Download 1 STA 291 Spring 2007 Lecture 15 - Wednesday, Feb

STA 291
Spring 2007
Lecture 15 - Wednesday, Feb 28
• 7 Binomial distribution (review)
• 8 Continuous Probability
Distributions
– 8.1 Probability Density Functions
– 8.2 Normal Distributions
STA 291 - Spring 2007 - Lect. 15
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Review
• Binomial distribution – two parameters:
n and
p
n is often known, but p
may be not.
The outcome (random variable) X can take
values 0,1, …, n
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A probability distribution is a model for the
population (that’s why we use the term
“parameters ”)
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Find the binomial probability
• Find probability by table
• Find probability by formula
• Find probability by look up a chart
• For n=10, p=0.68 we have
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Table
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P(0) = 0.000011259
P(1) = 0.00023925
P(2) = 0.0022879
P(3) = 0.012965
P(4) = 0.048212
P(5) = 0.12294
P(6) = 0.21771
P(7) = 0.26436
P(8) = 0.21066
P(9) = 0.099479
P(10) = 0.021139
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Formula
• The probability of observing k successes in n
independent trials is (here n=10, p=0.68 )
 10 
P ( X = k ) =   0.68k (1 −0.68)10 −k , k = 0,1,K , n,
k
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See probability chart at:
• http://people.hofstra.edu/faculty/Stefan_W
aner/RealWorld/stats/bernoulli.html
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Continuous probability distribution
• Outcome (of experiments) that is a
continuous numerical value – continuous
random variables.
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Continuous Probability Distributions
• For continuous distributions, we can not list
all possible values with probabilities (too
many)
• Instead, probabilities are assigned to intervals
of numbers (still too many out there)
• The probability of an individual number is 0
• Again, the probabilities have to be always
between 0 and 1
• The probability of the interval containing all
possible values equals 1
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For continuous Probabilities
• Formulas involve integrations
– OK but not easy
• Table – OK but do not contain every cases
• Chart – interactive chart is best. Or use
software ( like MS Excel)
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• Mathematically, a continuous probability
distribution corresponds to a (density)
function whose integral equals 1
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Continuous Probability
Distributions: Example
• Example: X=Weekly use of gasoline by
adults in North America (in gallons)
for example P(16<X<21)=0.34
• The probability that a randomly chosen
adult in North America uses between 16
and 21 gallons of gas per week is 0.34
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Continuous Probability Distributions
• Probabilities are assigned to intervals
• The probability of an individual number
is 0
• The probability of the interval
containing all possible values equals 1
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Continuous Distributions
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The Normal Probability Distribution
• Normal distribution is perfectly symmetric and
bell-shaped
• Characterized by two parameters:
mean µ and standard deviation s
• The 68%-95%-99.7% rule applies to the normal
distribution
• That is, the probability concentrated within 1
standard deviation of the mean is always 0.68
• The IQR = 4/3 s rule also applies
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Different Normal Distributions
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Applets for normal distributions
• From CD come with the book
• Normal distribution parameters applet
• Normal distribution areas applet
• Or
• http://psych.colorado.edu/~mcclella/java/n
ormal/handleNormal.html
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The Normal Probability Distribution
• Perfectly symmetric and bell-shaped
• Two parameters:
mean µ and standard deviation s
• The 68%-95%-99.7% rule applies to the normal
distribution
• Actually, to be precise, the more correct
percentages are 68.26895% - 95.44997% 99.73002%
• That is, the probability concentrated within 1
standard deviation of the mean is always
0.6826895 for a normal distribution
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Normal Distribution:
Example (female height)
• Assume that adult female height has a normal
distribution with mean µ=165 cm and standard
deviation s =8 cm
• With probability 0.68, a randomly selected adult
female has height between
µ - s = 157 cm and µ + s = 173 cm
• With probability 0.95, a randomly selected adult
female has height between
µ - 2s = 149 cm and µ + 2s = 181 cm
• Only with probability 1-0.997=0.003, a randomly
selected adult female has height below
µ - 3s = 141 cm or above µ + 3s = 189 cm
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Normal Distribution
• So far, we have looked at the probabilities
within one, two, or three standard
deviations from the mean
(µ + - s , µ + - 2s, µ + - 3s )
• How much probability is concentrated within
1.43 standard deviations of the mean?
• More general, how much probability is
concentrated within z standard deviations of
the mean?
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• All can be answered by using the applet.
• Or using the table.
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Normal Distribution Table
• Table 3 (page B-8) shows for different values
of z the probability between µ and µ + zs
• Probability that a normal random variable
takes any value between the mean, µ, and z
standard deviations above the mean
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• For z=1.43, the tabulated value is .4236
• The probability between µ and µ + 1.43s of a
normal distribution equals .4236
• Because of the perfect symmetry of the normal
distribution, also the probability between µ and
µ – 1.43s of a normal distribution equals .4236
• So, within (+ - )1.43 standard deviations of the
mean is how much probability?
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Verifying the Empirical Rule
• The 68%-95%-99.7% rule can be verified
using Table 3
• How much probability is within one (two,
three) standard deviation(s) of the mean?
• Note that the table only answers directly:
How much probability is between 0 and
one (two, three) standard deviation(s)
above the mean?
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Verifying the Empirical Rule (z=1)
• z=1.00, Table 3:
Between 0 and 1 standard deviations above
the mean is probability .3413
• Then, between 0 and 1 standard deviations
below the mean is also probability .3413
(symmetry)
• Therefore, within one standard deviation
from the mean is probability
.3413+.3413=.6828
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Verifying the Empirical Rule (z=2,3)
• z=2.00, Table 3:
Between 0 and 2 standard deviations above the
mean is probability ______
• Then, between 0 and 2 standard deviations below
the mean is also probability ______ (symmetry)
• Therefore, within two standard deviations from the
mean is probability _________________
• z=3.00: Table value:___________
• Probability within three standard deviations from the
mean:__________
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Application
• SAT scores are approximately normally
distributed with mean 500 and standard
deviation 100
• The 90th percentile of the SAT scores is
1.28 standard deviations above the mean
• µ + 1.28 s = 500 +1.28 ·100 = 628
• Find the 99th and the 5th percentile of SAT
scores
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• In fact for any normal probability
distributions, the 90th percentile is always
1.28 SD above the mean
the 95th percentile is ____ SD above mean
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More Examples
• For a normally distributed random
variable, verify that the probability between
µ – 1.96 s and µ + 1.96 s equals 0.95
µ – 0.67 s and µ + 0.67 s equals 0.5
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Homework 8
• Due Thursday next week (March 8 at 11
PM)
• Weekly online homework assignment
• Suggested problems from the textbook:
8.16, 8.18, 8.22, 8.26, 8.30,
8.34, 8.36, 8.55, 8.69
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Attendance Survey Question 15
• On a 4”x6” index card
– Please write down your name and
section number
– Today’s Question:
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