1. The Normal Distribution, continued

Transcription

1 Math 1125-Introductory Statistics Lecture 16 10/9/06 1. The Normal Distribution, continued Recall that the standard normal distribution is symmetric about z = 0, so the area to the right of zero is If you look at our table, when we get to z =3.00, P (0 z 3.00) = That means there is only about of area beyond that. So while the normal distribution has z going to ±, virtually the entire population lies between z = 3.00 and z = 3.00 (plus or minus three standard deviations). There is so little area out in the tail that after rounding to four decimal places, many of the table entries don t change. I ve left the repeated values blank so that it will be easier to see where they change. Note that P (z 3.89) = after rounding. This means that the area beyond this point is less than For us, this area is essentially zero. 2. Other areas not in the normal table Now, let s continue looking at how we compute probabilities in the standard normal distribution. Again, the normal curve is symmetric about z = 0, so probabilities on the left are exactly the same as those on the right. For example, (1) P ( 1.12 z 0) = P (0 z 1.12) = Draw a picture, if this is not obvious to you. Here are few more examples. I want to stress very strongly that it s much easier to figure out areas from a picture than it is to memorize all the different cases Figure 1. The area corresponding to P ( 1.54 z 2.11). Find P ( 1.54 z 2.11). The picture looks something like Figure 1. The area we re looking for is broken in two pieces by z = 0. The one on the right comes directly from the table. The one on the left goes with z = 1.54, but because the graph is symmetric, the 1

2 2 area is the same as the one for z = The picture tells us that we should add the two areas together. Therefore, (2) P ( 1.54 z 2.11) = P ( 1.54 z 0) + P (0 z 2.11) = = Figure 2. The area corresponding to P (0.45 z 2.11). Find P (0.45 z 2.11). In this case, both z values are positive (to the right of zero). We want the area between them. The area between z = 0 and z =2.11 is 0, and we can get this from the table. This area is broken into two pieces. One piece is the skinny area between z = 0 and z =0.45, which we know is from the table. We want the other piece. The picture tells us to subtract. (3) P (0.45 z 2.11) = P (0 z 2.11) P (0 z 0.45) = = Figure 3. The area corresponding to P ( 2.11 z 0.45). If we have an area on the left side of the graph, we do the same thing. For example, suppose we had P ( 2.11 z 0.45). The picture looks like Figure 3. You could work directly from the picture, or you could notice that if you take the mirror image, the area stays the same. Therefore, the answer is the same as with Figure 2.

3 3 3. Normally Distributed Populations It turns out that the mean for the standard normal distribution is µ = 0, and the standard deviation is σ = 1. Very few populations have this mean and standard deviation, of course, so we can t use the standard normal directly. It is very common, however, to find populations whose z-scores follow the standard normal very closely. These populations are said to be normally distributed. If we have a population that is normally distributed, and we also know its mean and standard deviation, then we can use the standard normal to compute probabilities. For a population with variable x, mean µ, and standard deviation σ, we can convert x s to z-scores using (4) z = x µ σ, and we ll assume that these z-scores belong to the standard normal distribution. In general, a naturally occuring population whose members are essentially the same with random variations is probably normally distributed, at least approximately. The weights of lions, for example, are probably not normally distributed, because you know that males and females tend towards different weights, and there might be variations among subspecies. The weights of males of one population of lions probably are normally distributed. 4. Computing probabilities Computing probabilities for normally distributed populations is pretty easy, if you understand how to find probabilities for the standard normal. I think once you ve seen an example, you ll find that it s straight forward. Let s say that we have a population of male lab rats whose weights are normally distributed with mean µ =3.2 ounces and standard deviation σ =0.5 ounces. If we were to take one of these rats at random, what is the probability that it weighs between 3.2 ounces and 4.0 ounces? We ll use x for the weights of these rats. First of all, let s convert these x-scores into z-scores. One relevant x-score is the number x =3.2. This is the mean weight. Plugging into the formula (5) z = x µ σ = = 0.5 =0. This shouldn t be too surprising. The mean will always convert to a z-score of z = 0. Equivalently, the mean will always lie at the center of the normal curve.

4 4 We re interested in x s that lie to the left of x =4.0, so we need to convert this to a z-score as well. (6) z = x µ σ = = =1.60. This x-score, x =4.0 is larger than the mean, so it will have a positive z-score, and it will lie to the right on the normal curve. As always, you should draw a picture, and the picture for this problem looks like Figure z x Figure 4. The area corresponding to P (3.2 x 4.0). We re interested in the probability of x-scores that lie to the right of x =3.2 and to the left of x = 4.0. Here s the probability. (7) P (3.2 x 4.0) = P (0 z 1.60) = We can interpret this probability as saying that about 44.52% of this population of rats weighs between 3.2 and 4.0 ounces. We ll talk about this more next time, but note that by giving three pieces of information, we can determine probabilities. That is, by saying what the mean is, what the standard deviation is, and that we have a normally distributed population, we ve described the population completely. 5. Quiz 16 Compute the following probabilities for the Standard Normal Distribution. Don t round. 1. P ( 1.88 z 0). 2. P ( z 1.88). 3. P ( 1.88 z 1.88). 4. P ( 0.83 z 2.87). 5. Suppose you have a normally distributed distribution with µ = 12.5 and σ = Find P (12.5 x 13.4).

5 5 6. Homework 16 For problems 1-7, compute the following probabilities for the Standard Normal Distribution. Don t round. 1. P ( 1.02 z 0). 2. P ( z 2.15). 3. P ( 1.28 z 1.38). 4. P ( 0.83 z 2.84). 5. P (0.83 z 2.84). 6. P (1.25 z 2.09). 7. P ( 1.55 z 1.00). For problems 8-10, assume that scores on a particular exam are normally distributed with µ = 500 and σ = 20. We want to find the probability that a random person taking the test will score between 500 and What is the z-score for x = 500? (Round to two decimal places.) 9. What is the z-score for x = 525? (Round to two decimal places.) 10. Find P (500 x 525). Bye.