Probability Distributions
Probability Distributions
Discrete
Probability Distributions
Continuous
Probability Distributions
Uniform
Normal
Chi-Sq
Bernoulli Binomial
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Pearson
Bernoulli Distribution
Consider only two outcomes: “success” or “failure”
Let P denote the probability of success
Let 1 – P be the probability of failure
Define random variable X:
x = 1 if success, x = 0 if failure
Then the Bernoulli probability distribution is
P(0) (1P) and P(1) P
3
Mean and Variance of a Bernoulli Random Variable
The mean is μx = P
μx E[X]xP(x)(0)(1P)(1)PP X
The variance is σ2x = P(1 – P)
σ2 E[(Xμ )2](xμ )2P(x)
xx (0P)2(1P)(1P)2PP(1P)
x X
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Ch. 4-4
4
as Prentice Hall
Developing the Binomial Distribution
The number of sequences with x successes in n independent trials is:
Cnx n! x!(nx)!
Wheren!=n·(n–1)·(n–2)·…·1 and 0!=1
These sequences are mutually exclusive, since no two can occur at the same time
Binomial Probability Distribution
A fixed number of observations, n
e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse
Two mutually exclusive and collectively exhaustive categories
e.g., head or tail in each toss of a coin; defective or not defective light bulb
Generally called “success” and “failure”
Probability of success is P , probability of failure is 1 – P
Constant probability for each observation
e.g., Probability of getting a tail is the same each time we toss the coin
Observations are independent
The outcome of one observation does not affect the outcome of
the other
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Possible Binomial Distribution Settings
A manufacturing plant labels items as either defective or acceptable
A firm bidding for contracts will either get a contract or not
A marketing research firm receives survey responses of “yes I will buy” or “no I will not”
New job applicants either accept the offer or reject it
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Ch. 4-7
Pearson
The Binomial Distribution
n!
P(x) x !(n x)!P (1- P)
X n X
Pprobability of x successes in n trials, with probability of success P on each
trial
x = number of ‘successes’ in sample, (x = 0, 1, 2, …, n)
n = sample size (number of independent trials or observations)
P = probability of “success”
Example: Flip a coin four times, let x = # heads:
n= 4
P = 0.5
1 – P = (1 – 0.5) = 0.5 x = 0, 1, 2, 3, 4
Ch. 4-8
Shape of Binomial Distribution
The shape of the binomial distribution depends on the values of P and n
Here, n = 5 and P = 0.1
Here, n = 5 and P = 0.5
n=5 P=0.1
0x 012345
P(x) .6
.4 .2
P(x) .6
.4 .2 0
0
n = 5
P = 0.5
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x 12345
Ch. 4-9
as Prentice Hall
Mean and Variance of a Binomial Distribution
Mean
μE(x)nP
Variance and Standard Deviation
σ2 nP(1-P) σ nP(1- P)
Where n = sample size
P = probability of success
(1 – P) = probability of failure
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Ch. 4-10
Pearson
Probability Distributions Continuous Random Variables
11
The Normal Distribution
‘Bell Shaped’
Symmetrical
Mean, Median and Mode are Equal
Location is determined by the mean, μ
Spread is determined by the standard deviation, σ
The random variable has an infinite theoretical range:
+ to
f(x)
σ
μ
Mean = Median = Mode
x
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The Normal Distribution
The normal distribution closely approximates the probability distributions of a wide range of random variables
Distributions of sample means approach a normal distribution given a “large” sample size
Computations of probabilities are direct and elegant
The normal probability distribution has led to good business decisions for a number of applications
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The Normal Distribution Shape
f(x)
Changing μ shifts the distribution left or right.
Changing σ increases or decreases the spread.
σ
μx
X~N(μ,σ2)
Given the mean μ and variance σ2 we define the
normal distribution using the notation
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Ch. 5-14
The Normal Probability Density Function
The formula for the normal probability density function is
f(x) 1 e(xμ)2 /2σ2 2π 2
pproximated by 2.71828 pproximated by 3.14159
variable, < x <
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h. 5-15
a a
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Cumulative Normal Distribution
For a normal random variable X with mean μ and variance σ2 , i.e., X~N(μ, σ2), the cumulative distribution function is
F(x0)P(Xx0) f(x)
P(Xx0) μx0 x
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Finding Normal Probabilities
The probability for a range of values is measured by the area under the curve
P(aXb)F(b)F(a)
aμb
x
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Finding Normal Probabilities
F(b)P(Xb) F(a)P(Xa)
aμb
x
aμb
P(aXb)F(b)F(a) Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
x
aμb
x
h. 5-18
C
The Standard Normal Distribution
Any normal distribution (with any mean
and variance combination) can be
transformed into the standardized
normal distribution (Z), with mean 0 and
f(Z)
variance 1
Z~N(0,1) 1 0
Z
Z
Xμ σ
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Need to transform X units into Z units by
Example
If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is
Z X μ 200 100 2.0 σ 50
This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.
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Comparing X and Z units
100 200 X 0 2.0 Z
(μ = 100, σ = 50)
(μ=0, σ=1)
Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z)
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f(x)
Finding Normal Probabilities
P(a X b) P a μ Z b μ σ σ
FbμFaμ σ σ
aμb aμ 0 bμ σσ
x
Z
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Probability as Area Under the Curve
The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below
f(X)
P( X μ) 0.5 0.5
μ
P(μ X ) 0.5
h. 5-23
P( X ) 1.0
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0.5
X
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Appendix Table 1
The Standard Normal Distribution table in the textbook (Appendix Table 1) shows values of the cumulative normal distribution function
For a given Z-value a , the table shows F(a)
(the area under the curve from negative infinity to a
)
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0
F(a)P(Za)
aZ
General Procedure for Finding Probabilities
To find P(a < X < b) when X is distributed normally:
Draw the normal curve for the problem in terms of X
Translate X-values to Z-values
Use the Cumulative Normal Table
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Finding Normal Probabilities
Suppose X is normal with mean 8.0 and standard deviation 5.0
Find P(X < 8.6)
8.0 8.6
X
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Finding Normal Probabilities
Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X
< 8.6) X μ 8.6 8.0
Z σ 5.0 0.12
(continue d)
8 8.6
P(X < 8.6)
X
μ=8 μ=0
σ = 10
σ = 1
0 0.12 Z
P(Z < 0.12)
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Ch. 5-27
Finding the X value for a Known Probability
Steps to find the X value for a known probability:
1. Find the Z value for the known probability 2. Convert to X units using the formula:
XμZσ
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h. 5-28
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Assessing Normality
Not all continuous random variables are normally distributed
It is important to evaluate how well the data is approximated by a normal distribution
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h. 5-29
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The Normal Probability Plot
Normal probability plot
► Arrange data from low to high values
► Find cumulative normal probabilities for all values
► Examine a plot of the observed values vs. cumulative probabilities (with the cumulative normal probability on the vertical axis and the observed data values on the horizontal axis)
► Evaluate the plot for evidence of linearity
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The Normal Probability Plot
Left-Skewed Right-Skewed
100
0
Data
100
0
Data
Nonlinear plots indicate a deviation from normality
Uniform
h. 5-31
100
0
Data
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Percent
Percent
Percent
C
h. 5-32
5.6
Let X1, X2, . . ., Xk be continuous random variables
Their joint cumulative distribution function,
F(x1, x2, . . ., xk)
simultaneously X1 is less than x1, X2 is less than x2, and so on; that is
Jointly Distributed Continuous Random Variables
F(x,x ,,x )P(X x X x X x )
defines the probability that
12k1122kk
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C
5.6
Let X1, X2, . . ., Xk be continuous random variables
Their joint cumulative distribution function,
F(x1, x2, . . ., xk)
simultaneously X1 is less than x1, X2 is less than x2, and so on; that is
Jointly Distributed Continuous Random Variables
F(x,x ,,x )P(X x X x X x )
defines the probability that
12k1122kk
h. 5-33
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C
Sampling Distribution of s2 (Normal Data)
Population variance (2) is a fixed (unknown) parameter based on the population of measurements
Sample variance (s2) varies from sample to sample (just as sample mean does)
Sums of squared indepedet standard Normal Distributions
Chi-Square distributions
► Positively skewed with positive density over (0,)
► Indexed by its degrees of freedom (df)
► Mean=df, Variance=2(df)
► Critical Values given in Table 7, pp. 1095-1096
Chi-Square Distributions
0.2 0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0
Chi-Square Distributions
df=4
df=10
df=20
f1(y) f2(y) f3(y) f4(y) f5(y)
df=30
df=50
0 10 20 30 40 50 60 70
X^2
f(X^2)
Chi-Square Distribution Critical Values
0.12
0.1
0.08
0.06
0.04
.025
0.02
.95
Chi-Square Distribution (df=10)
a
c 2(a ) df=10
0.995
2.156
0.990
2.558
0.975
3.247
0.950
3.940
0.900
4.865
0.100
15.987
0.050
18.307
0.025
20.483
0.010
23.209
0.005
25.188
0 .025
0 5 10 15 20 25 30 35 40
-0.02
3.247
X^2 20.48
f(X^2)
Chi-Square Critical Values (2-Sided Tests/CIs)
a/2
1-a
a/2
c2L
c2U
f(X^2)
F-Distributions
F distribution is the ratio of (y1/d1)/(y2/d2) where y1 and y2 are
Chi-sq distributions with d1 and d2 degrees of freedom
Take on positive density over the range (0 , )
Cannot take on negative values
Non-symmetric (skewed right)
Indexed by two degrees of freedom (d1 (numerator df) and d2 (denominator df))
Properties of F-Distributions F-Distributions
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
0
-0.1
f(5,5) f(5,10) f(10,20)
Density Function of F
0 1 2 3 4 5 6 7 8 9 10
F
0.7
0.6
0.5
0.4
0.3
0.2
0.1
.05
Critical Values of F (df1=5,df2=5)
.90
0
0 1 2 3 4 5 6 7 8 9 10
F
.05
F(.95,5,5)=1/F(.05,5
F(.05,5,5)=5.05
Density Function of F