Probability models 1 Bernouilli Process - UC3MGeometric distribution 1 Poisson Process Poisson...
Transcript of Probability models 1 Bernouilli Process - UC3MGeometric distribution 1 Poisson Process Poisson...
Estadística, Profesora: María Durbán1
1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
1 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
1 Bernouilli Process
Probability models
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1 Bernouilli Process
When an experiment has the following characteristics:
There are only two possible results: Acceptable (A)Defective (D)
The proportion of A and D is constant in the populationand remains constant regardless of the sample observed quantity
The observations are independent
Pr( )Pr( ) 1
D pA q p
== = −
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1 Bernouilli Process
When an experiment has the following characteristics:
There are only toew possible results: Aceptable (A)Defective (D)
The proportion of A and D is constant in the populationand remains constant regardless of the sampleobserved quantity
The observations are independent
Pr( )Pr( ) 1
D pA q p
== = −
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1 Bernouilli Process
Examples
Toss a coin
Observe if a manufactures piece is defective or not
Sex on a newborn baby
Transmission (right or wrong) of a bit in a digital channel
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1 Bernouilli Process
Bernouilli distribution
0 if A occurs 1 Pr( 0) 1 if A does not ocurr Pr( 1)
q p XX
p X→ = − = =
=→ = =
The probability function is:
1( ) (1 ) 0,1x xp x p p x−= − =
[ ]
[ ] 2 2
0 (1 ) 1
(0 ) (1 ) (1 ) (1 )
E X p p p
Var X p p p p p p
μ
σ
= = × − + × =
= = − − + − = −
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1 Bernouilli Process
Binomial distribution
X = Number of successes in n trials
X takes values 0,1,2,…,n
When a Bernoulli experiment with parameter p is repeated a fixed number of times, n, the number of successes follows a el Binomial distribution with parameters (n,p).
~ ( , )X B n p
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The probability function is:
1 Bernouilli Process
( ) (1 ) , 0,1, ,r n rnP X r p p r n
r−⎛ ⎞
= = − =⎜ ⎟⎝ ⎠
K
[ ][ ] (1 )
E X np
Var X np p
=
= −Estadística, Profesora: María Durbán
8
n=5
n=25 p=0.75 p=0.5 p=0.2
1 Bernouilli Process
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1 Bernouilli Process
Example
An electronic device has 40 integrated circuit. The probability that a circuit is defective is, and the circuits are independent. The device will work if there areNo defective circuits.
What is the probability that the device works?
X = Number of defective circuits among the 40
Pr( 0)X =
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1 Bernouilli Process
Example
An electronic device has 40 integrated circuit. The probability that a circuit isdefective is, and the circuits are independent. The device will work if there areNo defective circuits.
What is the probability that that the device works?
X = Number of defective circuits among the 40
Pr( 0)X =Experiment: Observe if a circuit is defective or not. It is repeated 40 times
The circuits are independent
The probability of being defective is constant, 0.01
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1 Bernouilli Process
Example
An electronic device has 40 integrated circuit. The probability that a circuit isdefective is, and the circuits are independent. The device will work if there areNo defective circuits.
What is the probability that that the device works?
X = Number of defective circuits among the 40
~ (40,0.01)X B
0 4040Pr( 0) 0.01 (1 0.01) 0.669
0X ⎛ ⎞
= = − =⎜ ⎟⎝ ⎠
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1 Bernouilli Process
Geometric distribution
The experiment has the following characteristics:
There are only two possible results
The probability of success is constant
The observations are independent
The experiment is repeated until the first success
X = Number of times that an experiment is repeated until a successoccurs.
~ ( )X Ge p
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1 Bernouilli Process
Geometric distribution
1,..... are Bernouilli iX i n=
1 2 3 4
1 0 0 0 1 Pr( 1)0 1 0 0 2 Pr( 2)0 0 1 0 3 Pr( 3)0 0 0 1 4 Pr( 4)
X X X X X
X X pX X qpX X qqpX X qqqp
↓ ↓ ↓ ↓ ↓⇒ = = =⇒ = = =⇒ = = =⇒ = = =
L
L
L
L
L
The probability function is:
1( ) (1 ) , 1, 2,rP X r p p r−= = − = KEstadística, Profesora: María Durbán
14
1 Bernouilli Process
Geometric distribution
[ ][ ] 2
1/
(1 ) /
E X p
Var X p p
=
= −
1 2 3 4
1 0 0 0 1 Pr( 1)0 1 0 0 2 Pr( 2)0 0 1 0 3 Pr( 3)0 0 0 1 4 Pr( 4)
X X X X X
X X pX X qpX X qqpX X qqqp
↓ ↓ ↓ ↓ ↓⇒ = = =⇒ = = =⇒ = = =⇒ = = =
L
L
L
L
L
1,..... are Bernouilli iX i n=
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1 Bernouilli Process
Geometric distribution
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1 Bernouilli Process
Example
The probability that a bit transmitted through a digital channel is received As an error is 0.1. If the transmissions are independent,
What is the average number of transmissions that have to be observed until the first error occurs?
X = Number of transmissions until the first error occurs
[ ] 1/ 1/ 0.1 10E X p= = =
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1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
1 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
2 Poisson Process
Probability models
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2 Poisson Process
When an experiment has the following characteristics:
We observe the occurrence of events in an interval
The probability that this happens in an interval
is the same for intervals of the same lengthis proportional to the length of the interval
The events occur independently. The number of events in an interval is independent of the number of events in another interval
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2 Poisson Process
X = Number of events in an interval of fixed length
Poisson distribution can be obtained as a limit of the Binomial distribution when
Poisson distribution
y 0n p→ ∞ →
npλ = → Average number of events in that interval
lim 1n
en
λλ −⎛ ⎞− =⎜ ⎟⎝ ⎠n → ∞
n → ∞
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The probability function is:
( ) , 0,1,!
reP X r rr
λλ−
= = = K
[ ] [ ]
[ ]
1
0 1
1 2 1 2
! ( 1)!
~ ( ) ~ ( ) independient ~ ( )
r reE X E X r er r
Var XX P Y P X Y P
λλλ λλ λ λ
λλ λ λ λ
− −∞ ∞−= → = = =
−
=
+ +
∑ ∑
Poisson distribution
2 Poisson Process
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2 Poisson Process
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Poisson distribution
2 Poisson Process
Examples
Number of faults in a millimeter of cable.
Number of phone calls in a switchboard per hour.
Number of mistakes in a page of a document
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2 Poisson Process
Example
What is the probability that no clients arrive in 3 minutes?
X = Number of clients per minute
Y = Number of clients in 3 minutes
~ ( 1)X P λ→ =
~ ( 3)Y P λ→ =
3 033Pr( 0)
0!eY e
−−= = =
The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.
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2 Poisson Process
Example
Opening the service point 8 hours, cost 6000 euros per day. What should be the minimum charge per client so that the service point is profitable?
Y = Number of clients in 8 hours ~ ( 60 8 480)Y P λ→ = × =
Profit = Charge x Y -6000
[ ]Expected Profit = Charge 6000 0 = Charge 480 6000 0
E Y× − >
× − >Charge > 12.5
The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.
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2 Poisson Process
The Exponential distribution is used to model
Time between phone callsTime between arivals at a service point Lifetime of a lamp
When the number of events follows a Poisson distribution, the time between events follows an Exponential distribution.
M
Distribución de exponencial
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2 Poisson Process
Exponential distribution
X = Number of events per time unit
T = Time until the first even occurs
We can calculate the distribution function:
0 0( ) (zero events in (0,t ))P T t P> =
~ ( )X P λ
X= Number of events per time unitY = Number of events in (0,t0) 0~ ( )Y P tλ
00( ) Pr( 0) tP T t Y e λ−> = = =
00 0( ) ( ) 1 tF t P T t e λ−= ≤ = −
~ ( )X P λ
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2 Poisson Process
~ ( )X P λ
( )( ) , 0tdF tf t e tdt
λλ −= = ≥
[ ][ ] 2
1/
1/
E X
Var X
λ
λ
=
=
If there are λ events, on average, per time unitThe average time between events is 1/λ
X = Number of events per time unit
T = Time until the first even occurs
Exponential distribution
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2 Poisson Process
0.1( ) 0.1 xf x e−=
0.5( ) 0.5 xf x e−=
2( ) 2 xf x e−=
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2 Poisson Process
Example
~ ( 1)X P λ→ =
~ ( 1)T Exp λ→ =
( )1 3 3Pr( 3) 1 Pr( 3) 1 (3) 1 1T T F e e− × −> = − ≤ = − = − − =
Pr(there are no clients in 3 minutes)=
The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.
What is the probability that no clients arrive in 3 minutes?
X = Number of clients per minute
Y = Time between clients
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2 Poisson Process
Property
1 2 1 2Pr(T > t +t / T > t ) = Pr( T > t )
1 22
1
( t +t )1 2 1 1 2
t1 1
Pr(T > t +t T > t ) Pr( T > t +t ) = Pr( T > t ) Pr( T > t )
te ee
λλ
λ
−−
−= =I
If there hasn’t been any clients in 4 minutes, what is the probability that there are no clients in the next 3 minutes?
3Pr( 7 | 4) Pr( 3) 1 (3)Y Y Y F e−> > = > = − =
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1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
2 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
Normal distribution
Probability models
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3 Normal distribution
Normal or Gaussian distribution describe a large amount of random processes
Measurement errorsNoise in a digital signalElectric current …
In many occasions others distributions can be approximated to a Normal distribution
It is the base of statistical inference
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3 Normal distribution
It is characterized by two parameters: the mean, μ, and the standard deviation, σ.
It can take any real value.
Its density function is:
2
2( )2
2
1( ) e2
[ ] [ ]
x
f x x
E X Var X
μσ
πσμ σ
− −
= − ∞ < < ∞
= =
( , )N μ σ
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It is bell shaped and symmetric with respect to the mean
μ
( )f x
3 Normal distribution
0.5 0.5
Mean, median and mode are the same.
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The effect ofThe effect of μμ andand σσ
How does the standard deviation affect the shape of f(x)?σ= 2
σ =3σ =4
μ = 10 μ = 11 μ = 12How does the mean affect the position of f(x)?
3 Normal distribution
It is a scalefactor
It is a translationfactor
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The probability is the areaunder the curve
c dX
f(X)
3 Normal distribution
Pr(c d)X≤ ≤It is not possible to compute the probability of an interval, simply by integrating the density function
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μ
σ
Density of X
Density of de X-μ
0
Density of (X-μ)/σ
1
3 Normal distribution
All Normal distributions can be transformed into N(0,1)
XX Z μσ−
→ =
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3 Normal distribution
~ (3, 2)X N
Pr( 6)X ≤3 6
0 1.56 3Pr Pr( 1.5)
2Z Z−⎛ ⎞≤ = ≤⎜ ⎟
⎝ ⎠
Samearea
~ (0,1)Z N
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3 Normal distribution
The distribution function of the standardized Normal has its ownnotation:
There exists certain boundaries to the function Q, that are used to calculate error boundaries of probabilities in communication systems
( ) Pr( ) ( )( ) ( ) 1 ( )
F x X x xQ x Pt X x x
φφ
= ≤ == > = −
( ) 1 ( )Q x Q x− = −
2
2
2
2
1( ) 02
1( ) 02
x
x
Q x e x
Q x e xxπ
−
−
≤ ≥
< ≥
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3 Normal distribution
The distribution function of the standardized Normal has its ownnotation:
There exists certain boundaries to the function Q, that are used to calculate error boundaries of probabilities in communication systems
( ) Pr( ) ( )( ) ( ) 1 ( )
F x X x xQ x Pt X x x
φφ
= ≤ == > = −
( ) 1 ( )Q x Q x− = −
2
2
2
2
1( ) 02
1( ) 02
x
x
Q x e x
Q x e xxπ
−
−
≤ ≥
< ≥
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3 Normal distribution
Example
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the probability that a semiconductor fails before 6000 hours?
What is the lifetime exceeded by 95.05% of semiconductors?
Pr( 6000)X <
Pr( ) 0.9505X a> =
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3 Normal distribution
Example
6000 7000Pr( 6000) Pr Pr( 1.66)600
X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠
1.66
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the probability that a semiconductor fails before 6000 hours?
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3 Normal distribution
Example
6000 7000Pr( 6000) Pr Pr( 1.66)600
X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠
1.66
1 Pr( 1.66)Z= − ≤
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the probability that a semiconductor fails before 6000 hours?
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1 Pr( 1.66)Z= − <
3 Normal distribution
Example
1 0.95150.0485
= −=
What is the probability that a semiconductor fails before6000 hours?
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3 Normal distribution
Example
7000Pr( ) 0.9505 Pr 0.9505600
aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠
a
0.95
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
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3 Normal distribution
Example
7000Pr( ) 0.9505 Pr 0.9505600
aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠
-b
0.95
-b Negative value
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
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3 Normal distribution
Example
( 7000)Pr( ) 0.9505 Pr 0.9505600
aX a Z − −⎛ ⎞> = → < =⎜ ⎟⎝ ⎠
b
0.95
b
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
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( 7000)Pr( ) Pr 0.9505600
aX a Z − −⎛ ⎞> = < =⎜ ⎟⎝ ⎠
What is the lifetime exceeded by 95% of semiconductors?
Example
3 Normal distribution
( 7000) 1.65600
6010
a
a
− −=
⇓=
95.05% of semiconductorslast more than 6010 hours
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Pr( -0.6 < Z < 1.83 )=
Pr( Z < 1.83 ) - Pr( Z -0.6 )
Pr ( Z <-0.6) =Pr ( Z >0.6 ) =1 - Pr (Z < 0.6 ) =1 – 0.7257 =
0.2743
Pr( Z < 1.83 ) =0.9664
= 0.7257 - 0.0336= 0.6921
1.83-0.6
3 Normal distribution
More Examples of how to compute probabilities
≤
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Normal distribution is important since, although some r.v. do not follow aNormal distribution, some statistics/estimators follow a Normal distribution.
3 Normal distribution
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3 Normal distribution
50 55 60 65 70
010
2030
4050
60
x
Example
Let X be an Uniform r.v. in [50,70].We have a sample of size 2000.
The sample has mean 59.9 andstandard deviation 17
The histogram does not resemble a Normal distribution with the same mean and standard deviation
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We randomly choose groups of 10 observations.
Compute the sample mean for each group
Means of each group are, more or less, similar to the mean of the original variable.
556970
565766
656263
595351
545954
695151
655354
605866
696060
596359
3ª2ª1ªGroups
59.4 58.5 61.1
3 Normal distribution
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55.22075556.160009
57.09926458.038518
58.97777359.917028
60.85628261.795537
62.73479263.674046
64.613301
aa$x
0
10
20
30
40
a
3 Normal distribution
The distribution of the sample means follow,approximately a Normal distribution.
The mean of this new variable is very similar to the mean of the original variable.
The observations of the new variable are closer to each other. The standard deviation is smaller, in this case, it is 1.92
xxx
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Let suppose that we have n in dependent variables Xi with means (μι) and standard deviations (σi) and they follow any distribution.
When n increases,
Central Limit Theorem
3 Normal distribution
2(0,1)i
i
YN
μ
σ
−≈∑
∑
1 2 nY X X X= + + +K
( )2~ ,i iY N μ σ∑ ∑the distribution of
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Central Limit Theorem
3 Normal distribution
It doesn’t matter what we measure, when we average over a large sample, we will have Normal distribution
Let suppose tah we have n in dependent variables Xi withmeans (μι) and standard deviations (σi) and they follow anyDistribution.
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1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
2 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
Probability models
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4 Normal as an approximation of other distributions
The Binomial is teh sum of Bernouilli variable tha take values equal to 0 o 1.
Binomial-Normal
1 2 nY X X X= + +K [ ][ ] (1 )
i
i
E X pVar X p p
=
= −C.L.T.
( ), (1 )Y N np np p≈ − 305
nnpq
>>
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5.000 7.625 10.250 12.875 15.500 18.125 20.750 23.375 26.000x
0.00
0.04
0.08
0.12
4 Normal as an approximation of other distributions
Binomial-Normal
( )15, 10.5N
50 0.310.5
n pnpq
= ==
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4 Normal as an approximation of other distributions
Example
A semiconductors manufacturer admits that 2% of the chips produced are defective. The chips are packed in lots of 2000.
A buyer will reject the lot if there 25 o more defective chips
What is the probability of rejecting the lot? Pr( 25)
25 40Pr6.26
Pr( 2.4)Pr( 2.4) 0.9918
X
Z
ZZ
≥
↓
−⎛ ⎞≥ =⎜ ⎟⎝ ⎠
≥ − =≤ =
~ (2000,0.02)30
40(1 ) 39.2
X Bnnpnp p
>=
− =
(40,6.26)X N≈
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4 Normal as an approximation of other distributions
Poisson-Normal
Poisson distribution appears as the limit of a Binomial distribution when the number of experiments tends to infinity.
We will approximate to a Normal distribution when λ is large (λ > 5)
( )~ ( )
,
X P
X N
λ
λ λ≈
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4 Normal as an approximation of other distributions
Poisson-Normal
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4 Normal as an approximation of other distributions
Example
The number of defects on a metal surface per squared meter follows a Poisson distribution with mean 100.
If we analyze a squared meter of surface. What is the probability of finding 95 or more defects?
Pr( 95)
95 100Pr Pr( 0.5)10
Pr( 0.5) 0.6915
X
Z Z
Z
≥
↓
−⎛ ⎞≥ = ≥ − =⎜ ⎟⎝ ⎠
≤ =