Download MA 1135 Lecture 07 - Other Normal Distribution Thursday, February

MA 1135 Lecture 07 - Other Normal Distribution
Thursday, February 12, 2015.
Objectives: Other Normal Distribution.
1. Normally Distributed Populations
What I’ve called the standard normal distribution has a population mean of µ = 0, and standard deviation σ = 1. Very few populations have this mean and standard deviation, of course, so we can’t use the
standard normal directly. If we adjust for the mean and standard deviation, however, it is very common to
find populations that follow the standard normal very closely. These populations are said to be normally
distributed.
For a population with variable x, mean µ, and standard deviation σ, we can convert “x-scores” to “z-scores”
using the formula
x−µ
(1)
z=
.
σ
A z-score essentially tells you how many standard deviations an x-score is above or below the mean. Positive
z-scores go with x-scores above the mean, and negative z-scores go with x-scores below the mean. A
population is said to be normally distributed, if the z-scores follow the standard normal distribution.
In general, a naturally occurring population whose members are essentially the same with random variations
is likely 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 may be normally
distributed.
2. 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
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.
(2)
z=
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.
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 picture
below.
(3)
z=
1
MA 1135 Lecture 07 - Other Normal Distribution
0
3.2
1.60
4.0
2
z
x
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.
(4)
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.
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.
Suppose we want to know what proportion of the population weighs more than 4.0 ounces, P (4.0 ≤ x). We
just found that an x-score of x = 4.0 converts to a z-score of z = 1.60, and our picture looks like
0
3.2
1.60
4.0
z
x
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 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.
(5)
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
x−µ
3.5 − 3.2
0.3
z=
=
=
= 0.60
σ
0.5
0.5
z=
(6)
The picture, therefore, looks like the picture below.
−1.40
2.5
0
3.2
0.60
3.5
z
x
MA 1135 Lecture 07 - Other Normal Distribution
3
Here, the computation goes like this. The picture tells us to add, so
(7)
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. Review of Process
Remember that if we have a normally distributed population of x’s with mean µ and standard deviation σ,
we can convert the x’s to z’s using
(8)
z=
x−µ
,
σ
and the z’s will follow the standard normal distribution, whose probabilities can be computed from our table.
4. Some examples
You should:
(1) Draw a picture with the mean in the middle.
(2) Convert relevant x’s to z-scores.
(3) Compute the probabilities using the table.
Example 1. Suppose water samples from wells in Mansfield supposedly have a mean iron concentration
µ = 0.19 mg/L (milligrams per liter) with a standard deviation of σ = 0.03. We’ll assume a normal
distribution.
What is the probability that a sample taken at random will fall between x = 0.19 and x = 0.24? That is, find
P (0.19 ≤ x ≤ 0.24). First, convert to z-scores.
(9)
x = 0.19 converts to z =
0.19 − 0.19
= 0.00
0.03
Our table goes to two decimal places, so always round your z-scores to two decimal places.
(10)
x = 0.24 converts to z =
0.24 − 0.19
= 1.67
0.03
0
0.19
1.67
0.24
z
x
This tells us that
(11)
P (0.19 ≤ x ≤ 0.24) = P (0 ≤ z ≤ 1.67) = 0.4525,
which comes directly from the table.
What is the probability that a sample will be higher than x = 0.24? That is, find P (0.24 ≤ x < ∞) =
P (0.24 ≤ x).
We’ve already found the z-scores, but the picture looks like
MA 1135 Lecture 07 - Other Normal Distribution
0
0.19
1.67
0.24
4
z
x
and we’re looking for P (0.24 ≤ x) = P (1.67 ≤ z), and so we’re going to subtract from 0.5000
(12)
P (0.24 ≤ x) = P (1.67 ≤ z) = 0.5000 − .4525 = 0.0475
Quiz 07A
1.
Suppose we have a normally distributed population with mean µ = 100 and standard deviation σ = 10.
Find P (90 ≤ x ≤ 110).
2.
Suppose we have a normally distributed population with mean µ = 50 and standard deviation σ = 2.50.
Find P (57 ≤ x).
3.
Suppose we have a population of x’s that are normally distributed with mean µ = 12.43 and standard
deviation σ = 1.07.
a.
Find P (12.43 ≤ x ≤ 13.00).
b.
Find P (14.00 ≤ x).
c.
Find P (x ≤ 10.00).
5. 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 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.
(13)
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 ≤ 3.89) = 0.5000, and we’re clearly rounding
to 0.5000 for every z-score bigger than z = 3.89. This gives us
(14)
P (x ≤ 4.1) = P (z ≤ −3.90) = 0.5000 − 0.5000 = 0.0000
So the probability is essentially zero, and so we would conclude that this sample came from a woman.
Standard tail areas
We’ll often be working our probabilities in reverse. For example, we’ll be interested in a tail that has 5% of
the area under the normal curve in it. In the picture below, we have 5% in each tail, which leaves 90% in
the middle.
MA 1135 Lecture 07 - Other Normal Distribution
5
90%
5%
5%
?
0
?
z
What z-score goes with 5% in the tail? This is backwards from what we’re doing. We’re starting with
probabilities/areas, and looking for a z-score. We can do that as follows. In this case, the middle 90% splits
up into two regions of 45%, which look like areas given in our table.
0.4500
5%
5%
0
z
Looking in the main body of our Standard Normal Table, you should be able to find 0.4495 and 0.4505,
which correspond to z-scores of z = 1.64 and z = 1.65. Either of these z-scores is fine, but since our desired
0.4500 lies exactly halfway between 0.4495 and 0.4505, it’s common to use z = 1.645. Since the normal
distribution is symmetric, the z-score that we want on the left is z = −1.645.
0.4500
5%
−1.645
5%
0
1.645 z
1% in the tails
Another common tail area is 1%, which puts 98% in the middle. Since half of 98% is 49% = 0.4900, our
picture looks like
0.4900
1%
1%
0
z
Looking in our normal table, you’ll find 0.4898 and 0.4901, which go with z-scores of z = 2.32 and z = 2.33.
Since 0.4901 is closer to 0.4900, we’ll just go with z = 2.33.
MA 1135 Lecture 07 - Other Normal Distribution
6
0.4900
1%
1%
0
−2.33
2.33 z
Quiz 07B
Find the z-scores that go with 10% in each tail.
Homework 07
1.
Consider any normally distributed population.
a.
The mean of a population (that is, an x-score of x = µ) will convert to a z-score of
b.
x-scores above the mean will lie on the
side of the normal curve.
c.
x-scores below the mean will lie on the
side of the normal curve.
d.
True or False, negative z-scores correspond to negative probabilities.
.
2.
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.
a.
What is the z-score for x = 500? (Round to two decimal places.)
b.
What is the z-score for x = 525? (Round to two decimal places.)
c.
Find P (500 ≤ x ≤ 525).
d.
Find P (x ≤ 450).
3.
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.
a.
What side of the normal curve will times faster than 12 seconds lie on?
b.
Find the z-score for x = 10.5 seconds.
c.
Find the probability that a person taken at random will complete the test faster than 10.5 seconds.
d.
Find the probability that a person taken at random will complete the test slower than 15.0 seconds.
e.
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.)
4.
Suppose you have a normally distributed population with µ = 12.5 and σ = 0.65. Find P (12.5 ≤ x ≤
13.4).
5.
Suppose you have a normally distributed population with µ = 0.0025 and σ = 0.0004. Find P (0.0020 ≤
x ≤ 0.0030).
6.
Consider a population with mean µ = 73.50 and standard deviation of σ = 7.33. Find the following
probabilities.
MA 1135 Lecture 07 - Other Normal Distribution
a.
P (73.50 ≤ x ≤ 84.00).
b.
P (90.00 ≤ x).
c.
P (x ≤ 50.00).
7.
Consider a population with µ = 500 and σ = 100.
a.
Find P (300 ≤ x ≤ 700).
b.
Find P (700 ≤ x).
c.
Find P (x ≤ 300).
8.
What z-scores correspond to having 0.5% = 0.0050 in each tail?
7
HW: 1a) 0. b) Right. c) Left. d) False.
2a) z = 0. b) z = 1.25. c) P (0 ≤ z ≤ 1.25) = 0.3944. d) P (z ≤ −2.50) = 0.5000 − 0.4938 = 0.0062.
3a) Left. b) z = −0.54. c) P (z ≤ −0.54) = 0.5000 − 0.2054 = 0.2946.
d) P (1.07 ≤ z) = 0.5000 − 0.3577 = 0.1423.
e) This is everyone not covered in Problem 12, so 1.000 − 0.1423 = 0.8577.
4) P (0 ≤ z ≤ 1.38) = 0.4162. 5) P (−1.25 ≤ z ≤ 1.25) = 0.3944 + 0.3944 = 0.7888.
6a) P (73.50 ≤ x ≤ 84.00) = P (0 ≤ z ≤ 1.43) = 0.4236.
b) P (90.00 ≤ x) = P (2.25 ≤ z) = 0.5000 − 0.4878 = 0.0122.
c) P (x ≤ 50.00) = P (z ≤ −3.21) = 0.5000 − 0.4993 = 0.0007.
7a) P (300 ≤ x ≤ 700) = P (−2.00 ≤ z ≤ 2.00) = 0.4772 + 0.4772 = 0.9544.
b) P (700 ≤ x) = P (2.00 ≤ z) = 0.5000 − 0.4772 = 0.0228.
c) P (x ≤ 300) = P (z ≤ −2.00) = 0.5000 − 0.4772 = 0.0228.
8) We’re looking for 99% in the middle, and split in two is 49.5% = 0.4950. We’re halfway between z = 2.57
and z = 2.58. Either of these is OK, or z = 2.575, if you want.
MA 1135 Lecture 07 - Other Normal Distribution
8
The Standard Normal Distribution
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.6
3.8
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
0.4990
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4991
0.4993
0.4998
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
0.4994
0.4995
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.4991
0.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.4992
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
0.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4994
0.4996
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.4992
0.4995
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.4993
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4996
0.4997
0.4997
0.4998
0.4999
0.5000
0.4999
0
z