ch 5 probability: thech 5 probability: the mathematics of ...mduan/stat3411/ch5_1.pdf ·...
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Ch 5 Probability: TheCh 5 Probability: The Mathematics of Randomness
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5.1.1 Random Variables and Their Distributions
• A random variable is a quantity that (prior to observation) can be thought of as dependent ) g pon chance phenomena.
Toss a coin 10 times
X=# of heads
T i til h dToss a coin until a head
X=# of tosses needed
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More random variablesMore random variablesToss a dieToss a die
X=points showing
Plant 100 seeds of pumpkinsX % germinatingX=% germinating
T li h b lbTest a light bulbX=lifetime of bulb
Test 20 light bulbsX=average lifetime of bulbs
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Types of Random VariableTypes of Random Variable
• A discrete random variable is one that has isolated or separated possible values.
(Counts, finite‐possible values)
• A continuous random variable is one that can be idealized as having an entire (continuous) interval of numbers as its set of possible values.
(Lifetimes, time, compression strength 0 infinity)
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Probability Distributiony• To specify a probability distribution for a random
variable is to give its set of possible values and (in one h ) l b bway or another) consistently assign numbers between
0 and 1 –called probabilities– as measures of the likelihood that be various numerical values will occur.likelihood that be various numerical values will occur.
• It is basically a rule of assigning probabilities to random y g g pevents.
Roll a die, X=# showing
x 1 2 3 4 5 6f(x) 1/6 1/6 1/6 1/6 1/6 1/6
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Probability DistributionProbability Distribution
h l f b bili di ib iThe values of a probability distribution must be numbers on the interval from 0 to 1.
The sum of all the values of a probability distribution must be equal to 1distribution must be equal to 1.
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ExampleExample
Inspect 3 parts. Let X be the # of parts with defect.
Find the probability distribution of X.
First, what are possible values of X?, p
(0, 1,2,3)
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When does X take on each value?When does X take on each value?
OOutcome X{NNN} 0 {NND, NDN, DNN} 1{NDD, DND, DDN} 2{NDD, DND, DDN} 2{DDD} 3
• If P(D)=0. 5, then P(N)=0.5 We have x 0 1 2 3
f(x)=P(X=x) 1/8 3/8 3/8 1/8
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Cumulative Distribution FunctionCumulative Distribution Function
l i di ib i ( ) f d• Cumulative distribution F(x) of a random variable X is
( ) ( )F x P X x= ≤
• Since (for discrete distributions) probabilities l l b l f f( ) fare calculated by summing values of f(x), for a
discrete distribution
( ) ( )F x P X x≤( ) ( )F x P X x= ≤
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“Defect” Example continuedDefect Example continued
W h dWe had f(0)=1/8, f(1)=f(2)=3/8, f(3)=1/8.
Therefore
0, 0;x<
• F(0)=f(0)=1/8;
• F(1)=f(0)+f(1)=1/8+3/8=1/2;
= 1/2,1/8,
F(x) 2;x11;x0
<≤<≤
• F(2)=f(0)+f(1)+f(2)=7/8;
• F(3)=f(0)+f(1)+f(2)+f(3)=1.
1,7/8,
3.x3;x2
≥<≤
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Summaries of Discrete Probability Distributions
Given a set of numbers, for example,
S= {1, 1, 1, 1, 3, 3, 4, 5, 5 , 6 }
To calculate the average of these 10 numbers, you can use
31030
106554331111
==++++++++++
Or you can get it this way
10(6)(1)(5)(2)(4)(1)(3)(2)(1)(4) ++++
3
)101(6)()
102(5)()
101(4)()
102(3)()
104(1)(
=
++++=
3.=
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Mean or ExpectationMean or Expectation
Th b l i h i h d f lThe above result is the weighted sum of x values:
x 1 3 4 5 6x 1 3 4 5 6Weights 4/10 2/10 1/10 2/10 1/10f(x) 0 4 0 2 0 1 0 2 0 1f(x) 0.4 0.2 0.1 0.2 0.1
The weights are actually the probability distribution for a randomThe weights are actually the probability distribution for a random variable X .
The weighted sum, 3, is called the mean or mathematical expectation of random variable X.
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DefinitionDefinition
Let X be a discrete random variable with probability distribution f(x). The mean or expected value of X is
∑==x
xf(x)E(X) µ
The expectation or mean of a random variable X is the long run average of the observations from thethe long run average of the observations from the random variable X.
The mean of X is not necessarily a possible value for X.
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Back to “defect” exampleBack to defect example
• What is the average number of defective items we expect to see?pX f(x)
0 1/80 1/8
1 3/8
2 3/8
3 1/8
1 3 3 1 3( ) ( ) 0 1 2 3E X x f xµ = = = × + × + × + × =∑( ) ( ) 0 1 2 38 8 8 8 2
E X x f xµ = = = × + × + × + × =∑
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VarianceVariance
Th i f th t f• The mean is a measure of the center of a random variable or distributionX: x= ‐1 0 1
f(x) ¼ ½ ¼
Y: y= ‐2 0 2 g(y) ¼ ½ ¼
Both variables have the same center: 0
Which has a larger variability?
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VarianceVariance
D i i f h ( )• Deviation from the center (mean)X−µX
The average of X−µX is zero!
• Consider the weighted average of (X−µX)2
E(X−0)2=(1)(1/4)+(0)(1/2)+(1)(1/4)=1/2 E(Y−0)2=(4)(1/4)+(0)(1/2)+(4)(1/4)=2
Y is more variable!
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Definition• Let X be a discrete random variable with probability
distribution f(x) and mean µdistribution f(x) and mean µ
h i f i• The variance of X is
f( ))(])E[(X 222 ∑ f(x)µ)(x]µ)E[(X σx
22 ∑ −=−=
The positive square root of the σ2 is called standard deviation of X, denoted by σ.
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Back to “defect” exampleBack to defect example
X f(x)
0 1/8
1 3/8
2 3/82 3/8
3 1/8
2 2 2( ) ( ) ( ) ( )Var X E x x f xσ µ µ= = − = −∑2 2 2 2
( ) ( ) ( ) ( )1 3 3 1(0 1.5) (1 1.5) (2 1.5) (3 1.5)8 8 8 8
Var X E x x f xσ µ µ= = − = −
= − × + − × + − × + − ×
∑( ) ( ) ( ) ( )
8 8 8 8
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1 3 3 12 2 2 2 2
2 2 2 2
1 3 3 1( ) 0 1 2 3 38 8 8 8
( ) ( )
E X = × + × + × + × =
2 2 2 2
2
( ) ( ) 3 1.5 0.75
0.75
Var X E Xσ µ
σ σ
= = − = − =
= =
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Binomial Distribution
I li d blIn many applied problems, we are interested in the probability that an event p ywill occur x times out of n.
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For examplepInspect 3 items. X=# of defects.
D d f t N t d f t S P(N) 5/6D=a defect, N=not a defect, Suppose P(N)=5/6
No defect: (x=0)( )NNN (5/6)(5/6)(5/6)
One defect: (x=1)NND (5/6)(5/6)(1/6)NND (5/6)(5/6)(1/6) NDN same DNN same
Two defect: (x=2)Two defect: (x=2)NDD (5/6)(1/6)(1/6)DND sameDDN sameDDN same
Three defect: (x=3)DDD (1/6)(1/6)(1/6)
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Binomial distribution• x f(x)
0 (5/6)3How many ways to0 (5/6)
1 3(1/6)(5/6)2
How many ways to choose x of 3 places for D
__ __ __
2 3(1/6)2(5/6)
3 (1/6)33 (1/6)xx
xf−
=
3
65
613
)(x
66
f [1‐P(D)]3‐ # of D[P(D)]# of D [1 P(D)]
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In general: n independent trialsg p
p probability of a success
# fx=# of successes
ways to choose x places for sn ways to choose x places for s,x
n
( ) (1 )x n xnf x p p
x−
= −
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• Roll a die 20 times. X=# of 6’s,n=20 p=1/6n=20, p=1/6
2020 1 5( )6 6
x x
f xx
− = 4 16
6 6
20 1 5( 4)
x
p x
= =
• Flip a fair coin 10 times X=# of heads
( 4)4 6 6
p x
Flip a fair coin 10 times. X=# of heads
1010 1101110 − xx
2110
21
2110
)(
=
=
xxxf
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More example
• Pumpkin seeds germinate with probability 0.93. Plant n=50 seeds
• X= # of seeds germinating
( ) ( ) xxxf −
= 5007093050
)( ( ) ( )x
xf
= 07.093.0)(
0 ( ) ( )248 07.093.04850
)48(
==XP
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Mean of Binomial DistributionMean of Binomial Distribution
X # f d f t i 3 it• X=# of defects in 3 itemsx f(x)0 (5/6)3
1 3(1/6)(5/6)2( / )( / )2 3(1/6)2(5/6) 3 (1/6)33 (1/6)
E t d l f X?Expected long run average of X?
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The mean of a binomial distribution
• Binomial distribution• Binomial distributionn= # of trials,
b bili f h i lp=probability of success on each trialX=# of successes
( ) (1 )x n xn ∑( ) (1 )x n xE x x p p npx
µ − = = − =
∑
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L ’ h k hi• Let’s check this. 33 3 1 5x x− 3
0
3 2 2 3
3 1 5( ) ( )6 6X
xE X x f x x
xµ
=
= = =
∑ ∑3 2 2 35 1 5 1 5 10* 1*3 2*3 3* 1/ 2
6 6 6 6 6 6 = + + + =
• Use binomial mean formula
( ) (1 ) 3*1/ 6 1/ 2x n xnE x x p p npµ −
= = − = = = ∑( ) (1 ) 3 1/ 6 1/ 2E x x p p npx
µ
∑
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More examplesMore examples
• Toss a die n=60 times, X=# of 6’s
known that p=1/6known that p 1/6
μ=μX =E(X)=np=(60)(1/6)=10
We expect to get 10 6’sWe expect to get 10 6 s.
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Variance for Binomial distributionVariance for Binomial distribution
• σ2=np(1-p) where n is # of trials and p is probability ofwhere n is # of trials and p is probability of a success.F h i l 3 1/6• From the previous example, n=3, p=1/6Then e
σ2=np(1-p)=3*1/6*(1-1/6)=5/12
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5 1 5 Poisson Distribution5.1.5 Poisson Distribution
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