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chapter 6 for Economics
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Probability Distributions
Statistics for Business and Economics
6th Edition
Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-2
Introduction to Probability Distributions
Random Variable Represents a possible numerical value from
a random experimentRandom Variables
Discrete Random Variable
ContinuousRandom Variable
Ch. 5 Ch. 6
Discrete Random Variables Can only take on a countable number of values
Examples:
Roll a die twiceLet X be the number of times 4 comes up (then X could be 0, 1, or 2 times)
Toss a coin 5 times. Let X be the number of heads(then X = 0, 1, 2, 3, 4, or 5)
Experiment: Toss 2 Coins. Let X = # heads.
T
T
Discrete Probability Distribution
4 possible outcomes
T
T
H
H
H H
Probability Distribution
0 1 2 x
x Value Probability
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
.50
.25
P
r
o
b
a
b
i
l
i
t
y
Show P(x) , i.e., P(X = x) , for all values of x:
P(x) 0 for any value of x
The individual probabilities sum to 1;
(The notation indicates summation over all possible x values)
Probability DistributionRequired Properties
=x
1P(x)
Cumulative Probability Function
The cumulative probability function, denotedF(x0), shows the probability that X is less than or equal to x0
In other words,
)xP(X)F(x 00 =
=
0xx0 P(x))F(x
Expected Value Expected Value (or mean) of a discrete
distribution (Weighted Average)
Example: Toss 2 coins, x = # of heads,
compute expected value of x:E(x) = (0 x .25) + (1 x .50) + (2 x .25)
= 1.0
x P(x)0 .251 .502 .25
==x
P(x)x E(x)
Variance and Standard Deviation
Variance of a discrete random variable X
Standard Deviation of a discrete random variable X
==x
22 P(x))(x
==x
222 P(x))(x)E(X
Standard Deviation Example
Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(x) = 1)
.707.50(.25)1)(2(.50)1)(1(.25)1)(0 222 ==++=
Possible number of heads = 0, 1, or 2
=x
2P(x))(x
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 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
Sequences of x Successes in n Trials
The number of sequences with x successes in n independent trials is:
Where n! = n(n 1)(n 2) . . . 1 and 0! = 1
These sequences are mutually exclusive, since no two can occur at the same time
x)!(nx!n!Cnx
=
P(x) = probability 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 trials or observations)
P = probability of success
P(x) nx ! n x
P (1- P)X n X!( )!====
Example: Flip a coin four times, let x = # heads:
n = 4
P = 0.51 - P = (1 - 0.5) = 0.5
x = 0, 1, 2, 3, 4
Binomial Probability Distribution
Example: Calculating a Binomial ProbabilityWhat is the probability of one success in five observations if the probability of success is 0.1?
x = 1, n = 5, and P = 0.1
.32805.9)(5)(0.1)(0
0.1)(1(0.1)1)!(51!
5!
P)(1Px)!(nx!
n!1)P(x
4
151
XnX
=
=
=
==
Binomial DistributionMean and Variance
Mean
Variance and Standard Deviation
nPE(x) ==
P)nP(1-2 =P)nP(1- =
Where n = sample sizeP = probability of success(1 P) = probability of failure
n = 5 P = 0.1
n = 5 P = 0.5
Mean
0.2.4.6
0 1 2 3 4 5x
P(x)
.2
.4
.6
0 1 2 3 4 5x
P(x)
0
0.5(5)(0.1)nP ===
0.67080.1)(5)(0.1)(1P)nP(1-
=
==
2.5(5)(0.5)nP ===
1.1180.5)(5)(0.5)(1P)nP(1-
=
==
Binomial CharacteristicsExamples
Using Binomial Tables
N x p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.5010 0
123456789
10
0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000
0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000
0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000
0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000
0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001
0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003
0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010
Examples: n = 10, x = 3, P = 0.35: P(x = 3|n =10, p = 0.35) = .2522n = 10, x = 8, P = 0.45: P(x = 8|n =10, p = 0.45) = .0229
The Poisson Distribution
Apply the Poisson Distribution when: You wish to count the number of times an event
occurs in a given continuous interval The probability that an event occurs in one subinterval
is very small and is the same for all subintervals The number of events that occur in one subinterval is
independent of the number of events that occur in the other subintervals
The average number of events per unit is (lambda)
Poisson Distribution Formula
where:x = number of successes per unit = expected number of successes per unite = base of the natural logarithm system (2.71828...)
x!eP(x)
x
=
Poisson Distribution Characteristics
Mean
Variance and Standard Deviation
E(x) ==
]2 == 2)[( XE =
where = expected number of successes per unit
Using Poisson Tables
X
0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90
01234567
0.90480.09050.00450.00020.00000.00000.00000.0000
0.81870.16370.01640.00110.00010.00000.00000.0000
0.74080.22220.03330.00330.00030.00000.00000.0000
0.67030.26810.05360.00720.00070.00010.00000.0000
0.60650.30330.07580.01260.00160.00020.00000.0000
0.54880.32930.09880.01980.00300.00040.00000.0000
0.49660.34760.12170.02840.00500.00070.00010.0000
0.44930.35950.14380.03830.00770.00120.00020.0000
0.40660.36590.16470.04940.01110.00200.00030.0000
Example: Find P(X = 2) if = .50
.07582!(0.50)e
!Xe)2X(P
20.50X
==
==
Graph of Poisson Probabilities
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P
(
x
)
X =
0.5001234567
0.60650.30330.07580.01260.00160.00020.00000.0000
P(X = 2) = .0758
Graphically: = .50
Poisson Distribution Shape
The shape of the Poisson Distribution depends on the parameter :
0.00
0.05
0.10
0.15
0.20
0.25
1 2 3 4 5 6 7 8 9 10 11 12
x
P
(
x
)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P
(
x
)
= 0.50 = 3.00
The Normal Distribution
Bell Shaped Symmetrical Mean, Median and Mode
are EqualLocation is determined by the mean,
Spread is determined by the standard deviation,
The random variable has an infinite theoretical range: + to
Mean = Median = Mode
x
f(x)
The Normal Probability Distribution
The formula for the normal probability distribution is
Where e = the mathematical constant approximated by 2.71828 = the mathematical constant approximated by 3.14159 = the population mean = the population standard deviationx = any value of the continuous variable, < x <
22 /2)(xe21f(x) =
The Standardized Normal Any normal distribution (with any mean and variance
combination) can be transformed into the standardized normal distribution (Z), with mean 0 and variance 1
1)N(0~Z ,
XZ =
Z
f(Z)
01
Example
If X is distributed normally with mean of 100and standard deviation of 50, the Z value for X = 200 is
2.050
100200
XZ ===