Chapter 7
Continuous Probability Distributions
1
Goals
1. Understand the difference between
discrete and continuous distributions
2. Compute the mean and the standard
deviation for a uniform distribution
3. Compute probabilities using the uniform
distribution
2
Goals
4. List the characteristics of the:
•
•
Normal probability distribution
Standard normal probability distribution
5. Define and calculate z values
6. Use the standard normal probability distribution
to find area:





Above the mean
Below the mean
Between two values
Above one value
Below one value
7. Use the normal distribution to approximate the
binomial probability distribution
3
Probability Distribution
 A listing of all the outcomes of an experiment
and the probability associated with each
outcome
 Probability distributions are useful for making
probability statements concerning the values of
a random variable
 Our goal is to find probability between two
values:
 Example: What is the probability that the daily water
usage will lie between 15 and 25 gallons? A: 68%
4
Probability Distribution
 Discrete Probability Distributions (Chapter 6)
 Based On Discrete Random Variables
 We looked at:
 Binomial Probability Distribution
 Continuous Probability Distribution (Chapter 7)
 Based On Continuous Random Variables:
 We will look at:
 Uniform Probability Distribution
 Normal Probability Distribution
5
Continuous Probability Distributions
 These continuous probability distributions
will be all about 
 Area!!!!
34.13
% area
34.13
% area
13.58
% area
13.58
% area
Possibilities within range
a-b
0.13%
area
2.15%
area
68.26% area
2.15%
area
95.44% area
0.13%
area
99.74% area
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
1400
1600
1800
0
μ
2000
1
μ +1 σ
2200
2
3
μ +2 σ μ +3 σ
2400
2600
z
x symbols
x in dollars
6
Continuous Probability Distribution:
 Uniform Probability Distribution
 Within the interval 15 to 25 minutes, the time it takes
to fill out a typical 1040EZ tax return at a VITA site
tends to follow a uniform distribution
 Random Variable is time (only possibilities within the
interval)
 Each value has same probability
 Normal Probability Distribution
 The weight distribution of a manufactured box of
cereal tends to follow a normal distribution
 Random Variable is box weight and will cover all possibilities
 Each value has different probability
7
Uniform Probability Distribution
1. Distributions shape is rectangular
2. Minimum value = a
3. Maximum value = b
 a and b imply a range
4. Height of the distribution is constant (uniform)
for all values between a and b
 Implies all values in range are equally likely
Minimum Value
Maximum Value
Mean of Uniform Distribution
Standard Deviation of Uniform Distribution
If a <= x <= b and 0 everywhere else
If a <= x <= b and 0 everywhere else
Area = Height * Base
=
=
=
=
=
=
=
a
b
(a+b)/2
(((b-a)^2)/12))^(1/2))
1/(b-a)
1/(b-a)
(1/(b-a))*(b-a)
= µ
= s
= P(x)
Height of Rectangle
=
1
8
Uniform Probability Distribution
P(x)
All Posibilities
1
b-a
Possibilities within range
a-b
Units
a
ab

2
b
< ------- b-a ------- >
(b  a)
s
12
1
P( x) 
, if a  x  b and 0 elsewhere
ba
1
Area  Height* Base 
(b  a)  1.00
(b  a)
9
2
Suppose the time
that you wait on the
telephone for a live
representative of
your phone company
to discuss your
problem with you is
uniformly distributed
between 5 and 25
minutes.
What is the mean wait time?
= a+b
2
= 5+25
= 15
2
What is the standard
deviation of the wait time?
(b-a)2
s=
12
2
(25-5)
=
= 5.77
12
10
What is the probability of waiting more than ten minutes?
The area from 10
to 25 minutes is
15 minutes.
Thus:
1
=
25-5
P(10 < wait time < 25) = height*base
=
1
(25-5) *15 = .75
1
20
P(x)
minutes
5
10
15
20
25
11
1
=
25-5
1
20
P(x)
minutes
5
10
15
20
25
What is the probability of waiting between 15 and 20
minutes?
The area from 15 P(15 < wait time < 20) = height*base
to 20 minutes is =
1
*5 = .25
5 minutes. Thus:
(25-5)
12
Normal Probability Distribution Is All About Area!
Total Area = 1.0
34.13
% area
34.13
% area
13.58
% area
13.58
% area
0.13%
area
2.15%
area
2.15%
area
μ + -3 σ μ + -2 σ μ + -1 σ
μ
μ +1 σ
μ +2 σ μ +3 σ
0.13%
area
x (values) symbols
68.26% area
95.44% area
99.74% area
13
Normal Probability Distribution Formula

1
P( x) 
e
s 2
( x )2
2s 2
Awesome!
No, that’s o.k., we can use Appendix or Excel functions!
x
s
µ
1
P( x ) 
e
s 2
=
=
=

( x )
2s 2
2
=
250
2
247
0.0648
1.5
0.064759
0.933193
250
0.933193
1.5
=(I3-I5)/I4
=NORMDIST(I3,I5,I4,FALSE)
=NORMDIST(I3,I5,I4,TRUE)
=NORMINV(K8,I5,I4)
=NORMSDIST((I3-I5)/I4)
=NORMSINV(K10)
14
Characteristics of a Normal
Probability Distribution
(And Accompanying Normal Curve)
 The normal curve is bell-shaped and has a single
peak at the exact center of the distribution
 The arithmetic mean, median, and mode of the
distribution are equal and located at the peak
 Thus half the area under the curve is above the mean
and half is below it
 The normal probability distribution is symmetrical
about its mean
 If we cut the normal curve vertically at this center
value, the two halves will be mirror images
15
Characteristics of a Normal
Probability Distribution
(And Accompanying Normal Curve)
 The normal probability distribution is asymptotic
 The curve gets closer and closer to the X-axis but
never actually touches it
 The “tails” of the curve extend indefinitely in both
directions
 The Location of a normal distribution is
determined by mean µ
 The dispersion of a normal distribution is
determined by the standard deviation s
Now Let’s Look At Some Pictures That Will Show
Relationships Amongst Various Means & Standard Deviations
16
Characteristics of a Normal Distribution
Normal
curve is
symmetrical
Mean, median, and
mode are equal
Theoretically,
curve
extends to
infinity
There Is A Family Of Normal Probability Distributions
17
Frequency
Frequency
Which One Is Normal?
Years
Frequency
Frequency
Years
Years
Years
18
Equal Means, Unequal Standard Deviations
σ = 3.1 years
μ = 20 years Length Of Service
19
Equal Means, Unequal Standard Deviations
σ = 3.1 years
σ = 3.9 years
μ = 20 years Length Of Service
20
Equal Means, Unequal Standard Deviations
σ = 3.1 years
σ = 3.9 years
σ = 5.0 years
μ = 20 years Length Of Service
21
Unequal Means, Equal Standard Deviations
σ = 1.6 Grams
μ = 283 Grams of
Super Yummy Cereal
σ = 1.6 Grams
μ = 301 Grams of
Super Neat Cereal
σ = 1.6
Grams
μ = 321 Grams of
Super Rad Cereal
22
Unequal Means, Unequal Standard Deviations
σ = 26 psi
σ = 41 psi
σ = 52 psi
μ = 2000 psi
μ = 2107 psi
μ = 2186 psi
This Family Of Normal Probability Distributions
Is Unlimited In Number!
Luckily, One Of The Family Members May Be Used In All
Circumstances Where The Normal Distribution Is Applicable
Standard Normal Distribution
23
Standard Normal Probability Distribution
 The standard normal distribution (z distribution )
is a normal distribution with a mean of 0 and a
standard deviation of 1
 The percentage of area between two z-scores in
any normal distribution is the same!
 Standard deviation & terms may be different, but area
will be the same!
 Normal distributions can be converted to the
standard normal distribution using z-values…
24
Define And Calculate z-values
 Any normal distribution can be converted, or “standardized”
to the standard normal distribution using z-values
 Z-values:
 Distance from the mean, measured in units of standard
deviation
X 
 The Formula Is:
z
Z-values are also called:
Standard normal value
Z score
Z statistic
Standard normal deviate
Normal deviate
z
x
μ
σ

s
Define Variables
= z-score
= particular value
= mean
= standard deviation
Remember Your Algebra
So That You Can Solve For
Any One Of The Variables
25
Standard Normal Probability Distribution
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
1400
1600
1800
0
μ
2000
1
μ +1 σ
2200
2
3
μ +2 σ μ +3 σ
2400
2600
z
x symbols
x in dollars
0 means that there is no deviation from the mean!
26
Convert Value From A Normal
Distribution To A Z-score Example 1
 The bi-monthly starting salaries of recent MBA
graduates follows the normal distribution with a
mean of $2,000 and a standard deviation of
$200
 What is the z-value for a salary of $2,200?
z 
X 
s
$2, 200  $2, 000

 1.00
$200
27
Convert Value From A Normal Distribution
To A Z-score, Example 2
 What is the z-value of $1,700?
X 
z 
s
$1, 700  $2, 000

 1.50
$200
 A z-value of 1 indicates that the value of $2,200 is one
standard deviation above the mean of $2,000 (2000 +
1*200)
 A z-value of –1.50 indicates that $1,700 is 1.5 standard
deviation below the mean of $2000 (2000 – 1.5*200)
Now we can look at a graph 
28
What is the probability that a
foreman’s salary will fall between
1,700 and $2,200?
29
But It Is Really All About Area Under The Curve
Remember:
 Areas Under the Normal Curve
• Empirical Rule (Normal Rule):
• About 68% of the observations will lie within 1 σ of
the mean
• About 95% of the observations will lie within 2 σ of
the mean
• Virtually all the observations will be within 3 σ of
the mean
Hints: Many statistical chores can be solved with this normal curve
Nevertheless, “The whole world does not fit into a normal curve”
30
Empirical Rule
34.13
% area
34.13
% area
13.58
% area
13.58
% area
0.13%
area
2.15%
area
(Normal Rule):
68.26% area
2.15%
area
95.44% area
0.13%
area
99.74% area
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
1400
1600
1800
0
μ
2000
1
μ +1 σ
2200
2
3
μ +2 σ μ +3 σ
2400
2600
z
x symbols
x in dollars
Between what two values do about 95% of the values occur?
What if you want to find the % of values that lie between z-scores 0 and 1.56?
31
Table In Appendix Or On Inside Cover
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
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
0.01
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
0.4987
Areas Under the Normal Curve
Example z = 1.96, then P(0 to z) = 0.4750
0.02
0.03
0.04
0.05
0.06
0.0080
0.0120
0.0160
0.0199
0.0239
0.0478
0.0517
0.0557
0.0596
0.0636
0.0871
0.0910
0.0948
0.0987
0.1026
0.1255
0.1293
0.1331
0.1368
0.1406
0.1628
0.1664
0.1700
0.1736
0.1772
0.1985
0.2019
0.2054
0.2088
0.2123
0.2324
0.2357
0.2389
0.2422
0.2454
0.2642
0.2673
0.2704
0.2734
0.2764
0.2939
0.2967
0.2995
0.3023
0.3051
0.3212
0.3238
0.3264
0.3289
0.3315
0.3461
0.3485
0.3508
0.3531
0.3554
0.3686
0.3708
0.3729
0.3749
0.3770
0.3888
0.3907
0.3925
0.3944
0.3962
0.4066
0.4082
0.4099
0.4115
0.4131
0.4222
0.4236
0.4251
0.4265
0.4279
0.4357
0.4370
0.4382
0.4394
0.4406
0.4474
0.4484
0.4495
0.4505
0.4515
0.4573
0.4582
0.4591
0.4599
0.4608
0.4656
0.4664
0.4671
0.4678
0.4686
0.4726
0.4732
0.4738
0.4744
0.4750
0.4783
0.4788
0.4793
0.4798
0.4803
0.4830
0.4834
0.4838
0.4842
0.4846
0.4868
0.4871
0.4875
0.4878
0.4881
0.4898
0.4901
0.4904
0.4906
0.4909
0.4922
0.4925
0.4927
0.4929
0.4931
0.4941
0.4943
0.4945
0.4946
0.4948
0.4956
0.4957
0.4959
0.4960
0.4961
0.4967
0.4968
0.4969
0.4970
0.4971
0.4976
0.4977
0.4977
0.4978
0.4979
0.4982
0.4983
0.4984
0.4984
0.4985
0.4987
0.4988
0.4988
0.4989
0.4989
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
0.08
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
0.09
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
0.4990
32
Use The Standard Normal Probability
Distribution To Find Area:
(Table On Inside Back Cover)
Computed
z-score
2.43
0.10
0.43
0.15
2.00
1.96
0.01
0.70
Area under
curve
.475
z
0
1.96
33
Example 1
 The daily water usage per person in New
Providence, New Jersey is normally
distributed
 Mean = 20 gallons
 Standard deviation = 5 gallons
 About 68% of those living in New
Providence will use how many gallons of
water?
 +/- 1 standard deviation will give us:
 About 68% of the daily water usage will lie
between 15 and 25 gallons
35
Example 2
 What is the probability that a person from
New Providence selected at random will
use between 20 and 24 gallons per day?
z 
z 
X 
s
X 
s
20  20

 0.00
5

24  20
 0.80
5
36
Use The Table In The Back Of The
Book And Look Up .80
 The area under a normal curve between a zvalue of 0 and a z-value of 0.80 is 0.2881
 We conclude that 28.81 percent of the residents
use between 20 and 24 gallons of water per day
 See the following diagram:
37
r
. 4
0
. 3
0
. 2
0
. 1
l
i
t r
b
u
i o
n
:

=
0
,
Area =.2881
“28.81% of the
residents use
between 20 and
24 gallons of
water per day”
f ( x
0
a
. 0
- 5
-4
-3 -2 -1
x
0
1
2
3
4
z
38
Example 3
 What percent of the population use
between 18 and 26 gallons per day?
z 
z 
X 
s
X 
s
18  20

 0.40
5
26  20

 1.20
5
39
Example 3
 The area associated with a z-value of –0.40
is .1554
 Because the curve is symmetrical. look up .40 on
the right
 The area associated with a z-value of 1.20 is
.3849
 .1554 + .3849 = .5403
 We conclude that 54.03 percent of the
residents use between 18 and 26 gallons of
water per day
40
Example 4
 Professor Mann has determined that the scores
in his statistics course are approximately
normally distributed with a mean of 72 and a
standard deviation of 5
 He announces to the class that the top 15
percent of the scores will earn an A
 What is the lowest score a student can earn and
still receive an A?
 .50 - .15 = .35  This is the area under the
curve!
 You must look into table and find the value closest to .35
41
Table In Appendix Or On Inside Cover
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
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
0.01
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
Areas Under the Normal Curve
Example z = 1.96, then P(0 to z) = 0.4750
0.02
0.03
0.04
0.05
0.06
0.0080
0.0120
0.0160
0.0199
0.0239
0.0478
0.0517
0.0557
0.0596
0.0636
0.0871
0.0910
0.0948
0.0987
0.1026
0.1255
0.1293
0.1331
0.1368
0.1406
0.1628
0.1664
0.1700
0.1736
0.1772
0.1985
0.2019
0.2054
0.2088
0.2123
0.2324
0.2357
0.2389
0.2422
0.2454
0.2642
0.2673
0.2704
0.2734
0.2764
0.2939
0.2967
0.2995
0.3023
0.3051
0.3212
0.3238
0.3264
0.3289
0.3315
0.3461
0.3485
0.3508
0.3531
0.3554
0.3686
0.3708
0.3729
0.3749
0.3770
0.3888
0.3907
0.3925
0.3944
0.3962
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
0.08
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
0.09
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
42
Solve for X, The Score You Need To
Get An A
X  72
1.04 
5
X  1.04 *5  72  77.2
 The result is the score that separates students
that earned an A from those that earned a B
 Those with a score of 77.2 or more earn an A
43
Example 5
The weekly incomes of shift foreman
in the glass industry are normally distributed
μ
σ
=
=
1000
100
=
mean weekly income of shift foreman in the glass industry
What is the
probability of
selecting a shift
foreman
whose salary is
between $790
& $1200?
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
700
800
900
0
μ
1000
1
μ +1 σ
1100
2
3
μ +2 σ μ +3 σ
1200
1300
z
x symbols
x in dollars
44
Example 5
 Find Z-scores
X 
790  1000
z 

 2.10
s
100
X 
1200  1000
z 

 2
s
100
 Look up area under the curve in the tables
45
Table In Appendix Or On Inside Cover
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
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
0.01
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
Areas Under the Normal Curve
Example z = 1.96, then P(0 to z) = 0.4750
0.02
0.03
0.04
0.05
0.06
0.0080
0.0120
0.0160
0.0199
0.0239
0.0478
0.0517
0.0557
0.0596
0.0636
0.0871
0.0910
0.0948
0.0987
0.1026
0.1255
0.1293
0.1331
0.1368
0.1406
0.1628
0.1664
0.1700
0.1736
0.1772
0.1985
0.2019
0.2054
0.2088
0.2123
0.2324
0.2357
0.2389
0.2422
0.2454
0.2642
0.2673
0.2704
0.2734
0.2764
0.2939
0.2967
0.2995
0.3023
0.3051
0.3212
0.3238
0.3264
0.3289
0.3315
0.3461
0.3485
0.3508
0.3531
0.3554
0.3686
0.3708
0.3729
0.3749
0.3770
0.3888
0.3907
0.3925
0.3944
0.3962
0.4066
0.4082
0.4099
0.4115
0.4131
0.4222
0.4236
0.4251
0.4265
0.4279
0.4357
0.4370
0.4382
0.4394
0.4406
0.4474
0.4484
0.4495
0.4505
0.4515
0.4573
0.4582
0.4591
0.4599
0.4608
0.4656
0.4664
0.4671
0.4678
0.4686
0.4726
0.4732
0.4738
0.4744
0.4750
0.4783
0.4788
0.4793
0.4798
0.4803
0.4830
0.4834
0.4838
0.4842
0.4846
0.4868
0.4871
0.4875
0.4878
0.4881
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
0.488446
0.08
0.031
0.071
0.110
0.148
0.184
0.219
0.251
0.282
0.310
0.336
0.359
0.381
0.399
0.416
0.430
0.442
0.453
0.462
0.469
0.476
0.481
0.485
0.488
Example 5
 The probability of selecting a shift foreman
whose salary is between $790 & $1200 is:
 .4772 + .4821 = .9593
47
Example 6
The weekly incomes of shift foreman
in the glass industry are normally distributed
μ
σ
=
=
1000
100
=
mean weekly income of shift foreman in the glass industry
What is the
probability of
selecting a shift
foreman
whose salary is
less than $790?
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
700
800
900
0
μ
1000
1
μ +1 σ
1100
2
3
μ +2 σ μ +3 σ
1200
1300
z
x symbols
x in dollars
48
Example 6
z 
X 
s
790  1000

 2.10
100
Look up the area
The area = .4821
.5 - .4821 = .0179
The probability of selecting a shift foreman
whose salary is less than $790 is .0179
49
50
Finding Area Under The Standard Normal
Distribution – It’s All About Area!
34.13
% area
34.13
% area
13.58
% area
13.58
% area
0.13%
area
2.15%
area
68.26% area
2.15%
area
95.44% area
0.13%
area
99.74% area
-3
-2
-1
μ + -3 σ μ + -2 σ μ + -1 σ
1400
1600
1800
0
μ
2000
1
μ +1 σ
2200
2
3
μ +2 σ μ +3 σ
2400
2600
z
x symbols
x in dollars51
Finding Area Under The Standard Normal
Distribution – Use Formulas & Tables
z 
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
X 
s
Areas Under the Normal Curve
Example z = 1.96, then P(0 to z) = 0.4750
0.02
0.03
0.04
0.05
0.06
0.0080
0.0120
0.0160
0.0199
0.0239
0.0478
0.0517
0.0557
0.0596
0.0636
0.0871
0.0910
0.0948
0.0987
0.1026
0.1255
0.1293
0.1331
0.1368
0.1406
0.1628
0.1664
0.1700
0.1736
0.1772
0.1985
0.2019
0.2054
0.2088
0.2123
0.2324
0.2357
0.2389
0.2422
0.2454
0.2642
0.2673
0.2704
0.2734
0.2764
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
52
Finding Area Under The Standard
Normal Distribution – Four Situations
1. If you wish to find the area between 0 and z (or
– z), then you can look up the value directly in
the table
2. If you wish to find the area beyond z (or –z),
then locate the probability of z in the table and
subtract it from .50
3. If you wish to find the area between two points
on different sides of the mean, determine the
z-values and add the corresponding areas
4. If you wish to find the area between two points
on the same side of the mean, determine the
z-values and subtract the smaller area from
the larger
53
Use The Normal Distribution To Approximate
The Binomial Probability Distribution


The normal distribution (a continuous
distribution) yields a good approximation
of the binomial distribution (a discrete
distribution) for large values of n
Let’s Remember the conditions that must
be met before we can run a binomial
experiment
54
For An Experiment To Be Binomial It Must
Satisfy The Following Conditions:
1.
A random variable (x)
counts the # of
successes in a fixed
number of trials (n)

2.
Trials make up the
experiment
Each trail must be
independent of the
previous trial

Outcome of one trial does
not affect the outcome of
any other trial
3. An outcome on each
trial of an experiment is
classified into one of
two mutually exclusive
categories:
1. Success
or
2. Failure
4. The probability of
success stays the same
for each trial (so does
the probability of failure)
55
Normal Approximating The Binomial
 The normal probability distribution is generally a
good approximation to the binomial probability
distribution when:
n
x

For Binomial Probability Distribution
= Fixed # of trials
= # of successes we want
= "pi" = Probability of success of each trial
Requirements for normal proabability distribution as an
aproximation for a binomial probability distribution
n*
>
5
n*(1-
>
5
56
Mean & Variance of the Binomial
Distribution
n

μ
σ
For Binomial Probability Distribution
= Fixed # of trials
= Probability of success of each trial
= Mean of binomial distribution
= Standard deviation of binomial distribution
  n
s  s  n (1   )
2
Empirical experiments have shown these to be acceptable estimates
57
Continuity Correction Factor

The continuity correction factor of .5 is used
to extend the continuous value of x one-half
unit in either direction
The correction compensates for estimating a
discrete distribution by a continuous
distribution
When you use discrete numbers, you have
“gaps” – you need to take an average that will
yield a number between



We will simply estimate by adding or subtracting
the value .5
58
Continuity Correction Factor
1. For the probability at least x occur, use the
area above (x - .5)
2. For the probability that more than x occur,
use the area above (x + .5)
3. For the probability that x or fewer occur, use
the area below (x + .5)
4. For the probability that fewer than x occur,
use the area below (x - .5)
59
Example
A recent study by a marketing research firm showed that
15% of American households owned a video camera.
For a sample of 200 homes, What is the probability that
less than 40 homes in the sample have video cameras?
Step 1: Binomial?
1. Fixed Trails = yes, n = 200
2. Independent = yes
3. S/F  Success = have video camera, Failure = don’t have
4.  Constant = .15
Example
Step 2 : Can we approximate binomial distribution with
a standard normal distribution?
  n  (.15)(200)  30  5
n(1   )  (.85)(200)  170  5
Step 3 : Calculate μ & σ for the binomial distribution
  n  (.15)(200)  30
s  n (1   )  (30)(1  .15)  5.0498
2
Step 4: Calculate z-value (.5 correction)
What is the probability that less than 40 homes in
the sample have video cameras?
 We use the correction factor, so X is 39.5
z
 The
X 
s
39.5  30.0

 1.88
5.0498
value of z is 1.88
Step 5: Look Up Area
 From
Appendix the area between 0 and 1.88 on the z
scale is .4699
 So
the area to the left of 1.88 is .5000 + .4699 = .9699
 The
likelihood that less than 40 of the 200 homes have a
video camera is about 97%
r
a
l
i
t r
b
u
i o
n
:

=
EXAMPLE 5
0
,
s2
=
1
Area = .5000+.4699 =.9699
. 4
0
. 3
0
. 2
0
. 1
f ( x
0
. 0
- 5
0
1
2
z=1.88
3
4
z
Summarize Chapter 7
1. Understand the difference between
discrete and continuous distributions
2. Compute the mean and the standard
deviation for a uniform distribution
3. Compute probabilities using the uniform
distribution
4. List the characteristics of the:
•
•
Normal probability distribution
Standard normal probability distribution
65
Summarize Chapter 7
1. Define and calculate z values
2. Use the standard normal probability
distribution to find area:





Above the mean
Below the mean
Between two values
Above one value
Below one value
3. Use the normal distribution to
approximate the binomial probability
distribution
66