Continuous Random Variables and Probability Distributions
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Transcript of Continuous Random Variables and Probability Distributions
1Chapter 4 Title and Outline
4Continuous Random Variables and Probability Distributions
4-1 Continuous Random Variables4-2 Probability Distributions and Probability Density Functions4-3 Cumulative Distribution Functions4-4 Mean and Variance of a Continuous Random Variable4-5 Continuous Uniform Distribution 4-6 Normal Distribution
4-7 Normal Approximation to the Binomial and Poisson Distributions4-8 Exponential Distribution4-9 Erlang and Gamma Distributions4-10 Weibull Distribution4-11 Lognormal Distribution4-12 Beta Distribution
CHAPTER OUTLINE
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Learning Objectives for Chapter 4After careful study of this chapter, you should be able to do the following:1. Determine probabilities from probability density functions.2. Determine probabilities from cumulative distribution functions, and cumulative
distribution functions from probability density functions, and the reverse.3. Calculate means and variances for continuous random variables.4. Understand the assumptions for some common continuous probability
distributions.5. Select an appropriate continuous probability distribution to calculate probabilities
for specific applications.6. Calculate probabilities, determine means and variances for some common
continuous probability distributions.7. Standardize normal random variables.8. Use the table for the cumulative distribution function of a standard normal
distribution to calculate probabilities.9. Approximate probabilities for some binomial and Poisson distributions.
Chapter 4 Learning Objectives 2
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Continuous Random Variables
3
The dimensional length of a manufactured part is subject to small variations in measurement due to vibrations, temperature fluctuations, operator differences, calibration, cutting tool wear, bearing wear, and raw material changes.
This length X would be a continuous random variable that would occur in an interval (finite or infinite) of real numbers.
The number of possible values of X, in that interval, is uncountably infinite and limited only by the precision of the measurement instrument.
Sec 4-1 Continuos Radom Variables
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Continuous Density FunctionsDensity functions, in contrast to mass functions,
distribute probability continuously along an interval.The loading on the beam between points a & b is the
integral of the function between points a & b.
4Sec 4-2 Probability Distributions & Probability Density Functions
Figure 4-1 Density function as a loading on a long, thin beam. Most of the load occurs at the larger values of x.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
A probability density function f(x) describes the probability distribution of a continuous random variable. It is analogous to the beam loading.
Sec 4-2 Probability Distributions & Probability Density Functions 5
Figure 4-2 Probability is determined from the area under f(x) from a to b.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Probability Density Function
Sec 4-2 Probability Distributions & Probability Density Functions 6
For a continuous random variable , a is a function such that
(1) 0 means that the function is always non-negative.
(2)
probability density functio
( ) 1
(3)
n
area
X
f x
f x dx
P a X b f x dx
under from to
(4) 0 means there is no area exactly at .
b
a
f x dx a b
f x x
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
HistogramsA histogram is graphical display of data showing a series of adjacent
rectangles. Each rectangle has a base which represents an interval of data values. The height of the rectangle creates an area which represents the relative frequency associated with the values included in the base.
A continuous probability distribution f(x) is a model approximating a histogram. A bar has the same area of the integral of those limits.
Sec 4-2 Probability Distributions & Probability Density Functions 7
Figure 4-3 Histogram approximates a probability density function.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Area of a Point
Sec 4-2 Probability Distributions & Probability Density Functions 8
1 2
1 2 1 2 1 2 1 2
1 2
If is a continuous random variable, for any and , (4-2)
which implies that 0.
From another perspective:As approaches , the area or probability beco
X x xP x X x P x X x P x X x P x X x
P X x
x x
1 2
mes smaller and smaller. As becomes , the area or probability becomes zero. x x
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-1: Electric CurrentLet the continuous random variable X denote the
current measured in a thin copper wire in milliamperes (mA). Assume that the range of X is 0 ≤ x ≤ 20 and f(x) = 0.05. What is the probability that a current is less than 10mA?
Answer:
Sec 4-2 Probability Distributions & Probability Density Functions 9
10
0
20
5
10 0.5 0.5
Another example,
5 20 0.5 0.75
P X dx
P X dx
Figure 4-4 P(X < 10) illustrated.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-2: Hole DiameterLet the continuous random variable X denote the diameter of a
hole drilled in a sheet metal component. The target diameter is 12.5 mm. Random disturbances to the process result in larger diameters. Historical data shows that the distribution of X can be modeled by f(x)= 20e-20(x-12.5), x ≥ 12.5 mm. If a part with a diameter larger than 12.60 mm is scrapped, what proportion of parts is scrapped?
Answer:
Sec 4-2 Probability Distributions & Probability Density Functions 10
20 12.5
12.6
Figure 4-5 12.60 20 0.135xP X e dx
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Cumulative Distribution Functions
Sec 4-3 Cumulative Distribution Functions 11
The of a continuous random
cumulative distribution variable is,
for (4-
function
3)x
X
F x P X x f u du x
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-3: Electric CurrentFor the copper wire current measurement in
Exercise 4-1, the cumulative distribution function (CDF) consists of three expressions to cover the entire real number line.
Sec 4-3 Cumulative Distribution Functions 12
0 x < 0 F (x ) = 0.05x 0 ≤ x ≤ 20
1 20 < x
Figure 4-6 This graph shows the CDF as a continuous function.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-4: Hole Diameter
For the drilling operation in Example 4-2, F(x) consists of two expressions. This shows the proper notation.
Sec 4-3 Cumulative Distribution Functions 13
20 12.5
12.5
20 12.5
0 for 12.5
20
1 for 12.5
xu
x
F x x
F x e du
e x
Figure 4-7 This graph shows F(x) as a continuous function.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Density vs. Cumulative Functions
• The probability density function (PDF) is the derivative of the cumulative distribution function (CDF).
• The cumulative distribution function (CDF) is the integral of the probability density function (PDF).
Sec 4-3 Cumulative Distribution Functions 14
Given , as long as the derivative exists.
dF xF x f x
dx
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exercise 4-5: Reaction Time
• The time until a chemical reaction is complete (in milliseconds, ms) is approximated by this CDF:
• What is the PDF?
• What proportion of reactions is complete within 200 ms?
Sec 4-3 Cumulative Distribution Functions 15
0.010 for 01 for 0 xx
xF x e
0.01 0.010 0 for 0
1 0.01 for 0 xx xdF x d xf x e edx dx
2200 200 1 0.8647P X F e
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mean & Variance
Sec 4-4 Mean & Variance of a Continuous Random Variable 16
mean
expected value
Suppose is a continuous random variable with probability density function . The or
of , denoted as or , is
(4-
va
4)
The rianc
Xf x
X E X
E X xf x dx
2
22 2 2
2
of , denoted as or , is
The of i
e
standard deviat s ion .
X V X
V X x f x dx x f x dx
X
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-6: Electric Current
For the copper wire current measurement in Exercise 4-1, the PDF is f(x) = 0.05 for 0 ≤ x ≤ 20. Find the mean and variance.
Sec 4-4 Mean & Variance of a Continuous Random Variable 17
2020 2
0 0
203202
0 0
0.05 102
0.05 1010 33.33
3
xE X x f x dx
xV X x f x dx
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mean of a Function of a Random Variable
Sec 4-4 Mean & Variance of a Continuous Random Variable 18
If is a continuous random variable with a probability density function ,
(4-5)
Xf x
E h x h x f x dx
Example 4-7: In Example 4-1, X is the current measured in mA. What is the expected value of the squared current?
20
2 2
0
2020 32 2
0 0
0.050.05 133.33 mA3
E h x E X x f x dx
xx dx
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-8: Hole Diameter
For the drilling operation in Example 4-2, find the mean and variance of X using integration by parts. Recall that f(x) = 20e-20(x-12.5)dx for x ≥ 12.5.
Sec 4-4 Mean & Variance of a Continuous Random Variable 19
20 12.5
12.5 12.5
20 12.520 12.5
12.5
2 2
12.5
20
12.5 0.05 12.55 mm20
12.55 0.0025 mm and 0.05 mm
x
xx
E X xf x dx x e dx
exe
V X x f x dx
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Continuous Uniform Distribution• This is the simplest continuous distribution
and analogous to its discrete counterpart.• A continuous random variable X with
probability density functionf(x) = 1 / (b-a) for a ≤ x ≤ b (4-
6)
Sec 4-5 Continuous Uniform Distribution 20
Figure 4-8 Continuous uniform PDF
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mean & Variance
• Mean & variance are:
• Derivations are shown in the text. Be reminded that b2 - a2 = (b + a)(b - a)
Sec 4-5 Continuous Uniform Distribution 21
22 and (4-7)2 12a b b a
E X V X
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-9: Uniform CurrentLet the continuous random variable X denote the
current measured in a thin copper wire in mA. Recall that the PDF is F(x) = 0.05 for 0 ≤ x ≤ 20.
What is the probability that the current measurement is between 5 & 10 mA?
Sec 4-5 Continuous Uniform Distribution 22
10
5
5 10 0.05 5 0.05 0.25P x dx
Figure 4-9
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Continuous Uniform CDF
Sec 4-5 Continuous Uniform Distribution 23
1
The CDF is completely described as0
1
x
a
x aF x dub a b a
x aF x x a b a a x b
b x
Figure 4-6 (again) Graph of the Cumulative Uniform CDF
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Distribution• The most widely used distribution is the normal
distribution, also known as the Gaussian distribution.• Random variation of many physical measurements are
normally distributed.• The location and spread of the normal are independently
determined by mean (μ) and standard deviation (σ).
Sec 4-6 Normal Distribution 24
Figure 4-10 Normal probability density functions
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Probability Density Function
Sec 4-6 Normal Distribution 25
2
22
A random variable with probability density function
1 -
normal random variable
(4-8)2
is a with parameters , where - , and 0. Also,
x
X
f x e x
E X
2
2
and (4-9)
and the notation , is used to denote the distribution.
Note that f cannot be intergrated analytically, so F is expressed through numerical in
V X
N
XX
tegration with Excel or Minitab, and written as Appendix A, Table III.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-10: Normal ApplicationAssume that the current measurements in a strip of wire
follows a normal distribution with a mean of 10 mA & a variance of 4 mA2. Let X denote the current in mA.
What is the probability that a measurement exceeds 13 mA?
Sec 4-6 Normal distribution 26
Figure 4-11 Graphical probability that X > 13 for a normal random variable with μ = 10 and σ2 = 4.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Empirical RuleP(μ – σ < X < μ + σ) = 0.6827
P(μ – 2σ < X < μ + 2σ) = 0.9545P(μ – 3σ < X < μ + 3σ) = 0.9973
Sec 4-6 Normal Distribution 27
Figure 4-12 Probabilities associated with a normal distribution – well worth remembering to quickly estimate probabilities.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Standard Normal Distribution
A normal random variable withμ = 0 and σ2 = 1
Is called a standard normal random variable and is denoted as Z. The cumulative distribution function of a standard normal random variable is denoted as:
Φ(z) = P(Z ≤ z) = F(z)Values are found in Appendix Table III and by
using Excel and Minitab.
Sec 4-6 Normal Distribution 28
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-11: Standard Normal Distribution
Assume Z is a standard normal random variable.Find P(Z ≤ 1.50). Answer: 0.93319
Find P(Z ≤ 1.53). Answer: 0.93699Find P(Z ≤ 0.02). Answer: 0.50398
Sec 4-6 Normal Distribution 29
Figure 4-13 Standard normal PDF
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-12: Standard Normal Exercises
1. P(Z > 1.26) = 0.1038
2. P(Z < -0.86) = 0.195
3. P(Z > -1.37) = 0.915
4. P(-1.25 < 0.37) =
0.5387
5. P(Z ≤ -4.6) ≈ 0
6. Find z for P(Z ≤ z) =
0.05, z = -1.65
7. Find z for (-z < Z < z) =
0.99, z = 2.58
Sec 4-6 Normal Distribution 30
Figure 4-14 Graphical displays for standard normal distributions.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Standardizing
Sec 4-6 Normal Distribution 31
2Suppose is a normal random variable with mean and variance .
Then, (4-11)
where is a standard normal random variabl , and
is the z-value a
e
obt
XX xP X x P P Z z
Z
xz
inedby X.
The probability is obtained by using Appendix Table III
standa
with
rdizin
.
g
xz
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-14: Normally Distributed Current-1From a previous example with
μ = 10 and σ = 2 mA, what is the probability that the current measurement is between 9 and 11 mA?
Answer:
Sec 4-6 Normal Distribution 32
9 10 10 11 109 112 2 2
0.5 0.5
0.5 0.50.69146 0.30854 0.38292
xP X P
P z
P z P z
Figure 4-15 Standardizing a normal random variable.
0.38292 = NORMDIST(11,10,2,TRUE) - NORMDIST(9,10,2,TRUE)Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-14: Normally Distributed Current-2
Determine the value for which the probability that a current measurement is below this value is 0.98.
Answer:
Sec 4-6 Normal Distribution 33
14.107 = NORMINV(0.98,10,2)Using Excel
10 102 2
10 0.982
2.05 is the closest value.2 2.05 10 14.1 mA.
X xP X x P
xP Z
zz
Figure 4-16 Determining the value of x to meet a specified probability.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-15: Signal Detection-1Assume that in the detection of a digital signal, the background
noise follows a normal distribution with μ = 0 volt and σ = 0.45 volt. The system assumes a signal 1 has been transmitted when the voltage exceeds 0.9. What is the probability of detecting a digital 1 when none was sent? Let the random variable N denote the voltage of noise.
Sec 4-6 Normal Distribution 34
0 0.90.9 20.45 0.45
1 0.97725 0.02275
NP N P P Z
0.02275 = 1 - NORMDIST(0.9,0,0.45,TRUE)Using Excel
This probability can be described as the probability of a false detection.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-15: Signal Detection-2Determine the symmetric bounds about 0 that include 99% of all
noise readings. We need to find x such that P(-x < N < x) = 0.99.
Sec 4-6 Normal Distribution 35
0.45 0.45 0.45
2.58 2.580.45 0.45
2.58 0.45 0 1.16
x N xP x N x P
x xP Z P Z
x
Figure 4-17 Determining the value of x to meet a specified probability.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-15: Signal Detection-3Suppose that when a digital 1 signal is transmitted, the mean of
the noise distribution shifts to 1.8 volts. What is the probability that a digital 1 is not detected? Let S denote the voltage when a digital 1 is transmitted.
Sec 4-6 Normal Distribution 36
1.8 0.9 1.80.90.45 0.45
2 0.02275
SP S P
P Z
0.02275 = NORMDIST(0.9, 1.8, 0.45, TRUE)Using Excel
This probability can be interpreted as the probability of a missed signal.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-16: Shaft Diameter-1The diameter of the shaft is normally distributed with μ = 0.2508
inch and σ = 0.0005 inch. The specifications on the shaft are 0.2500 ± 0.0015 inch. What proportion of shafts conform to the specifications? Let X denote the shaft diameter in inches.
Answer:
Sec 4-6 Normal Distribution 37
0.2485 0.2515
0.2485 0.2508 0.2515 0.25080.0005 0.0005
4.6 1.4
1.4 4.6
0.91924 0.0000 0.91924
P X
P Z
P Z
P Z P Z
0.91924 = NORMDIST(0.2515, 0.2508, 0.0005, TRUE) - NORMDIST(0.2485, 0.2508, 0.0005, TRUE)Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-16: Shaft Diameter-2Most of the nonconforming shafts are too large, because the process
mean is near the upper specification limit. If the process is centered so that the process mean is equal to the target value, what proportion of the shafts will now conform?
Answer:
Sec 4-6 Normal Distribution 38
0.2485 0.2515
0.2485 0.2500 0.2515 0.25000.0005 0.0005
3 3
3 30.99865 0.00135 0.99730
P X
P Z
P Z
P Z P Z
0.99730 = NORMDIST(0.2515, 0.25, 0.0005, TRUE) - NORMDIST(0.2485, 0.25, 0.0005, TRUE)Using Excel
By centering the process, the yield increased from 91.924% to 99.730%, an increase of 7.806%
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Approximations
• The binomial and Poisson distributions become more bell-shaped and symmetric as their means increase.
• For manual calculations, the normal approximation is practical – exact probabilities of the binomial and Poisson, with large means, require technology (Minitab, Excel).
• The normal is a good approximation for the:– Binomial if np > 5 and n(1-p) > 5.– Poisson if λ > 5.
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 39
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Approximation to the Binomial
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 40
Suppose we have a binomial distribution with n = 10 and p = 0.5. Its mean and standard deviation are 5.0 and 1.58 respectively.
Draw the normal distribution over the binomial distribution.
The areas of the normal approximate the areas of the bars of the binomial with a continuity correction.
Figure 4-19 Overlaying the normal distribution upon a binomial with matched parameters.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-17: In a digital comm channel, assume that the number of bits
received in error can be modeled by a binomial random variable. The probability that a bit is received in error is 10-5. If 16 million bits are transmitted, what is the probability that 150 or fewer errors occur? Let X denote the number of errors.
Answer:
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 41
150 1600000016000000 5 5
0
150 10 1 10x x
xx
P X C
0.2280 = BINOMDIST(150,16000000,0.00001,TRUE)Using Excel
Can only be evaluated with technology. Manually, we must use the normal approximation to the binomial.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Approximation Method
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 42
If is a binomial random variable with parameters and ,
(4-12)1
is approximately a standard normal rand
X n pX npZnp p
continuity correctiom variable. To approximate a binomial
probability with a normal distribution, a is applied as follows:
0.50.51
and
0.50
o
.
n
51
x npP X x P X x P Znp p
x npP X x P X x P Znp p
The approximation is good for 5 and 1 5. Refer to Figure 4-19 to
see the rationale for adding and subtracting the 0.5 continuity correction.
np n p
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-18: Applying the Approximation
The digital comm problem in the previous example is solved using the normal approximation to the binomial as follows:
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 43
5 5
150 150.5
160 150.5 160
160 1 10 160 1 10
9.5 0.75104 0.226312.6491
P X P X
XP
P Z P
0.2263 = NORMDIST(150.5, 160, SQRT(160*(1-0.00001)), TRUE)-0.7% = (0.2263-0.228)/0.228 = percent error in the approximation
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-19: Normal Approximation-1Again consider the transmission of bits. To judge how well the
normal approximation works, assume n = 50 bits are transmitted and the probability of an error is p = 0.1. The exact and approximated probabilities are:
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 44
50 50 50 49 50 2 480 1 22 0.9 0.1 0.9 0.1 0.9 0.112
5 2.5 5250 0.1 0.9 50 0.1 0.9
1.18 0.119
P X C C C
XP X P
P Z
0.1117 = BINOMDIST(2,50,0.1,TRUE)0.1193 = NORMDIST(2.5, 5, SQRT(5*0.9), TRUE)
6.8% = (0.1193 - 0.1117) / 0.1117 = percent error
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-19: Normal Approximation-2
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 45
8 9 8.5
8.5 5 1.65 0.052.12
5 4.5 5.5
4.5 5 5.5 52.12 2.120.24 0.24
0.24 0.24 0.19
P X P X P X
P Z P Z
P X P X
P Z
P Z
P Z P Z
0.1849 = BINOMDIST(5,50,0.1,FALSE)0.1863 = NORMDIST(5.5, 5, SQRT(5*0.9), TRUE) - NORMDIST(4.5, 5, SQRT(5*0.9), TRUE)
0.8% = (0.1863 - 0.1849) / 0.1849 = percent error
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Reason for the Approximation LimitsThe np > 5 and n(1-p) > 5
approximation rule is needed to keep the tails of the normal distribution from getting out-of-bounds.
As the binomial mean approaches the endpoints of the range of x, the standard deviation must be small enough to prevent overrun.
Figure 4-20 shows the asymmetric shape of the binomial when the approximation rule is not met.
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 46
Figure 4-20 Binomial distribution is not symmetric as p gets near 0 or 1.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Approximation to Hypergeometric
Recall that the hypergeometric distribution is similar to the binomial such that p = K / N and when sample sizes are small relative to population size.
Thus the normal can be used to approximate the hypergeometric distribution also.
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 47
hypergeometric ≈ binomial ≈ normaldistribution distribution distribution
n / N < 0.1 np < 5n (1-p ) < 5
Figure 4-21 Conditions for approximatine hypergeometric and binomial with normal probabilities
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Normal Approximation to the Poisson
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 48
If is a Poisson random variable with and ,
(4-13)
is approximately a standard normal random variable. The same
X E X V X
XZ
continuity correction used for the binomial distribution can also be applied. The approximation is good for 5
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-20: Normal Approximation to PoissonAssume that the number of asbestos particles in a square meter
of dust on a surface follows a Poisson distribution with a mean of 100. If a square meter of dust is analyzed, what is the probability that 950 or fewer particles are found?
Sec 4-7 Normal Approximation to the Binomial & Poisson Distributions 49
1000950
0
1000950 ... too hard manually!!
950.5 1000950.51000
1.57 0.058
x
x
eP Xx
P X P Z
P Z
0.0578 = POISSON(950,1000,TRUE)0.0588 = NORMDIST(950.5, 1000, SQRT(1000), TRUE)
1.6% = (0.0588 - 0.0578) / 0.0578 = percent error
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exponential Distribution• The Poisson distribution defined a random variable as the
number of flaws along a length of wire (flaws per mm).• The exponential distribution defines a random variable as the
interval between flaws (mm’s between flaws – the inverse).
Sec 4-8 Exponential Distribution 50
0
Let denote the number of flaws in mm of wire.If the mean number of flaws is per mm,
has a Poisson distribution with mean .
=P 00!
1 , 0, the CDF.
Now differenti
xx
x
X x
N x
e xP X x N e
F x P X x e x
ating:
, 0, the PDF.xf x e x
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exponential Distribution Definition
The random variable X that equals the distance between successive events of a Poisson process with mean number of events λ > 0 per unit interval is an exponential random variable with parameter λ. The probability density function of X is:
f(x) = λe-λx for 0 ≤ x < (4-14)
Sec 4-8 Exponential Distribution 51
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exponential Distribution Graphs
Sec 4-8 Exponential Distribution 52
Figure 4-22 PDF of exponential random variables of selected values of λ.
The y-intercept of the exponential probability density function is λ.
The random variable is non-negative and extends to infinity.
F(x) = 1 – e-λx is well-worth committing to memory – it is used often.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exponential Mean & Variance
Sec 4-8 Exponential Distribution 53
22
If the random variable has an exponential distribution with parameter ,
1 1 and (4-15)
X
E X V X
Note that, for the:• Poisson distribution, the mean and variance are the same.• Exponential distribution, the mean and standard deviation are the same.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-21: Computer Usage-1In a large corporate computer network, user log-ons to the
system can be modeled as a Poisson process with a mean of 25 log-ons per hour. What is the probability that there are no log-ons in the next 6 minutes (0.1 hour)? Let X denote the time in hours from the start of the interval until the first log-on.
Sec 4-8 Exponential Distribution 54
25 0.125
0.1
0.1 25
1 0.1 0.082
xP X e dx e
F
Figure 4-23 Desired probability.0.0821 = 1 - EXPONDIST(0.1,25,TRUE)
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-21: Computer Usage-2
Continuing, what is the probability that the time until the next log-on is between 2 and 3 minutes (0.033 & 0.05 hours)?
Sec 4-8 Exponential Distribution 55
0.0525
0.033
0.0525
0.033
0.033 0.05 25
0.152
0.05 0.033 0.152
x
x
P X e
e
F F
0.148 = EXPONDIST(3/60, 25, TRUE) - EXPONDIST(2/60, 25, TRUE)(difference due to round-off error)
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-21: Computer Usage-3• Continuing, what is the interval of time such that the
probability that no log-on occurs during the interval is 0.90?
• What is the mean and standard deviation of the time until the next log-in?
Sec 4-8 Exponential Distribution 56
25 0.90, 25 ln 0.900.10536 0.00421 hour 0.253 minute
25
xP X x e x
x
1 1 0.04 hour 2.4 minutes25
1 1 0.04 hour 2.4 minutes25
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Characteristic of a Poisson Process
• The starting point for observing the system does not matter.
• The probability of no log-in in the next 6 minutes [P(X > 0.1 hour) = 0.082], regardless of whether:– A log-in has just occurred or– A log-in has not occurred for the last hour.
• A system may have different means:– High usage period , e.g., λ = 250 per hour– Low usage period, e.g., λ = 25 per hour
Sec 4-8 Exponential Distribution 57
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-22: Lack of Memory Property
• Let X denote the time between detections of a particle with a Geiger counter. Assume X has an exponential distribution with E(X) = 1.4 minutes. What is the probability that a particle is detected in the next 30 seconds?
• No particle has been detected in the last 3 minutes. Will the probability increase since it is “due”?
– No, the probability that a particle will be detected depends only on the interval of time, not its detection history.
Sec 4-8 Exponential Distribution 58
0.5 1.40.5 0.5 1 0.30P X F e 0.300 = EXPONDIST(0.5, 1/1.4, TRUE)
Using Excel
3 3.5 3.5 3 0.0353.5 3 0.303 1 3 0.117
P X F FP X X
P X F
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Lack of Memory Property
• Areas A+B+C+D=1• A = P(X < t2)
• A+B+C = P(X<t1+t2)
• C = P(X<t1+t2 X>t1)
• C+D = P(X>t1)
• C/(C+D) = P(X<t1+t2|X>t1)• A = C/(C+D)
Sec 4-8 Exponential Distribution 59
Figure 4-24 Lack of memory property of an exponential distribution.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Exponential Application in Reliability• The reliability of electronic components is often modeled
by the exponential distribution. A chip might have mean time to failure of 40,000 operating hours.
• The memoryless property implies that the component does not wear out – the probability of failure in the next hour is constant, regardless of the component age.
• The reliability of mechanical components do have a memory – the probability of failure in the next hour increases as the component ages. The Weibull distribution is used to model this situation.
Sec 4-8 Exponential Distribution 60
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Erlang & Gamma Distributions
• The Erlang distribution is a generalization of the exponential distribution.
• The exponential models the interval to the 1st event, while the Erlang models the interval to the rth event, i.e., a sum of exponentials.
• If r is not required to be an integer, then the distribution is called gamma.
• The exponential, as well as its Erlang and gamma generalizations, is based on the Poisson process.
Sec 4-9 Erlang & Gamma Distributions 61
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-23: Processor FailureThe failures of CPUs of large computer systems are often
modeled as a Poisson process. Assume that units that fail are repaired immediately and the mean number of failures per hour is 0.0001. Let X denote the time until 4 failures occur. What is the probability that X exceed 40,000 hours?
Let the random variable N denote the number of failures in 40,000 hours. The time until 4 failures occur exceeds 40,000 hours iff the number of failures in 40,000 hours is ≤ 3.
Sec 4-9 Erlang & Gamma Distributions 62
43
0
40,000 3
40,000 0.0001 4 failure in 40,000 hours
43 0.433!
k
k
P X P N
E N
eP Nk
0.433 = POISSON(3, 4, TRUE)
Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Erlang Distribution
Generalizing from the prior exercise:
Sec 4-9 Erlang & Gamma Distributions 63
1
0
1
1!
Now differentiating :
for 0 and 1, 2,...1 !
kxr
k
r r x
e xP X x F x
kF x
x ef x x rr
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Gamma Function
The gamma function is the generalization of the factorial function for r > 0, not just non-negative integers.
Sec 4-9 Erlang & Gamma Distributions 64
1
0
1 2
, for 0 (4-17)
Properties of the gamma function1 1 recursive property
1 ! factorial function
1 0! 1
1 2 1.77 useful if manual
r xr x e dx r
r r r
r r
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Gamma Distribution
The random variable X with a probability density function:
has a gamma random distribution with parameters λ > 0 and r > 0. If r is an positive integer, then X has an Erlang distribution.
Sec 4-9 Erlang & Gamma Distributions 65
1
, for 0 (4-18)r r xx ef x xr
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mean & Variance of the Gamma
• If X is a gamma random variable with parameters λ and r,
μ = E(X) = r / λ and σ2 = V(X) = r / λ2 (4-19)
• r and λ work together to describe the shape of the gamma distribution.
Sec 4-9 Erlang & Gamma Distributions 66
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Gamma Distribution GraphsThe λ and r parameters are often
called the “shape” and “scale”, but may take on different meanings.
Different parameter combinations change the distribution.
The distribution becomes symmetric as r (and μ) increases.
Sec 4-9 Erlang & Gamma Distributions 67
Figure 4-25 Gamma probability density functions for selected values of λ and r.
Name Text Excel MinitabScale λ β = 1 / λ 1 / λShape r α r
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-24: Gamma Application-1The time to prepare a micro-array slide for high-output genomics is a Poisson process with a
mean of 2 hours per slide. What is the probability that 10 slides require more than 25 hours?
Let X denote the time to prepare 10 slides. Because of the assumption of a Poisson process, X has a gamma distribution with λ = ½, r = 10, and the requested probability is P(X > 25).
Using the Poisson distribution, let the random variable N denote the number of slides made in 10 hours. The time until 10 slides are made exceeds 25 hours iff the number of slides made in 25 hours is ≤ 9.
Sec 4-9 Erlang & Gamma Distributions 68
12.59
0
25 9
25 1 2 12.5 slides in 25 hours
12.59 0.2014
!
k
k
P X P N
E N
eP N
k
0.2014 = POISSON(9, 12.5, TRUE)
Using Excel
0.2014 = 1 - GAMMADIST(25,10,2,TRUE)Using Excel
25 10 9 0.5
0
0.525 110
xx eP X dx
Using the gamma distribution, the same result is obtained.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-24: Gamma Application-2What is the mean and standard deviation of the time to
prepare 10 slides?
Sec 4-9 Erlang & Gamma Distributions 69
22
10 20 hours0.5
10V 40 hours0.25
10 40 6.32 hours
rE X
rX
SD X
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-24: Gamma Application-3
Sec 4-9 Erlang & Gamma Distributions 70
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00X
Dens
ity
31.4
0.95
0
Distribution PlotGamma, Shape=10, Scale=2, Thresh=0
The slides will be completed by what length of time with 95% probability? That is: P(X ≤ x) = 0.95
31.41 = GAMMAINV(0.95, 10, 2)Using ExcelMinitab: Graph > Probability Distribution Plot >
View Probability
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Chi-Squared Distribution
• The chi-squared distribution is a special case of the gamma distribution with– λ = 1/2 – r = ν/2 where ν (nu) = 1, 2, 3, …– ν is called the “degrees of freedom”.
• The chi-squared distribution is used in interval estimation and hypothesis tests as discussed in Chapter 7.
Sec 4-9 Erlang & Gamma Distributions 71
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Weibull Distribution
• The Weibull distribution is often used to model the time until failure for physical systems in which failures:– Increase over time (bearings)– Decrease over time (some semiconductors)– Remain constant over time (subject to external
shock)• Parameters provide flexibility to reflect an item’s
failure experience or expectation.
Sec 4-10 Weibull Distribution 72
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Weibull PDF
Sec 4-10 Weibull Distribution 73
1
The random variable with probability density function
for 0 (4-20)
is a Weibull random variable with scale parameter 0 and shape parameter 0.
The c
x
X
f x x e x
2
2 2 2
umulative density function is:
1 (4-21)
11
2 11 1 (4-21a)
xF x e
E X
V X
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Weibull Distribution Graphs
Sec 4-10 Weibull Distribution 74
Figure 4-26 Weibull probability density function for selected values of δ and β.
Added slide
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-25: Bearing Wear• The time to failure (in hours) of a bearing in a mechanical
shaft is modeled as a Weibull random variable with β = ½ and δ = 5,000 hours.
• What is the mean time until failure?
• What is the probability that a bearing will last at least 6,000 hours? (error in text solution)
Sec 4-10 Weibull Distribution 75
5000 1 1 2 5000 1.5
5000 0.5 4,431.1 hours
E X
4,431.1 = 5000 * EXP(GAMMALN(1.5))Using Excel
0.56000
5000
1.0954
6,000 1 6,000
0.334
P X F e
e
0.334 = 1 - WEIBULL(6000, 1/2, 5000, TRUE)Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Lognormal Distribution• Let W denote a normal random variable with mean of θ and
variance of ω2, i.e., E(W) = θ and V(W) = ω2
• As a change of variable, let X = eW = exp(W) and W = ln(X)• Now X is a lognormal random variable.
Sec 4-11 Lognormal Distribution 76
2
2 2 2
ln2
2 2
exp ln
ln x -θ ln x -θ=P Z =Φ for 0
ω ω
0 for 0
1 for 02
and 1
x
F x P X x P W x P W x
x
x
f x e xx
E X e V X e e
(4-22)
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Lognormal Graphs
Sec 4-11 Lognormal Distribution 77
Figure 4-27 Lognormal probability density functions with θ = 0 for selected values of ω2.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-27: Semiconductor Laser-1The lifetime of a semiconductor laser has a lognormal distribution
with θ = 10 and ω = 1.5 hours. • What is the probability that the lifetime exceeds 10,000 hours?
Sec 4-11 Lognormal Distribution 78
10,000 1 exp 10,000
1 ln 10,000
ln 10,000 101
1.5
1 0.5264 0.701
P X P W
P W
1 - NORMDIST(LN(10000), 10, 1.5, TRUE) = 0.701
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-27: Semiconductor Laser-2• What lifetime is exceeded by 99% of lasers?
• What is the mean and variance of the lifetime?
Sec 4-11 Lognormal Distribution 79
exp ln
ln 101 0.99
1.5
1 0.99 therefore 2.33
ln 102.33 and exp 6.505 668.48 hours
1.5
P X x P W x P W x
x
z z
xx
-2.3263 = NORMSINV(0.99)6.5105 = -2.3263 * 1.5 + 10 = ln(x)672.15 = EXP(6.5105)
(difference due to round-off error)
2 2
2 2 2 2
2 10 1.5 2
2 2 10 1.5 1.5
exp 11.125 67,846.29
1 1
exp 22.25 exp 2.25 1 39,070,059,886.6
197,661.5
E X e e
V X e e e e
SD X
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Beta DistributionA continuous distribution that is flexible, but bounded
over the [0, 1] interval is useful for probability models. Examples are:– Proportion of solar radiation absorbed by a material.– Proportion of the max time to complete a task.
Sec 4-12 Beta Distribution 80
11
The random variable with probability density function
= 1 for 0 1
is a beta random variable with parameters 0 and 0.
X
f x x x x
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Beta Shapes Are Flexible
Distribution shape guidelines:1. If α = β, symmetrical
about x = 0.5.2. If α = β = 1, uniform.3. If α = β < 1, symmetric
& U- shaped.4. If α = β > 1, symmetric
& mound-shaped.5. If α ≠ β, skewed.
Sec 4-12 Beta Distribution 81
Figure 4-28 Beta probability density functions for selected values of the parameters α and β.
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-27: Beta Computation-1Consider the completion time of a large commercial real estate
development. The proportion of the maximum allowed time to complete a task is a beta random variable with α = 2.5 and β = 1. What is the probability that the proportion of the max time exceeds 0.7? Let X denote that proportion.
Sec 4-12 Beta Distribution 82
111
0.7
11.5
0.7
12.5
0.7
2.5
0.7 1
3.5
2.5 1
2.5 1.5 0.52.51.5 0.5 1
1 0.7 0.59
P X x x dx
x dx
x
0.590 = 1 - BETADIST(0.7,2.5,1,0,1)Using Excel
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Example 4-27: Beta Computation-2
Sec 4-12 Beta Distribution 83
This Minitab graph illustrates the prior calculation. FIX
2.5
2.0
1.5
1.0
0.5
0.0X
Dens
ity
0.7
0.590
0 1
Beta: alpha=2.5, beta=1Example 4-28
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mean & Variance of the Beta Distribution
Sec 4-12 Beta Distribution 84
22 1
E X
V X
If X has a beta distribution with parameters α and β,
22 2
2.5 2.5 0.712.5 1 3.5
2.5 1 2.5 0.0453.5 4.52.5 1 2.5 1 1
Example 4-28: In the prior example, α = 2.5 and β = 1. What are the mean and variance of this distribution?
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Mode of the Beta Distribution If α >1 and β > 1, then the beta distribution is mound-shaped and has an interior peak,
called the mode of the distribution. Otherwise, the mode occurs at an endpoint.
Sec 4-12 Beta Distribution 85
General formula:1Mode
2for 0 and 0.
1.00.80.60.40.20.0
2.5
2.0
1.5
1.0
0.5
0.0
X
Dens
ity
11.1
beta
Distribution PlotBeta Distribution: alpha=2.5
case alpha betaExample 4-28 2.25 1 1.00 = (2.5-1) / (2.5+1.0-2)Alternate 2.25 1.1 0.94 = (2.5-1) / (2.5+1.1-2)
mode
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Extended Range for the Beta Distribution
The beta random variable X is defined for the [0, 1] interval. That interval can be changed to [a, b]. Then the random variable W is defined as a linear function of X:
W = a + (b –a)XWith mean and variance:
E(W) = a + (b –a)E(X)V(W) = (b-a)2V(X)
Sec 4-12 Beta Distribution 86
© John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.
Important Terms & Concepts of Chapter 4
Beta distributionChi-squared distributionContinuity correctionContinuous uniform distributionCumulative probability distribution for
a continuous random variableErlang distributionExponential distributionGamma distributionLack of memory property of a
continuous random variableLognormal distributionMean for a continuous random variableMean of a function of a continuous
random variableNormal approximation to binomial &
Poisson probabilitiesNormal distributionProbability density functionProbability distribution of a continuous
random variableStandard deviation of a continuous
random variableStandardizingStandard normal distributionVariance of a continuous random
variableWeibull distribution
Chapter 4 Summary 87