from the same sample. No difference. Makes sense, when you think about it. All we've done is put the raw scores into a different, but mathematically related, form; so of course the two distributions would be identical. Only the labels across the bottom would be different. Also, the mean of a distribution of z-scores is always going to be 0. Makes sense too if you remember that we put the scores in terms of deviation from the mean. Remember that the mean of deviations from the mean equals 0? And when we convert scores to standard scores, we set the mean equal to 0. One more thing about distributions of standard scores; their variance is always equal to 1. When we convert scores to z-scores, we set up a standard deviation as equal to 1, don't we? Variance is standard deviation squared; and 1²=1. So variance will be equal to 1.

Why are we interested in z-scores? z-scores enable us to do something your junior high teachers told you was impossible--compare apples to oranges. They can put two unlike things into similar terms so that they can be compared. Here's an example:

The ACT test is widely used in the center of the country to measure the achievement of high school juniors and seniors. Many colleges use ACT scores to predict a student's chances of success with college work. Nationwide, the mean on the ACT test is 18, with a standard deviation of 6. The SAT test is widely used for the same purpose, only mostly in the eastern and western US. On each section of the SAT, the nationwide mean is 500, with a standard deviation of 100.

Eastern University's highly selective creative writing program only accepts students who scored at least 650 on the verbal section of the SAT. Eliot, a South Dakota senior, has applied to the program, but has taken the ACT, not the SAT test. His English score on the ACT was 28. The University has to decide whether to accept Eliot in the writing program. How will they know what to do?

A good solution is to convert a score of 650 on the SAT into a z-score, then convert Eliot's 28 on the ACT into a z-score too. Then by comparing the two z-scores, the University should be able to make a fair determination whether Eliot's ACT is high enough for acceptance. Go ahead and do the math:

So Eliot's z-score on the ACT test was higher than the minimum SAT z-score required to enter Eastern's writing program. Assuming he looks good in other ways, it looks as though Eliot is on his way to Eastern University next year. And we've found a useful way to compare apples to oranges.

Now we can put these two concepts--the normal distribution and standard scores--together to make a new one, the standard normal distribution. This is simply a normal distribution of raw scores converted to standard (z) scores. This has in common with all normal distributions of z-scores a mean of 0, a variance of 1, and the predictable symmetrical shape. The shape means we can predict with perfect accuracy the proportion (percentage) of scores under the curve between any two points.

Finding Proportion of the Area Between 2 Points: For example, 0.3413 of scores are found between the mean and a z-score of +1 (1 standard deviation above the mean) and 0.1359 of scores are found between a z-score of +1 and a z score of +2. So what proportion of scores is found between the mean and a z-score of +2? The problem is pictured below with the areas of interest shaded in green and yellow:

So the total area (proportion of scores) between the mean and z=+2 is the total of the two areas shaded: the green has an area of 0.3413 and the yellow has an area of 0.1359. 0.3413 + 0.1359 = 0.4772.

Using the Table of Areas: This is fine for z-scores that are whole numbers for which we have the areas; but what do we do if we want to know the area between the mean and z=-2.16? Well, to figure out the areas under the curve for every possible z-score requires the use of calculus, a branch of mathematics that is well beyond the scope of this course; however there is a solution. Someone who really loves calculus has already done all the hard work for us, and the results appear in the back of your textbook, Appendix B, Table 1, Table of Areas Under the Normal Curve, beginning on page 399.

Take a look at the table. You can see it is several pages long, and each page has three sets of columns. Each set of columns has three columns in it: Column (A) lists z-scores; Column (B) gives the area between the mean and the z-score listed on that line; Column (C) gives the area out in the tail of the curve beyond the z-score listed on that line. Note that there are no negative z-scores listed. This is because the normal distribution is symmetrical, so you can find the areas for a z-score of -1 under z=1. You will want to remember that you look up negative scores on the same line as you find their positive counterparts; the areas will be the same whether the z-score is positive or negative.

You will also want to remember that the entire area under the curve is 1, so the area above the mean is 0.5 and the area below the mean is also 0.5. The sum of these two areas is 1, the total area under the curve. This means that the area listed for a particular z-score in Column (B) plus the area listed in Column (C) should equal 0.5. Check this out; choose any line in the table and sum the numbers in Columns (B) and (C)--it works.

Now, let's return to our question: what is the area between the mean and z=-2.16? The area of interest is roughly indicated below:

This is easy now; look up 2.16 (Remember, negative z-scores are read just as positive ones are.) in Column (A). Now you must decide whether you want the area in Column (B) or in Column (C). Since Column (B) gives the area between the mean and z, it gives the area we want here: 0.4846.

[What would we have found if we'd looked in Column (C) instead? The number there is 0.0154, which subtracted from the 0.5 area on the left side of the mean, gives us 0.4846--precisely the same answer you found in Column (B) above. These things are related.]

Suppose we want to know the proportion of scores that fell below z=-2.16. Now is the time we're interested specifically in Column (C). OR we could subtract 0.4846 from 0.5 to get 0.0154.

More Complicated Situations: Now let's ask ourselves what proportion of scores is found above z=-2.16. The area in question is marked in green below:

Now we need two areas: the area between the mean and z=-2.16 and the area above the mean. We've already looked up the first one; it was 0.4846. The second area represents half the area under the curve, or 0.5. So the total area above z=-2.16 is the total of 0.4846 and 0.5, or 0.9846.

[There's another way to find this area. It is to find the area in the tail which is NOT shaded in green, then subtract it from 1, which is the total area under the curve. To find this area, look up z=-2.16 in the (C) column. We've already done this; the area is 0.0154.

1 - 0.0154 = 0.9846. Same answer, different method. Either method is valid; use whichever one makes more intuitive sense to you.]

Finding Areas that Stretch Across the Mean: This last example provided us with a procedure we can use whenever finding a proportion of cases, some of which fall above the mean and some of which fall below the mean.

Here's another example: What proportion of cases fall between z=-2.41 and z=+1.73? The problem is illustrated below, with the area below the mean shown in yellow and the area above the mean shown in green:

What you did on the problem above was to find the area in question above the mean and add it to the area in question below the mean. This gave you the total area. The same procedure should work here. How do you find the area shaded in green above?

If you said, "Look in the Table of Areas for z=1.73 in Column (B)," then you're on the right track. We look in Column (B) because the area we're interested in is between the mean and z. The area given in the table is 0.4582. And how will we find the area shaded in yellow above?

Same sort of answer, isn't it? This time you look in the Table of Areas for z=2.41. (Remember, there are no negative z-scores listed, so you just look up the positive counterpart to the score you want.) The area in Column (B)--because once again, we're interested in the area between the mean and z--is 0.4920.

Now what? Does it make sense that we have to add the yellow and green areas together to find the area between these 2 z-scores? It should. So we add 0.4582 + 0.4920 to get 0.9502.

So the rule is, whenever you want to find the area between two z-scores which fall on opposite sides of the mean, you find the area from the mean to each, then add these areas together. If you have trouble remembering how this works, remember that drawing a quick sketch of the situation, like our sketches above, will help you to visualize the area you're interested in.

Finding the Area Between 2 z-scores on the Same Side of the Mean: Now, let's figure out how to find the area between two positive z-scores or between two negative z-scores, in other words, when the 2 z-scores are on the same side of the mean. For example, what is the area between z=+1.37 and z=+2.43? The problem is illustrated below:

I can see two possible ways to go about this. One has more steps than the other. Let's try the easier one first. You can look up z=+1.37 in the Table of Areas and find that the area between the mean and this z-score is 0.4147. That's the area above the mean up to where the yellow section begins. Then you can look up z=+2.43 and find that the area between the mean and this z-score is 0.4925. That's the area above the mean up to the end of the yellow section. Does it make sense that the yellow section is the difference between these two values? If so, then subtract 0.4925 - 0.4147 to get 0.0778.

This would mean that the rule is, to find the area between two z-scores on the same side of the mean, look up their areas in the (B) column and subtract.

Another procedure that will get you the same place and that might make more intuitive sense has more steps. Here it is important to remember that the total area above the mean is 0.5. If we subtract from that 0.5 the two unshaded (white) areas around the yellow area, what's left should be the yellow area. We already know the size of the white area below the yellow; it is 0.4147. The white area above the yellow is found in Column (C) for z=+2.43; it is 0.0075. Now we subtract both of these areas from 0.5: 0.5 - 0.0075 = 0.4925; and 0.4925 - 0.4147 = 0.0778. (We could have added these two white areas together to get 0.4222, then subtracted this total from 0.5 to get 0.0778.) Once again, the same answer. Often in math, there are two (or more) equally valid ways to find a correct answer. It doesn't matter which you use, as long as you end up in the right place.

Finding Scores at Given Percentile Ranks: So what use is all of this anyhow? Don't tell me you haven't wondered this once or twice by now. Let's see how we can use standard scores and the standard normal distribution to help us figure out something someone might actually want to know.

Let's say you own a grapefruit grove. Yields on your trees are normally distributed with a mean yield of 100# fruit and a standard deviation of 7#. Now let's say your county extension agent tells you that trees producing below the 25^th percentile are not earning enough to cover the costs of their cultivation and should be replaced. What you need to know now is what yield is at the 25^th percentile, so you'll know which trees to remove. Finding the z-score associated with the 25^th percentile will enable us to figure this out.

Let's make a picture to see if this helps us find a place to start:

Note that the line we placed to mark the 25^th percentile is not placed precisely; since we're not going to measure, but simply use this picture to get an idea what we have to do, it doesn't matter much exactly where the line goes. The only important thing is to get it on the correct side of the mean. Since the 25^th percentile is below the mean, we placed the line to the left of the mean. Now the orange area represents the trees we want to replace, those producing below the 25^th percentile.

What does the 25^th percentile represent? I hope you remember from Chapter 2 that this is the yield at or below which we find 25% of trees. This means we know the areas on our diagram:

So what is the z-score marking the orange area? If you remember that Column (B) on the Table of Areas designates the area between the mean and z, then it will make sense to you that finding 0.2500 in Column (B) will tell us the z-score at the 25^th percentile. [You can do the same sort of thing by looking for the size of the orange area--also 0.2500--in Column (C) on the table.]

Scanning the table, we do not find exactly 0.2500 listed in either Column (B) or Column (C). In Column (B) we see an area of 0.2486 for z=-0.67 and an area of 0.2517 for z=-0.68. The best we can do here is determine which is closer to 0.2500; it is (by a little) 0.2486, or z=-0.67. So we declare the z-score associated with the 25^th percentile to be -0.67.

So what yield is this? We can use the z-score formula to find a raw score (X):

You would remove and replace trees yielding less than about 95 pounds of fruit annually.

The z-score associated with any percentile rank can be found in similar manner. When the percentile rank is above 50, then the method varies slightly. Suppose you wanted to find the yield at the 94^th percentile. Here's a picture:

Since the 94^th percentile is the yield at or below which 94% of trees scored, the areas on the above picture can help us to identify the z-score we need. Looking through the (B) column in the Table of Areas, we find 0.4394 at z=+1.55 and 0.4406 at z=+1.56. Since these are exactly equidistant from 0.4400, we can make an executive decision and declare the appropriate z-score to be either +1.55 or +1.56. Let's use +1.55. We can now identify the yield at this z-score:

So trees at the 94^th percentile produce about 111 pounds of fruit annually.

Using Proportions to Find Numbers of Cases: Let's say you're organizing student volunteers. Suppose that volunteer hours tend to be normally distributed with a mean of 39 hours worked and a standard deviation of 3.75 hours. The Brotherhood/Sisterhood Agency needs 10 students who can work at least 45 hours each. Because of training considerations, they don't accept students who work fewer hours. Before you sign up to provide volunteers for the agency, you need to estimate whether you're going to be able to meet its needs.

The question is, how many students from your pool of 68 volunteers are likely to work at least 45 hours? What do you need to know to find the number?

In order to find a number, you'll need to find the proportion of your 68 students who are likely to work 45 hours. To find that, you'll need to know the z-score associated with 45 hours of work. And to find that:

Now we can find the proportion of students who are at or above z=+1.6; looking in the (C) column, we find 0.0548. Once we know the proportion of our students who meet the criterion, we can easily find out the number of students that represents:

(0.0548)(68) = 3.11

This isn't close to the 10 volunteers that Brotherhood/Sisterhood needed, so you'll most likely have to regretfully decline.

You've finished Chapter 4. If you haven't gone through the Comprehension Checks in the chapter, I encourage you to do so. Once you've worked your way through these and checked your answers, try the Review Exercises at the end of the chapter. Remember, the answers for these are provided in your book too. This gives you many opportunities to practice and to check your understanding.

When you've finished all of the practice problems in the textbook, request a Chapter 4 worksheet via e-mail. It will be sent along with answers, so that you may check your work. When you feel ready, request the test.