Download Math 1125-Introductory Statistics — Lecture 17 10/11/06 1. Normally

Math 1125-Introductory Statistics — Lecture 17
10/11/06
1. Normally Distributed Populations
We left off last time saying that normally distributed populations are populations whose
z-scores are distributed like the Standard Normal.
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
(1)
z=
x−µ
,
σ
and we’ll assume that these z-scores belong to the standard normal distribution.
2. Computing probabilities
We computed one probability last time. Let’s do a few more.
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 more than 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
(2)
z=
x−µ
3.2 − 3.2
0
=
=
= 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.
We’re interested in x’s that lie above x = 4.0, so we need to convert this to a z-score as well.
(3)
z=
x−µ
4.0 − 3.2
0.8
=
=
= 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 1.
We’re interested in the probability of x-scores to the right of x = 4.0. The weights of our
rats surely do not go to infinity, but a true normal distribution does. The probabilities
1
2
0
3.2
1.60
4.0
z
x
Figure 1. The area corresponding to P (4.0 ≤ x < ∞).
beyond about three or four standard deviations is practically zero, however, so computing
the probability out to infinity is convenient and reasonable. Here’s the probability.
(4)
P (4.0 ≤ x < ∞) = P (1.60 ≤ z < ∞) = 0.5000 − 0.4452 = 0.0548
We can interpret this probability as saying that about 5.48% of this population of rats
weighs more than 4.0 ounces.
Using this same population, we can ask for the probability that one of these rats weighs
between 2.5 and 3.5 ounces. Since x = 2.5 is below the mean, we know that we’ll get a
negative z-score for it, and it will lie on the left side of the graph. The x = 3.5 is above the
mean, so it will lie to the right. We can compute the z-scores as we did before.
2.5 − 3.2
−0.7
x−µ
=
=
= −1.40
σ
0.5
0.5
(5)
x−µ
3.5 − 3.2
0.3
z=
=
=
= 0.60
σ
0.5
0.5
The picture, therefore, looks like Figure 2.
z=
−1.40
2.5
0
3.2
0.60
3.5
z
x
Figure 2. The area corresponding to P (2.5 ≤ z ≤ 3.5).
Here, the computation goes like this. The picture tells us to add, so
(6)
P (2.5 ≤ z ≤ 3.5) = P (−1.40 ≤ z ≤ 0.60) = 0.4192 + 0.2257 = 0.6449
It looks like almost two-thirds of the population lies between these two weights.
3. Some statistical reasoning
Suppose Chemical X is known to be present in the hair of men. It is also in women’s
hair, but usually in much smaller quantities. If the concentration of Chemical X in men
3
is normally distributed with mean 12.3 ppm (parts per million) and standard deviation 2.1
ppm, would it be reasonable to conclude that a hair sample with a Chemical X concentration
of 4.1 ppm came from a woman?
In statistics, a standard quantity to look at would be the probability that a man would have
concentration this low or lower. In other words, what is P (x ≤ 4.1)? We need to convert
x = 4.1 to a z-score.
(7)
z=
4.1 − 12.3
= −3.90
2.1
This z-score is not in our table. Our table runs out at P (0 ≤ z ≤ 3.19) = 0.4993, and we’re
well beyond that. We can’t find the precise probability, but we know that
(8)
P (x ≤ 4.1) = P (z ≤ −3.90) < P (z ≤ −3.19) = 0.5000 − 0.4993 = 0.0007
That is, the probability that this is a man’s hair sample with concentration this low is less
than 0.07%. This is so unlikely, that we should conclude that this sample came from a
woman.
4. Quiz 17
Suppose that we have a dexterity test, and the completion times are normally distributed
with µ = 5.3 seconds and σ = 0.7 seconds. Find the probability of a random person getting
a time . . .
1.
Faster than 3.5 seconds.
2.
Faster than 6.2 seconds.
3.
Slower than 6.2 seconds.
4.
Between 5.0 and 6.0 seconds.
5.
Slower than 7.0 seconds.
4
5. Homework 17
For all problems, write z-scores out to two decimal places, and probabilities out to four.
For problems 1-4, consider any population.
1.
The mean of a population (that is, an x-score of x = µ) will convert to a z-score of
.
2.
x-scores above the mean will lie on the
side of the normal curve.
3.
x-scores below the mean will lie on the
side of the normal curve.
4.
True or False, negative z-scores correspond to negative probabilities.
For problems 5-9, consider the population of male lab rats discussed in these lecture notes
(µ = 3.2 ounces and σ = 0.5 ounces).
5.
Find the z-score for a rat that weighs x = 2.0 ounces.
6.
Find the probability that a rat taken at random weighs less than 2.0 ounces.
7.
Find the probability that one of these rats weighs between 2.0 ounces and 3.2 ounces.
8.
Find the probability that one of these rats weighs between 2.0 ounces and 4.4 ounces.
9.
Find the probability that one of these rats weighs more than 4.4 ounces.
For problems 10-15, suppose we have a particular standardized test whose scores are normally distributed with mean µ = 800 and standard deviation σ = 100. We’ll use x for the
scores on the test.
10.
Find the z-score for x = 800. Remember to give z-scores out to two decimal places.
11.
Find the z-score for x = 600.
12.
Find the probability that a test taker chosen at random will score below 600. Remember
that I want probabilities taken out to four decimal places.
13.
Find the probability that a random test taker will score above 1100.
14.
Find the probability that a random test taker will score between 600 and 1000.
15.
Find the probability that a random test taker will score between 900 and 1000.
5
For problems 16-20, suppose that the times to complete a particular dexterity test are
normally distributed with mean µ = 12.0 seconds and standard deviation σ = 2.8 seconds.
16.
What side of the normal curve will times faster than 12 seconds lie on?
17.
Find the z-score for x = 10.5 seconds.
18.
Find the probability that a person taken at random will complete the test faster than
10.5 seconds.
19.
Find the probability that a person taken at random will complete the test slower than
15.0 seconds.
20.
Find the probability that a person taken at random will complete the test faster than
15.0 seconds. (Be careful, and be sure to look at a correct picture. It should be an easy
problem, however.)
Bye.