Download Probability Distributions - NYU Stern School of Business

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Probability Distributions /1
Source:
New York University
Stern School of Business
ELEMENTARY STATISTICS: A Brief
Version
Allan G. Bluman
McGraw-Hill, 2000
ISBN: 0-07-237288-5
Probability Distributions
Prof. Ian Giddy
New York University
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 2
Outline
6-1
6-2l 6-1 Introduction
6-2 Probability Distributions
l 6-3 Mean, Variance, and
Expectation
l 6-4 The Binomial Distribution
l
Chapter 6
Probability
Distributions
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 3
Objectives
l
random variable.
Find the mean, variance, and expected
value for a discrete random variable.
Find the exact probability for X successes in
n trials of a binomial experiment.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 4
Objectives
6-3l Construct a probability distribution for a
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 5
6-4l Find the mean, variance, and standard
deviation for the variable of a binomial
distribution.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 6
Giddy/NYU
Probability Distributions /2
6-2 Probability Distributions
6-2 Probability Distributions
6-5
6-6
If a variable can assume only a specific
number of values, such as the outcomes
for the roll of a die or the outcomes for the
toss of a coin, then the variable is called a
discrete variable.
variable
l Discrete variables have values that can be
counted.
lA
variable is defined as a
characteristic or attribute that can
assume different values.
l A variable whose values are
determined by chance is called a
random variable.
variable
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 7
6-2 Probability Distributions
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 8
6-2 Probability Distributions -
6-7
6-8
H
If a variable can assume all values in the
interval between two given values then the
variable is called a continuous variable.
Example - temperature between 680 to
780.
l Continuous random variables are obtained
from data that can be measured rather
than counted.
l
Copyright ©1999 Ian H. Giddy
6-2 Probability Distributions Coins
Probability Distributionbs 9
Tossing Two
6-9
Tossing Two Coins
H
T
Second Toss
H
T
First Toss
T
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 10
6-2 Probability Distributions Coins
Tossing Two
6-10
Sample Space
l From
the tree diagram, the sample
space will be represented by HH,
HT, TH, TT.
l If X is the random variable for the
number of heads, then X assumes
the value 0, 1, or 2.
2
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 11
TT
Number of Heads
0
TH
1
HT
HH
Copyright ©1999 Ian H. Giddy
2
Probability Distributionbs 12
Giddy/NYU
Probability Distributions /3
6-2 Probability Distributions Coins
Tossing Two
6-2 Probability Distributions
6-11
6-12
OUTCOME
X
0
PROBABILITY
P(X)
1/4
1
2/4
2
1/4
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 13
6-2 Probability Distributions -Graphical Representation
l
A probability distribution consists of the
values a random variable can assume
and the corresponding probabilities of
the values. The probabilities are
determined theoretically or by
observation.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 14
6-3 Mean, Variance, and Expectation for Discrete
Variable
6-13
6-14
The mean of the random variable of a
probability distribution is
µ = X ⋅ P ( X ) + X ⋅ P ( X ) + ... + X ⋅ P( X )
= ∑ X ⋅ P( X )
where X , X ,..., X are the outcomes and
P( X ), P ( X ), ... , P( X ) are the corresponding
probabilities.
Experiment: Toss Two Coins
PROBABILITY
1
1
0.5
.25
1
1
0
1
2
1
3
NUMBER OF HEADS
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 15
2
2
2
2
n
n
n
n
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 16
6-3 Mean for Discrete Variable Example
6-3 Mean for Discrete Variable Example
6-15
6-16
l
Find the mean of the number of spots that
appear when a die is tossed. The
probability distribution is given below.
XX
11
22
33
44
55
µ = ∑ X ⋅ P( X )
= 1 ⋅ (1 / 6) + 2 ⋅ (1 / 6) + 3 ⋅ (1 / 6) + 4 ⋅ (1 / 6)
+ 5 ⋅ (1 / 6) + 6 ⋅ (1 / 6)
.
= 21 / 6 = 35
66
That
That is,
is, when
when aa die
die isis tossed
tossed many
many times,
times,
the
the theoretical
theoretical mean
mean will
will be
be 3.5.
3.5.
P(X)
P(X) 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 17
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 18
Giddy/NYU
Probability Distributions /4
6-3 Mean for Discrete Variable Example
6-3 Mean for Discrete Variable Example
6-17
6-18
l
In a family with two children, find the mean
number of children who will be girls. The
probability distribution is given below.
µ = ∑ X ⋅ P( X )
= 0 ⋅ (1 / 4) + 1⋅ (1 / 2) + 2 ⋅ (1 / 4)
= 1.
XX
00
11
22
That
That is,
is, the
the average
average number
number of
of
girls
girls in
in aa two-child
two-child family
family isis 1.
1.
P(X)
P(X) 1/4
1/4 1/2
1/2 1/4
1/4
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 19
6-3 Formula for the Variance of a
Probability Distribution
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 20
6-3 Formula for the Variance of a
Probability Distribution
6-19
6-20
l
The formula for the variance of a
probability distribution is
The variance of a probability distribution is
found by multiplying the square of each
outcome by its corresponding probability,
summing these products, and subtracting
the square of the mean.
σ = ∑ [ X ⋅ P ( X )] − µ .
2
2
2
The standard deviation of a
probability distribution is
σ= σ .
2
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 21
6-3 Variance of a Probability
Distribution - Example
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 22
6-3 Variance of a Probability
Distribution - Example
6-21
6-22
l
Copyright ©1999 Ian H. Giddy
The probability that 0, 1, 2, 3, or 4
people will be placed on hold when they
call a radio talk show with four phone
lines is shown in the distribution below.
Find the variance and standard
deviation for the data.
Probability Distributionbs 23
X
0
1
2
3
4
P(X) 0.18 0.34 0.23 0.21 0.04
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 24
Giddy/NYU
Probability Distributions /5
6-3 Variance of a Probability
Distribution - Example
6-3 Variance of a Probability
Distribution - Example
6-23
6-24
X⋅ P(X) X2⋅ P(X)
X
P(X)
0
0.18
0
0
1
0.34
0.34
0.34
2
0.23
0.46
0.92
3
0.21
0.63
1.89
l
4
0.04
0.16
0.64
l
l
σσ22 ==
2
3.79
3.79 –– 1.59
1.592
== 1.26
1.26
l
µ = 1.59 ΣX2⋅ P(X)
=3.79
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 25
6-3 Expectation
Now, µ = (0)(0.18) + (1)(0.34) + (2)(0.23) +
(3)(0.21) + (4)(0.04) = 1.59.
2
Σ X P(X) = (02)(0.18) + (12)(0.34) +
(22)(0.23) + (32)(0.21) + (42)(0.04) = 3.79
1.592 = 2.53 (rounded to two decimal places).
σ 2 = 3.79 – 2.53 = 1.26
σ=
= 1.12
1.26
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 26
6-3 Expectation - Example
6-25
6-26
The expected value of a discrete
l
random variable of a probability
distribution is the theoretical average
of the variable. The formula is
µ = E ( X ) = ∑ X ⋅ P( X )
The symbol E ( X ) is used for the
expected value.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 27
6-3 Expectation - Example
A ski resort loses $70,000 per season
when it does not snow very much and
makes $250,000 when it snows a lot.
The probability of it snowing at least 75
inches (i.e., a good season) is 40%.
Find the expected profit.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 28
6-4 The Binomial Distribution
6-27
6-28
Profit, X 250,000
P(X)
l
Copyright ©1999 Ian H. Giddy
0.40
–70,000
0.60
The expected profit = ($250,000)(0.40)
+ (–$70,000)(0.60) = $58,000.
Probability Distributionbs 29
A binomial experiment is a probability
experiment that satisfies the following four
requirements:
l Each trial can have only two outcomes or
outcomes that can be reduced to two
outcomes. Each outcome can be
considered as either a success or
a failure.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 30
Giddy/NYU
Probability Distributions /6
6-4 The Binomial Distribution
6-4 The Binomial Distribution
6-29
6-30
l
There must be a fixed number of trials.
l The outcomes of each trial must be
independent of each other.
l The probability of success must remain the
same for each trial.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 31
6-4 The Binomial Distribution
The outcomes of a binomial experiment
and the corresponding probabilities of
these outcomes are called a binomial
distribution.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 32
6-4 Binomial Probability Formula
6-31
6-32
Notation for the Binomial Distribution:
P(S) = p, probability of a success
l P(F) = 1 – p = q, probability of a failure
l n = number of trials
l X = number of successes.
l
In a binomial experiment, the probability of
exactly X successes in n trials is
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 33
6-4 Binomial Probability - Example
P( X ) =
n!
p Xq n − X
(n − X )! X !
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 34
6-4 Binomial Probability - Example
6-33
6-34
If a student randomly guesses at five
multiple-choice questions, find the
probability that the student gets exactly
three correct. Each question has five
possible choices.
l Solution: n = 5, X = 3, and p = 1/5. Then,
P(3) = [5!/((5 – 3)!3! )](1/5)3(4/5)2 0.05.
l
l
A survey from Teenage Research
Unlimited (Northbrook, Illinois.) found that
30% of teenage consumers received their
spending money from part-time jobs. If
five teenagers are selected at random, find
the probability that at least three of them
will have part-time jobs.
≈
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 35
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 36
Giddy/NYU
Probability Distributions /7
6-4 Binomial Probability - Example
6-4 Binomial Probability - Example
6-35
6-36
l Solution:
l
n = 5, X = 3, 4, and 5, and p
= 0.3.
Then, P(X ≥ 3) = P(3) + P(4) + P(5) =
0.1323 + 0.0284 + 0.0024 = 0.1631.
l NOTE: You can use Table B in the
textbook to find the Binomial
probabilities as well.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 37
A report from the Secretary of Health and
Human Services stated that 70% of singlevehicle traffic fatalities that occur on weekend
nights involve an intoxicated driver. If a
sample of 15 single-vehicle traffic fatalities
that occurred on a weekend night is selected,
find the probability that exactly 12 involve a
driver who is intoxicated.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 38
6-4 Mean, Variance, Standard Deviation for the Binomial
Distribution - Example
6-4 Binomial Probability - Example
6-37
6-38
l Solution:
n = 15, X = 12, and
p = 0.7. From Table B,
P(X =12) = 0.170
l
l
l
l
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 39
A coin is tossed four times. Find the mean,
variance, and standard deviation of the
number of heads that will be obtained.
Solution: n = 4, p = 1/2, and q = 1/2.
µ = n⋅p = (4)(1/2) = 2.
σ 2 = n⋅p⋅q = (4)(1/2)(1/2) = 1.
σ = = 1.
1
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 40
Outline
7-1
7-2
7-1 Introduction
l 7-2 Properties of the Normal
Distribution
l 7-3 The Standard Normal
Distribution
l 7-4 Applications of the Normal
Distribution
l
Chapter 7
The Normal
Distribution
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 41
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 42
Giddy/NYU
Probability Distributions /8
Outline
Objectives
7-3
l 7-5
The Central Limit Theorem
l 7-6 The Normal Approximation to the
Binomial Distribution
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 43
Objectives
7-4
Identify distributions as symmetric or
skewed.
l Identify the properties of the normal
distribution.
l Find the area under the standard
normal distribution given various z
values.
l
Copyright ©1999 Ian H. Giddy
Objectives
7-5
7-6
l
l
Find probabilities for a normally
distributed variable by transforming it
into a standard normal variable.
l Find specific data values for given
percentages using the standard normal
distribution.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 44
Probability Distributionbs 45
7-2 Properties of the Normal
Distribution
Use the Central Limit Theorem to solve
problems involving sample means.
l Use the normal approximation to
compute probabilities for a binomial
variable.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 46
7-2 Mathematical Equation for the
Normal Distribution
7-7
7-8
The mathematical equation for the normal distribution:
l
l
Copyright ©1999 Ian H. Giddy
Many continuous variables have distributions
that are bell-shaped and are called
approximately normally distributed variables.
The theoretical curve, called the normal
distribution curve,
curve can be used to study many
variables that are not normally distributed but
are approximately normal.
Probability Distributionbs 47
y=
e
−( x−µ )2 2σ
2
σ 2π
where
e ≈ 2.718
π ≈ 314
.
µ = population mean
σ = population standard deviation
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 48
Giddy/NYU
Probability Distributions /9
7-2 Properties of the Normal
Distribution
7-2 Properties of the
Theoretical Normal Distribution
7-9
7-10
The shape and position of the normal
distribution curve depend on two
parameters, the mean and the standard
deviation.
l Each normally distributed variable has its
own normal distribution curve, which
depends on the values of the variable’s
mean and standard deviation.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 49
7-2 Properties of the
Theoretical Normal Distribution
The normal distribution curve is
bell-shaped.
l The mean, median, and mode are equal
and located at the center of the
distribution.
l The normal distribution curve is
unimodal (single mode).
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 50
7-2 Properties of the
Theoretical Normal Distribution
7-11
7-12
l
The curve is symmetrical about the
mean.
l The curve is continuous.
l The curve never touches the x-axis.
l The total area under the normal
distribution curve is equal to 1.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 51
The area under the normal curve that lies
within
one standard deviation of the mean is
approximately 0.68 (68%).
ü two standard deviations of the mean is
approximately 0.95 (95%).
ü three standard deviations of the mean is
approximately 0.997 (99.7%).
ü
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 52
7-3 The Standard Normal
Distribution
7-2 Areas Under the Normal Curve
7-13
7-14
The standard normal distribution is a
normal distribution with a mean of 0 and a
standard deviation of 1.
l All normally distributed variables can be
transformed into the standard normally
distributed variable by using the formula
for the standard score:
(see next slide)
l
68%
95%
µ −3σ
−3σ
Copyright ©1999 Ian H. Giddy
99.7%
µ −2σ
−2σ µ −1σ
−1σ µ µ +1σ
+1σ µ +2σ
+2σ µ +3σ
+3σ
Probability Distributionbs 53
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 54
Giddy/NYU
Probability Distributions /10
7-3 The Standard Normal
Distribution
7-3 Area Under the Standard Normal Curve Example
7-15
7-16
z=
value − mean
standard deviation
Find the area under the standard
normal curve between z = 0 and
z = 2.34 ⇒ P(0 ≤ z ≤ 2.34).
2.34)
l Use your table at the end of the text to
find the area.
l The next slide shows the shaded area.
l
or
z=
X −µ
σ
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 55
7-3 Area Under the Standard
- Example
Normal Curve
7-17
Copyright ©1999 Ian H. Giddy
7- 3 Area Under the Standard
- Example
Probability Distributionbs 56
Normal Curve
7-18
Find the area under the standard normal
curve between z = 0 and
z = –1.75 ⇒ P(–1.75 ≤ z ≤ 0).
0)
l Use the symmetric property of the normal
distribution and your table at the end of the
text to find the area.
l The next slide shows the shaded area.
l
0.4904
0
2.34
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 57
7-3 Area Under the Standard
- Example
Normal Curve
7-19
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 58
7-3 Area Under the Standard Normal Curve Example
7-20
l Find
0.4599
−1.75
Copyright ©1999 Ian H. Giddy
the area to the right of z = 1.11
⇒ P(z > 1.11).
1.11)
l Use your table at the end of the text to
find the area.
l The next slide shows the shaded area.
0.4599
0
1.75
Probability Distributionbs 59
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 60
Giddy/NYU
Probability Distributions /11
7-3 Area Under the Standard
- Example
Normal Curve
7-21
7-3 Area Under the Standard
- Example
Normal Curve
7-22
Find the area to the left of z = –1.93
⇒ P(z < –1.93).
1.93)
l Use the symmetric property of the
normal distribution and your table at the
end of the text to find the area.
l The next slide shows the area.
l
0.5000
−0.3665
0.1335
0.3665
0
1.11
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 61
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 62
7-3 Area Under the Standard Normal Curve Example
7-23
7-3 Area Under the Standard
Curve - Example
Normal
7-24
Find the area between z = 2 and
z = 2.47 ⇒ P(2 ≤ z ≤ 2.47).
2.47)
l Use the symmetric property of the
normal distribution and your table at the
end of the text to find the area.
l The next slide shows the area.
l
0.5000
−0.4732
0.0268
0.0268
0.4732
−1.93
0
1.93
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 63
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 64
7-3 Area Under the Standard Normal Curve Example
7-3 Area Under the Standard Normal Curve - Example
7-25
7-26
Find the area between z = 1.68 and
z = –1.37 ⇒ P(–1.37 ≤ z ≤ 1.68).
1.68)
l Use the symmetric property of the
normal distribution and your table at the
end of the text to find the area.
l The next slide shows the area.
l
0.4932
−0.4772
0.4932
0.0160
0.4772
0
Copyright ©1999 Ian H. Giddy
2 2.47
Probability Distributionbs 65
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 66
Giddy/NYU
Probability Distributions /12
7-3 Area Under the Standard Normal Curve Example
7-3 Area Under the Standard Normal Curve - Example
7-27
7-28
0.4535
+0.4147
the area to the left of z = 1.99 ⇒
P(z < 1.99).
1.99)
l Use your table at the end of the text to
find the area.
l The next slide shows the area.
l Find
0.8682
0.4147
0.4535
−1.37
0
1.68
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 67
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 68
7-3 Area Under the Standard Normal Curve Example
7-29
7-3 Area Under the Standard Normal Curve - Example
7-30
l Find
the area to the right of
z = –1.16 ⇒ P(z > –1.16).
–1.16)
l Use your table at the end of the text to
find the area.
l The next slide shows the area.
0.5000
+0.4767
0.9767
0.5000
0.4767
0
1.99
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 69
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 70
RECALL: The Standard Normal
Distribution
7-3 Area Under the Standard Normal Curve - Example
7-31
7-32
z=
0.5000
+ 0.3770
value − mean
standard deviation
0.8770
or
0.377 0.5000
−1.16
Copyright ©1999 Ian H. Giddy
z=
0
Probability Distributionbs 71
Copyright ©1999 Ian H. Giddy
X −µ
σ
Probability Distributionbs 72
Giddy/NYU
Probability Distributions /13
7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
7-33
7-34
Each month, an American household
generates an average of 28 pounds of
newspaper for garbage or recycling.
Assume the standard deviation is 2
pounds. Assume the amount generated is
normally distributed.
l If a household is selected at random, find
the probability of its generating:
l
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 73
l
l
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 74
7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
7-35
7-36
Between 27 and 31 pounds per month.
First find the z-value for 27 and 31.
z1
= [X –µ]/σ = [27 – 28]/2 = –0.5;
z2
= [31 – 28]/2 = 1.5
l Thus, P(–0.5 ≤ z ≤ 1.5) = 0.1915 + 0.4332
= 0.6247.
l
0.5000
−0.3643
l
0.1357
0
1.1
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 75
7-4 Applications of the Normal
Distribution - Example
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 76
7-4 Applications of the Normal
Distribution - Example
7-37
7-38
l
0.4332
0.1915
0.1915
+ 0.4332
0.6247
0
−0.5
Copyright ©1999 Ian H. Giddy
More than 30.2 pounds per month.
First find the z-value for 30.2.
z =[X –µ]/σ = [30.2 – 28]/2 = 1.1.
Thus, P(z > 1.1) = 0.5 – 0.3643 = 0.1357.
That is, the probability that a randomly
selected household will generate more than
30.2 lbs. of newspapers is 0.1357 or 13.57%.
1.5
Probability Distributionbs 77
Copyright ©1999 Ian H. Giddy
The American Automobile Association reports
that the average time it takes to respond to an
emergency call is 25 minutes. Assume the
variable is approximately normally distributed
and the standard deviation is 4.5 minutes. If
80 calls are randomly selected, approximately
how many will be responded to in less than
15 minutes?
Probability Distributionbs 78
Giddy/NYU
Probability Distributions /14
7-4 Applications of the Normal
Distribution - Example
7-39
7-4 Applications of the Normal
Distribution - Example
7-40
First find the z-value for 15 is
z=
[X –µ]/σ = [15 – 25]/4.5 = –2.22.
l Thus, P(z < –2.22) = 0.5000 – 0.4868
= 0.0132.
l The number of calls that will be made in
less than 15 minutes = (80)(0.0132)
=
1.056 ≈ 1.
l
−2.22
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 79
7-4 Applications of the Normal
Distribution - Example
0.5000
− 0.4868
0.0132
0.0132
0
2.22
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 80
7- 4 Applications of the Normal
Distribution - Example
7-41
7-42
l
An exclusive college desires to accept only
the top 10% of all graduating seniors
based on the results of a national
placement test. This test has a mean of
500 and a standard deviation of 100. Find
the cutoff score for the exam. Assume the
variable is normally distributed.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 81
Work backward to solve this problem.
Subtract 0.1 (10%) from 0.5 to get the area
under the normal curve for accepted
students.
l Find the z value that corresponds to an
area of 0.4000 by looking up 0.4000 in the
area portion of Table E. Use the closest
value, 0.3997.
l
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 82
7-4 Applications of the Normal
Distribution - Example
7- 4 Applications of the Normal
Distribution - Example
7-43
7-44
X− µ
Substitute in the formula z =
σ
and solve for X.
l The z-value for the cutoff score (X) is z =
[X –µ]/σ = [X – 500]/100 = 1.28. (See next
slide).
l Thus, X = (1.28)(100) + 500 = 628.
l The score of 628 should be used as a
cutoff score.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 83
0.1
0.4
0
Copyright ©1999 Ian H. Giddy
X = 1.28
Probability Distributionbs 84
Giddy/NYU
Probability Distributions /15
7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
7-45
7-46
l
NOTE: To solve for X, use the following
formula: X = z⋅σ + µ.
l
l Example: For a medical study, a
researcher wishes to select people in the
middle 60% of the population based on
blood pressure. (Continued on the next
slide).
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 85
(Continued)-- If the mean systolic blood
pressure is 120 and the standard deviation
is 8, find the upper and lower readings that
would qualify people to participate in the
study.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 86
7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
7-47
7-48
(continued)
l
l
l
Note that two values are needed, one above the
mean and one below the mean. The closest z
values are 0.84 and – 0.84 respectively.
X = (z)(σ) + µ = (0.84)(8) + 120 = 126.72.
The other X = (–0.84)(8) + 120 = 113.28.
See next slide.
i.e. the middle 60% of BP readings is between
113.28 and 126.72.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 87
7-5 Distribution of Sample Means
0.2
−0.84
0.3 0.3
0
0.2
0.84
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 88
7-5 Distribution of Sample Means
7-49
7-50
l Distribution
of Sample means: A
sampling distribution of sample
means is a distribution obtained by
using the means computed from
random samples of a specific size
taken from a population.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 89
l Sampling
error is the difference
between the sample measure and the
corresponding population measure
due to the fact that the sample is not
a perfect representation of the
population.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 90
Giddy/NYU
Probability Distributions /16
7-5 Properties of the Distribution of
Sample Means
7-5 Properties of the Distribution of
Sample Means - Example
7-51
7-52
The mean of the sample means will be the
same as the population mean.
l The standard deviation of the sample
means will be smaller than the standard
deviation of the population, and it will be
equal to the population standard deviation
divided by the square root of the sample
size.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 91
Suppose a professor gave an 8-point quiz
to a small class of four students. The
results of the quiz were 2, 6, 4, and 8.
Assume the four students constitute the
population.
l The mean of the population is
µ = ( 2 + 6 + 4 + 8)/4 = 5.
5
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 92
7-5 Graph of the Original
7-53
7-5 Properties of the Distribution of
Sample Means - Example
l
l
l
7-54
The standard deviation of the population is
2
2
2
2

ó =2.236.
= { (2 − 5 ) + (6 − 5 ) + (4 − 5 ) + (8 − 5)  /4}

The graph
of the distribution of the scores is
uniform and is shown on the next slide.
Next we will consider all samples of size 2
taken with replacement.
Copyright ©1999 Ian H. Giddy
7-55
Probability Distributionbs 93
7-5 Properties of the Distribution of
Sample Means - Example
Copyright ©1999 Ian H. Giddy
Distribution
Copyright ©1999 Ian H. Giddy
7-56
Sample
Mean
Sample
Mean
2, 2
2
6, 2
4
2, 4
3
6, 4
5
2, 6
4
6, 6
6
2, 8
5
6, 8
7
4, 2
3
8, 2
5
4, 4
4
8, 4
6
4, 6
5
8, 6
7
4, 8
6
8, 8
8
Probability Distributionbs 95
Probability Distributionbs 94
7-5 Frequency Distribution of the
Sample Means - Example
X-bar 2
(mean)
f
1
Copyright ©1999 Ian H. Giddy
3
4
5
6
7
8
2
3
4
3
2
1
Probability Distributionbs 96
Giddy/NYU
Probability Distributions /17
7-5 Mean and Standard Deviation of
Means
7-5 Graph of the Sample Means
7-57
the Sample
7-58
D IS T R IB U T IO N O F S AM P LE M E AN S
(AP PR O XIM AT ELY N O R M AL)
Mean of Sample Means
2 + 3+...+8 80
µ =
= =5
16
16
which is the same as the
population mean. Thus µ = µ.
Frequency
4
3
X
2
1
0
2
3
4
5
6
SAM PLE M E AN S
Copyright ©1999 Ian H. Giddy
7
8
X
Probability Distributionbs 97
7-5 Mean and Standard Deviation of
Means
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 98
the Sample
7-5 The Standard Error of the
7-59
Mean
7-60
The standard deviation of the sample
means is
(2 − 5) + (3 − 5) +...+(8 − 5)
16
= 1581
. .
σ =
X
2
This is the same as
2
The standard deviation of the sample
means is called the standard error of
the mean. Hence
σ
σ =
.
n
2
σ
.
2
Copyright ©1999 Ian H. Giddy
X
Probability Distributionbs 99
7-5 The Central Limit Theorem
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 100
7-5 The Central Limit Theorem
7-61
7-62
l
Copyright ©1999 Ian H. Giddy
As the sample size n increases, the shape
of the distribution of the sample means
taken from a population with mean µ and
standard deviation of σ will approach a
normal distribution. As previously shown,
this distribution will have a mean µ and
standard deviation
σ / √n .
Probability Distributionbs 101
The central limit theorem can be used
to answer questions about sample means
in the same manner that the normal distribution
can be used to answer questions about
individual values. The only differenceis that
a new formula must be used for the z - values.
It is
X −µ
.
z=
σ/ n
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 102
Giddy/NYU
Probability Distributions /18
7-5 The Central Limit Theorem Example
7-5 The Central Limit Theorem Example
7-63
7-64
l
A.C. Neilsen reported that children between the
ages of 2 and 5 watch an average of 25 hours of
TV per week. Assume the variable is normally
distributed and the standard deviation is 3 hours.
If 20 children between the ages of 2 and 5 are
randomly selected, find the probability that the
mean of the number of hours they watch TV is
greater than 26.3 hours.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 103
7-5 The Central Limit Theorem Example
The standard deviation of the sample
means is σ/ √n = 3/ √20 = 0.671.
l The z-value is z = (26.3 - 25)/0.671= 1.94.
l Thus P(z > 1.94) = 0.5 – 0.4738 = 0.0262.
That is, the probability of obtaining a
sample mean greater than 26.3 is 0.0262
= 2.62%.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 104
7-5 The Central Limit Theorem Example
7-65
7-66
l
0.5000
− 0.4738
0.0262
0
Copyright ©1999 Ian H. Giddy
The average age of a vehicle registered in
the United States is 8 years, or 96 months.
Assume the standard deviation is 16
months. If a random sample of 36 cars is
selected, find the probability that the mean
of their age is between 90 and 100
months.
1.94
Probability Distributionbs 105
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 106
7-5 The Central Limit Theorem Example
7-5 The Central Limit Theorem Example
7-67
7-68
The standard deviation of the sample
means is σ/ √n = 16/ √36 = 2.6667.
l The two z-values are
z1
= (90 – 96)/2.6667 = –2.25 and
z2
= (100 – 96)/2.6667 = 1.50.
l Thus
P(–2.25 ≤ z ≤ 1.50) = 0.4878 + 0.4332
= 0.921 or 92.1%.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 107
0.4878
−2.25
Copyright ©1999 Ian H. Giddy
0.4332
0
1.50
Probability Distributionbs 108
Giddy/NYU
Probability Distributions /19
7-6 The Normal Approximation to
Distribution
the Binomial
7-69
7-6 The Normal Approximation to
Distribution
the Binomial
7-70
l
The normal approximation to the
binomial is appropriate when np ≥ 5 and
nq ≥ 5.
l In addition, a correction for continuity
may be used in the normal
approximation.
l
The normal distribution is often used to
solve problems that involve the binomial
distribution since when n is large (say,
100), the calculations are too difficult to do
by hand using the binomial distribution.
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 109
7-6 The Normal Approximation to
Distribution
the Binomial
7-71
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 110
7-6 The Normal Approximation to the Binomial
Distribution - Example
7-72
l
l
Copyright ©1999 Ian H. Giddy
A correction for continuity is a correction
employed when a continuous distribution is
used to approximate a discrete distribution.
The continuity correction means that for any
specific value of X, say 8, the boundaries of X
in the binomial distribution (in this case 7.5
and 8.5) must be used.
Probability Distributionbs 111
7-6 The Normal Approximation to the Binomial
Distribution - Example
7-73
l
Copyright ©1999 Ian H. Giddy
Prevention magazine reported that 6% of
American drivers read the newspaper
while driving. If 300 drivers are selected at
random, find the probability that exactly 25
say they read the newspaper while driving.
Probability Distributionbs 112
7-6 The Normal Approximation to the Binomial
Distribution - Example
7-74
l
l
Copyright ©1999 Ian H. Giddy
Here p = 0.06, q = 0.94, and n = 300.
Check for normal approximation: np =
(300)(0.06) = 18 and
nq = (300)(0.94) = 282. Since both
values are at least 5, the normal
approximation can be used.
Probability Distributionbs 113
(continued) µ = np = (300)(0.06) = 18
and
σ = √npq = √ (300)(0.06)(0.94) = 4.11.
l So P(X = 25) = P(24.5 ≤ X ≤ 25.5).
l z1 = [24.5 – 18]/4.11 = 1.58 and
z2= [25.5 – 18]/4.11 = 1.82.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 114
Giddy/NYU
Probability Distributions /20
7-6 The Normal Approximation to the Binomial
Distribution - Example
7-75
(continued) P(24.5 ≤ X ≤ 25.5)
=
P(1.58 ≤ z ≤ 1.82)
= 0.4656 – 0.4429 = 0.0227.
l Hence, the probability that exactly 25
people read the newspaper while
driving is 2.27%.
l
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 115
www.giddy.org
Ian H. Giddy
NYU Stern School of Business
Tel 212-998-0332; Fax 212-995-4233
[email protected]
http://www.giddy.org
Copyright ©1999 Ian H. Giddy
Probability Distributionbs 116