Transcript Statistics Chapter 6

Overview
This chapter will deal with the
construction of
probability distributions
Probability Distributions will describe
what will probably happen instead of
what actually did happen.
1
Random Variables
2
Definitions
 Random Variable
a variable (typically represented by x) that has a
single numerical value, determined by chance,
for each outcome of a procedure
Probability Distribution
a graph, table, or formula that gives the
probability for each value of the random variable
3
Definitions
Discrete random variable
has either a finite number of values or countable
number of values, where ‘countable’ refers to the
fact that there might be infinitely many values,
but they result from a counting process.
Continuous random variable
has infinitely many values, and those values can
be associated with measurements on a
continuous scale with no gaps or interruptions.
4
Requirements for
Probability Distribution
 P(x) = 1
where x assumes all possible values
0  P(x)  1
for every value of x
5
Mean, Variance and Standard Deviation
of a Probability Distribution
Formula 6-1
µ =  [x • P(x)]
Formula 6-2
2
 =  [(x - µ) • P(x)]
2
6
The probability that a car insurance company
pays $200,000 is .01, $100,000 is .02 and
$50,000 is .04.
.93
1. Draw the probability density function.
2. Determine the minimum value the insurance
company should charge to break even.
.04
.02
0 $50000
$100000
.01
$200000
µA= ($200000)(.01)+($100000)(.02)+($50000)(.04)+(0)(.93) = $6000
7
Page 243 #5
Amount Payment
92
68
51
17
522 None
Percent
$10,000.00
$25,000.00
$60,000.00
$90,000.00
12.27%
9.07%
6.80%
2.27%
69.60%
.70
1. Draw the probability density function.
2. Determine the minimum value the insurance
company should charge to break even.
.12
.09
0 10
25
.07
60
.02
90
µA= ($10000)(.12)+($25000)(.09)+($60000)(.07)+ (.022) ($90000)+ (0)(.93) = $9450
8
Definition
Expected Value
The average value of outcomes
E =  [x • P(x)]
9
What is the expected value for a $1.00 ticket that pays off a $500.00
prize if 1000 tickets are sold?
E =  [x • P(x)]
Event
Win
Lose
10
E =  [x • P(x)]
Event
x
Win
$499
Lose
- $1
11
E =  [x • P(x)]
Event
x
P(x)
Win
$499
0.001
Lose
- $1
0.999
12
E =  [x • P(x)]
Event
x
P(x)
x • P(x)
Win
$499
0.001
0.499
Lose
- $1
0.999
- 0.999
13
E =  [x • P(x)]
Event
x
P(x)
x • P(x)
Win
$499
0.001
0.499
Lose
- $1
0.999
- 0.999
E = -$.50
14
Definitions
 Binomial Probability Distribution
1. The experiment must have a fixed number of
trials.
2. The trials must be independent. (The outcome of
any individual trial doesn’t affect the
probabilities in the other trials.)
3. Each trial must have all outcomes classified
into two categories.
4. The probabilities must remain constant for each
trial.
15
Notation for Binomial Probability
Distributions
n =
fixed number of trials
r = specific number of successes in n trials
p = probability of success in one of n trials
q = probability of failure in one of n trials
(q = 1 - p )
P(r) = probability of getting exactly r
success among n trials
Be sure that r and p both refer to the same category being
called a success.
16
Binomial Probability
Formula
17
Binomial Probability
Formula
 P(x) =
n!
•
(n - r )! r!
pr •
n-r
q
18
Method 1
Binomial Probability
Formula
 P(r) =
n!
•
(n - r )! r!
 P(r) = nCr • pr
•
pr •
n-r
q
qn-r
for calculators with nCr key.
19
Example: Find the probability of getting exactly
3 correct responses among 5 different requests
from AT&T directory assistance. Assume in
general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n=5
r=3
p = 0.90
q = 0.10
20
Example: Find the probability of getting exactly
3 correct responses among 5 different requests
from AT&T directory assistance. Assume in
general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n=5
r=3
p = 0.90
q = 0.10
Using the binomial probability formula to solve:
P(3) =
5C3
• 0.93 • 0.12 =.0729
21
Example: Using n = 5 and p = 0.90, find the
following:
a) The probability of exactly 3 successes
b) The probability of at least 3 successes
a) P(3) = 0.073
b) P(at least 3) = P(3 or 4 or 5)
= P(3) or P(4) or P(5)
= 0.073 + 0.328 + 0.590
= 0.991
22
For Binomial Distributions:
• Formula 4-6 µ
=n•p
• Formula 4-7  = n • p • q
2
or =
n•p•q
23
Example:
Find the mean and standard
deviation for the number of girls in
groups of 14 births.
• n = 14
• p = 0.5
• q = 0.5
• Using the binomial distribution
formulas:
24
Example:
•
•
•
•
•
Find the mean and standard
deviation for the number of girls in
groups of 14 births.
n = 14
p = 0.5
q = 0.5
Using the binomial distribution formulas:
µ = (14)(0.5) = 7 girls
  (14)(.5)(.5) = 1.9 girls
25
Reminder
• Maximum usual values = µ + 2

• Minimum usual values = µ - 2 
26
Example:
Determine whether 68 girls among
100 babies could easily occur by chance.
•
•

•
•
•
For this binomial distribution,
µ = 50 girls
= 5 girls
µ + 2  = 50 + 2(5) = 60
µ - 2  = 50 - 2(5) = 40
The usual number girls among 100 births would be from 40
to 60. So 68 girls in 100 births is an unusual result.
27
Poisson Random Variable
Poisson process is the random occurrence
of a series of events over time (or space) in
such a way that:
1. The number of successes can be any whole number.
2. The probability of success is equally spread out.
λ
e λ
P(x  r) 
r!
r
the mean number of successes
r the desired number of successes

28
Example: On an average day, 6 people visit
the barbershop
a. What is the probability that exactly 4 will visit the barbershop?
b. What is the probability that 2 or less will visit the barbershop?
c. What is the probability that more than 3 will visit the barbershop?
e-6  64
a. P(4) 
 .1338
4!
e-6  60 e-6  61 e-6  62
b. P(X  2) 


 .0024  .0149  .0446  .0619
0!
1!
2!
 e-6  60 e-6  61 e-6  62 e-6  63 
  .849
c. P(X  3)  1  



1!
2!
3! 
 0!
29
Example: A 100-foot of pipe has 20 leaks. A 25-foot
section is chosen at random
a. What is the probability it has exactly 3 leaks?
b. What is the probability it has at most 2 leaks?
c. What is the probability it has more than 3 leaks?
=5
e-5  53
a. P(3) 
 .1404
3!
e-5  50 e-5  51 e-5  52
b. P(X  2) 


 .0067  .0337  .0842  .1246
0!
1!
2!
 e-5  50 e-5  51 e-5  52 e-5  53 
  .735
c. P(X  3)  1  



1!
2!
3! 
 0!
30
Overview
 Continuous random variable
 Normal distribution
31
Overview
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
µ
Score
32
Overview
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
µ
Score
y=
e
1
2

x-µ
2
(  )
2p
33
Definitions
 Uniform Distribution
a probability distribution in which the
continuous random variable values are
spread evenly over the range of
possibilities; the graph results in a
rectangular shape.
34
Definitions
 Density Curve (or probability
density function)
the graph of a continuous
probability distribution
35
Definitions
 Density Curve (or probability density
function)
the graph of a continuous
probability distribution
1. The total area under the curve must
equal 1.
2. Every point on the curve must have a
vertical height that is 0 or greater.
36
Because the total area under
the density curve is equal to 1,
there is a correspondence
between area and probability.
37
Times in First or Last Half Hours
38
A continuous random variable,T, has a uniform
(rectangular) probability distribution over the interval
[2,10].
a. Graph the probability distribution.
b. Find P(x>6)
c. Find P(5<x<8)
a. A = b*h, 1=b*8, h = .125
b. P(x>6) A = b*h = 4*.125 = .5
c. P(5<x<8) A = b*h = 3*.125 = .375
2
10
39
A continuous random variable,T, has a uniform
(triangular) probability distribution over the interval
[0,6].
a. Graph the probability distribution.
b. Find P(x > 4)
c. Find P(1 < x < 4)
(0, 1/3)
(1, 5/18)
(2, 2/9)
(3, 1/6)
(4, 1/9)
(5, 1/18)
0
6
a. A = ½ b*h, 1= ½* 6*h, h = 1/3
b. P(x > 4)
A = ½ b*h = ½ *2*(1/9) = 1/9
c. P(1 < x < 4) A = ½ b1*h1 - ½ b2*h2 =
½* 5*(5/18) – ½ *2*(1/9) =
25/36 – 1/9 = 21/36 =
7/12
40
Heights of Adult Men and Women
Women:
µ = 63.6
 = 2.5
Figure 5-4
Men:
µ = 69.0
 = 2.8
63.6
69.0
Height (inches)
41
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
42
Table A-2
Formulas and Tables card
 Appendix A2
43
Table A-2
Standard Normal Distribution
=1
µ=0
0
x
z
44
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Area found in
Table A-2
Area = 0.3413
0.4429
-3
-2
-1
0
1
2
3
0
z = 1.58
Score (z )
Figure 5-5
Figure 5-6
45
Table A-2 Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
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
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
46
To find:
z Score
the distance along horizontal scale of the
standard normal distribution; refer to the
leftmost column and top row of Table A-2
Area
the region under the curve; refer to the
values in the body of Table A-2
47
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
48
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
P ( 0 < x < 1.58 ) =
0
1.58
49
Table A-2 Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
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
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
50
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
51
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
The probability that the chosen
thermometer will measure freezing water
between 0 and 1.58 degrees is 0.4429.
52
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
There is 44.29% of the thermometers with
readings between 0 and 1.58 degrees.
53
Using Symmetry to Find the Area
to the Left of the Mean
Because of symmetry, these areas are equal.
Figure 5-7
(a)
(b)
0.4925
0.4925
0
z = - 2.43
0
Equal distance away from 0
z = 2.43
NOTE: Although a z score can be negative, the area
under the curve (or the corresponding probability)
can never be negative.
54
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water, and if one thermometer is randomly
selected, find the probability that it reads freezing water
between -2.43 degrees and 0 degrees.
Area = 0.4925
P ( -2.43 < x < 0 ) = 0.4925
-2.43
0
The probability that the chosen thermometer
will measure freezing water between -2.43
and 0 degrees is 0.4925.
55
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
56
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
68% within
1 standard deviation
34%
x-s
34%
x
x+s
57
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
13.5%
x - 2s
13.5%
x-s
x
x+s
x + 2s
58
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x - 3s
x - 2s
13.5%
x-s
x
x+s
x + 2s
x + 3s
59
Probability of Half of a Distribution
0.5
0
60
Finding the Area to the Right of z = 1.27
Value found
in Table A-2
0.3980
0
This area is
0.5 - 0.3980 = 0.1020
z = 1.27
61
Finding the Area Between z = 1.20 and z = 2.30
0.4893 (from Table A-2 with z = 2.30)
Area A is 0.4893 - 0.3849 =
0.1044
0.3849
A
0
z = 1.20 z = 2.30
62
Notation
P(a < z < b)
denotes the probability that the z score is between a
and b
P(z > a)
denotes the probability that the z score is greater
than a
P (z < a)
denotes the probability that the z score is less
than a
63
Figure 5-10
Interpreting Area Correctly
64
Interpreting Area Correctly
‘greater than
‘at least
x’
x’
‘more than
Subtract
from
0.5
Add to
0.5
x’
‘not less than
x’
0.5
x
x
65
Interpreting Area Correctly
‘greater than
‘at least
x’
Add to
0.5
x’
‘more than
Subtract
from
0.5
x’
‘not less than
x’
0.5
x
Add to
0.5
x
‘less than
‘at most
x’
x’
‘no more than
x’
‘not greater than
Subtract
from
0.5
x’
0.5
x
x
66
Interpreting Area Correctly
‘greater than
‘at least
x’
Add to
0.5
x’
‘more than
Subtract
from
0.5
x’
‘not less than
x’
0.5
x
Add to
0.5
x
‘less than
‘at most
x’
x’
‘no more than
x’
‘not greater than
Subtract
from
0.5
x’
0.5
x
x
Add
C
‘between
x1
and
Use
A=C-B
x2’
A
x1
x2
B
x1 x2
67
Finding a z - score when given a probability
Using Table A-2
1. Draw a bell-shaped curve, draw the centerline, and
identify the region under the curve that corresponds to
the given probability. If that region is not bounded by
the centerline, work with a known region that is
bounded by the centerline.
2. Using the probability representing the area bounded by
the centerline, locate the closest probability in the body
of Table A-2 and identify the corresponding z score.
3. If the z score is positioned to the left of the centerline,
make it a negative.
68
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
z
0
( z score will be positive )
Finding the 95th Percentile
69
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
Finding the 95th Percentile
70
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
z
0
(z score will be negative)
Finding the 10th Percentile
71
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
Finding the 10th Percentile
72
Other Normal Distributions
0
  1
If
or
(or both), we will convert
values to standard scores using Formula 6-7,
then procedures for working with all normal
distributions are the same as those for the
standard normal distribution.
z=
x-µ

73
Converting to Standard Normal
Distribution
P
(a)

x
74
Converting to Standard Normal
Distribution
x-
z=

P
P
(a)

x
(b)
0
z
75
Probability of Weight between 143
pounds and 201 pounds
z=
x = 143
σ = 29
143
201
201 - 143
29
= 2.00
Weight
z
0
2.00
76
Probability of Weight between 143
pounds and 201 pounds
Value found
in Table A-2
x = 143
σ = 29
143
201
Weight
z
0
2.00
77
Probability of Weight between 143
pounds and 201 pounds
0.4772
x = 143
σ = 29
143
201
Weight
z
0
2.00
78
Probability of Weight between 143
pounds and 201 pounds
There is a 0.4772 probability
of randomly selecting a
woman with a weight between
143 and 201 lbs.
x = 143
σ = 29
143
201
Weight
z
0
2.00
79
Probability of Weight between 143
pounds and 201 pounds
OR - 47.72% of women have
weights between 143 lb and
201 lb.
x = 143
σ = 29
143
201
Weight
z
0
2.00
80
Cautions to keep in mind
1. Don’t confuse z scores and areas.
Z scores are distances along the horizontal scale,
but areas are regions under the normal curve.
Table A-2 lists z scores in the left column and
across the top row, but areas are found in the
body of the table.
2. Choose the correct (right/left) side of the graph.
3. A z score must be negative whenever it is located
to the left of the centerline of 0.
81
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
Finding the 95th Percentile
82
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
Finding the 10th Percentile
83
Procedure for Finding Values
Using Table A-2 and Formula 6-7
1. Sketch a normal distribution curve, enter the given probability or
percentage in the appropriate region of the graph, and identify the
value(s) being sought.
x
2. Use Table A-2 to find the z score corresponding to the region bounded
by x and the centerline of 0. Cautions:

Refer to the BODY of Table A-2 to find the closest area, then identify
the corresponding z score.

Make the z score negative if it is located to the left of the centerline.
3. Using Formula 5-2, enter the values for µ, , and the z score found in
step 2, then solve for x.
x = µ + (z • )
(Another form of Formula 6-7)
4. Refer to the sketch of the curve to verify that the solution makes sense
in the context of the graph and the context of the problem.
84
Finding P10 for Weights of Women
10%
90%
40%
x=?
143
50%
Weight
85
Finding P10 for Weights of Women
0.10
0.40
0.50
x=?
143
-1.28
0
Weight
86
Finding P10 for Weights of Women
x = 143 + (-1.28 • 29) = 105.88
0.10
0.40
0.50
x=?
143
-1.28
0
Weight
87
Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates
the lowest 10% from the highest 90%.
0.10
0.40
x = 106
-1.28
0.50
143
Weight
0
88
Forgot to make z score negative???
x = 143 + (1.28 • 29) = 180
0.10
0.40
0.50
x=?
143
1.28
0
Weight
89
Forgot to make z score negative???
x = 143 + (1.28 • 29) = 180
0.10
0.40
x = 180
1.28
0.50
143
Weight
0
90
Forgot to make z score negative???
UNREASONABLE ANSWER!
0.10
0.40
x = 180
1.28
0.50
143
Weight
0
91
REMEMBER!
Make the z score negative if the
value is located to the left (below)
the mean. Otherwise, the z score
will be positive.
92
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In
reality, the amount of cheese is normally distributed with a mean value of .5
lbs with a standard deviation of .025 lbs.
a.
What is the probability that the amount of cheese on a medium pizza is
between .525 and .550 lbs?
b. What is the probability that a medium pizza has less than .47 lbs of
cheese?
c. If the pizza has more cheese than 95% of the other pizzas, how many
pounds of cheese does it have?
a. z 
xμ
σ
z
.525  .5
1
.025
z
.550  .5
2
.025
.4772-.3413=.1359
93
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In
reality, the amount of cheese is normally distributed with a mean value of .5
lbs with a standard deviation of .025 lbs.
a.
What is the probability that the amount of cheese on a medium pizza is
between .525 and .550 lbs?
b. What is the probability that a medium pizza has less than .47 lbs of
cheese?
c. If the pizza has more cheese than 95% of the other pizzas, how many
pounds of cheese does it have?
b.
z
xμ
σ
z
.47  .5
 -1.2
.025
.5-.3849=.1150
94
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In
reality, the amount of cheese is normally distributed with a mean value of .5
lbs with a standard deviation of .025 lbs.
a.
What is the probability that the amount of cheese on a medium pizza is
between .525 and .550 lbs?
b. What is the probability that a medium pizza has less than .47 lbs of
cheese?
c. If the pizza has more cheese than 95% of the other pizzas, how many
pounds of cheese does it have?
c. z 
xμ
σ
z = 1.645
x = μ + zσ
x = .5 + (1.645)(.025)
x =.5411
95
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
a.
What is the probability that the mileage of a randomly selected car is
between 31.2 mi/gal and 33 mi/gal?
b. What is the probability that a car selected at random exceeds 32 mi/gal?
c. What is the miles per gallon if the car is in the bottom 20%?
a.
xμ
31.2  30
z
z
1
σ
1.2
z
33  30
 2.5
1.2
. 4938-.3413=.1525
96
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
a.
What is the probability that the mileage of a randomly selected car is
between 31.2 mi/gal and 33 mi/gal?
b. What is the probability that a car selected at random exceeds 32 mi/gal?
c. What is the miles per gallon if the car is in the bottom 20%?
b. z 
xμ
σ
z
32  30
 1.67
1.2
. 5 - .4525 = .0475
97
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
a.
What is the probability that the mileage of a randomly selected car is
between 31.2 mi/gal and 33 mi/gal?
b. What is the probability that a car selected at random exceeds 32 mi/gal?
c. What is the miles per gallon if the car is in the bottom 20%?
c. z 
xμ
σ
x = μ + zσ
x = 30 – (.84)(1.2) = 28.99
98