Download Math 1125-Introductory Statistics — Lecture 16 10/9/06 1. The

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 0.5000. If you look at our table, when we get to z = 3.00, P (0 ≤ z ≤ 3.00) =
0.4987. That means there is only about 0.0013 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) = 0.5000 after rounding. This means that the
area beyond this point is less than 0.00005. 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) = 0.3686
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.
.4382 .4826
−1.54
0
2.11
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
area is the same as the one for z = +1.54. The picture tells us that we should add the two
areas together. Therefore,
P (−1.54 ≤ z ≤ 2.11) = P (−1.54 ≤ z ≤ 0) + P (0 ≤ z ≤ 2.11)
(2)
= 0.4382 + 0.4826 = 0.9208.
.4826
.1736
0
2.11
0.45
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 0.4826, 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 0.1736 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) = 0.4826 − 0.1736 = 0.3090
.4826
.1736
−2.11
0
−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. 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=
3.2 − 3.2
0
x−µ
=
=
= 0.
σ
0.5
.5
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
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=
4.0 − 3.2
0.8
x−µ
=
=
= 1.60.
σ
0.5
0.5
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 4.
0
3.2
1.60
4.0
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) = 0.4452
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 = 0.65. Find
P (12.5 ≤ x ≤ 13.4).
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 525.
8.
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.