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Continuous probability distributions
 Examples:
 The probability that the average daily temperature in Georgia
during the month of August falls between 90 and 95 degrees is
__________
 The probability that a given part will fail before 1000 hours of use is
__________
 In general, __________
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ETM 620 - 09U
Understanding continuous distributions
 The probability that the average
daily temperature in Georgia
during the month of August falls
between 90 and 95 degrees is
-5
-3
-1
1
3
5
 The probability that a given part
will fail before 1000 hours of
use is
0
2
5
10
15
20
25
30
ETM 620 - 09U
Continuous probability distributions
 The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R

 f ( x)dx  1

P ( a  X  b) 
b
 f ( x)dx
a
 The cumulative distribution, F(x)
x
F ( x )  P(X  x ) 
 f (t)dt

 Expectation and variance
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μ = E(X) = ______________
σ2 = ________________
ETM 620 - 09U
An example
f(x) =
{
x,
2-x,
0,
0<x<1
1≤x<2
elsewhere
1st – what does the function look like?
a) P(X < 1.2) = ___________________
b) P(0.5 < X < 1) = ___________________
c) E(X) = -∞∫∞ x f(x) dx = ________________________
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ETM 620 - 09U
Continuous probability distributions
 Many continuous probability distributions, including:
 Uniform
 Exponential
 Gamma
 Weibull
 Normal
 Lognormal
 Bivariate normal
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Uniform distribution
 Simplest – characterized by the interval endpoints, A and B.
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f ( x; A,B) 
BA
=0
A≤x≤B
elsewhere
 Mean and variance:
AB

2
6
and
2
(
B

A
)
2 
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ETM 620 - 09U
Example
A circuit board failure causes a shutdown of a computing
system until a new board is delivered. The delivery time X is
uniformly distributed between 1 and 5 days.
What is the probability that it will take 2 or more days for
the circuit board to be delivered?
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Gamma & Exponential distributions
 Recall the Poisson Process
 Number of occurrences in a given interval or region
 “Memoryless” process
 Sometimes we’re interested in the time or area until a certain
number of events occur.
 For example
An average of 2.7 service calls per minute are received at a
particular maintenance center. The calls correspond to a Poisson
process.
 What is the probability that up to a minute will elapse before 2 calls arrive?
 How long before the next call?
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ETM 620 - 09U
Gamma distribution
 The density function of the random variable X with gamma
distribution having parameters α (number of occurrences) and β
(time or region, 1/λ).
x
1
f ( x)  
x  1e 
 ( )
x > 0.
( n)  ( n  1)!
Gamma Distribution
1
f(x)
0.8
  
  
2
0.6
0.4
2
0.2
0
0
2
4
6
8
x
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ETM 620 - 09U
Exponential distribution
 Special case of the gamma distribution with α = 1.
f (x) 
1

x
e
x > 0.
 Describes the time until or time between Poisson events.
μ=β
σ2 = β2
0
10
5
10
15
20
25
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ETM 620 - 09U
Example
An average of 2.7 service calls per minute are received at a
particular maintenance center. The calls correspond to a Poisson
process.
What is the probability that up to a minute will elapse before 2
calls arrive?
β = ________
α = ________
P(X ≤ 1) = _________________________________
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ETM 620 - 09U
Example (cont.)
What is the expected time before the next call arrives?
β = ________
α = ________
μ = _________________________________
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ETM 620 - 09U
Another example
 Example 6-6, page 136.
 Note: in Excel, use =GAMMADIST(x,alpha,beta,cumulative) to find the
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probability associated with a specific value of x. Use
=GAMMAINV(probability,alpha,beta) to find the value of x given a
probability.
ETM 620 - 09U
Wiebull Distribution
 Used for many of the same applications as the gamma and
exponential distributions, but
 does not require memoryless property of the exponential
f ( x; ,  )   x
F( x)  1  e
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 1 x 
e
, x0
x 
ETM 620 - 09U
Example
Designers of wind turbines for power generation are interested in
accurately describing variations in wind speed, which in a certain
location can be described using the Weibull distribution with α =
0.02 and β = 2. A designer is interested in determining the
probability that the wind speed in that location is between 3 and 7
mph.
P(3 < X < 7) = ___________________________
 In Excel, =WEIBULL(x,alpha,beta,cumulative)
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ETM 620 - 09U
Normal distribution
 The “bell-shaped curve”
 Also called the Gaussian distribution
 The most widely used distribution in statistical analysis
 forms the basis for most of the parametric tests we’ll perform later
in this course.
 describes or approximates most phenomena in nature, industry, or
research
 Random variables (X) following this distribution are called
normal random variables.
 the parameters of the normal distribution are μ and σ (sometimes μ
and σ2.)
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ETM 620 - 09U
Normal distribution
 The density function of the normal random variable X, with
mean μ and variance σ2, is
1
n( x; , ) 
2 
( x   )2
2
2

e
, all x.
Normal Distribution
P(x)
(μ = 5, σ = 1.5)
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
12
x
 In Excel, =NORMDIST(x,mean,standard_dev,cumulative) and
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=NORMINV(probability,mean,standard_dev)
ETM 620 - 09U
Standard normal RV …
 Note: the probability of X taking on any value between x1 and
x2 is given by:
x2
x2

P( x1  X  x 2 )  n( x; , )dx 
x1

x1
1
2 
( x   ) 2
2
e 2 dx
 To ease calculations, we define a normal random variable
Z
X 

where Z is normally distributed with μ = 0 and σ2 = 1 and,
z
P( Z  z) 
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

x
1
e 2 dx  ( Z )
2
ETM 620 - 09U
Standard normal distribution
 Table II, pg. 601-602: “Cumulative Standard Normal Distribution”
 In Excel, =NORMSDIST(z) and =NORMSINV(probability)
Standard Normal Distribution
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
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Examples
 P(Z ≤ 1) = Φ(1) =
-5
0
5
-5
0
5
-5
0
 P(Z ≥ -1) =
 P(-0.45 ≤ Z ≤ 0.36) =
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Your turn …
 Use Excel to determine (draw the picture!)
1. P(Z ≤ 0.8) =
2. P(Z ≥ 1.96) =
3. P(-0.25 ≤ Z ≤ 0.15) =
4. P(Z ≤ -2.0 or Z ≥ 2.0) =
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The normal distribution “in reverse”

Example:
Given a normal distribution with μ = 40 and σ = 6, find the
value of X for which 45% of the area under the normal curve is
to the left of X.
1)
If Φ(k) = 0.45,
k = ___________
2)
Z = _______
-5
0
5
X = _________
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ETM 620 - 09U
Normal approximation to the binomial
 If n is large and p is not close to 0 or 1,
or
if n is smaller but p is close to 0.5, then
the binomial distribution can be approximated by the normal
distribution using the transformation:
Z
( X  0.5)  np
npq
 NOTE: add or subtract 0.5 from X to be sure the value of interest
is included (draw a picture to know which)
 Look at example 7-10, pg. 156
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ETM 620 - 09U
Another example
The probability of a patient making a full recovery from a rare
heart ailment is 0.4. If 100 people are diagnosed with this
ailment, what is the probability that fewer than 30 of them
recover?
p = 0.4
n = 100
μ = ____________
σ = ______________
if x = 30, then z = _____________________
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and, P(X < 30) = P (Z < _________) = _________
ETM 620 - 09U
Your Turn
Refer to the previous example,

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DRAW THE
PICTURE!!
1.
What is the probability that more than 50 survive?
2.
What is the probability that exactly 45 survive?
ETM 620 - 09U
Lognormal Distribution
• When the random variable Y = ln(X) is normally distributed with
mean μ and standard deviation σ, then X has a lognormal
distribution with the density function,
f ( x; , ) 
 e
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2 x
e
1
2
2
[ln(
x
)


]
2
, x0
  2 /2
 e
2
1

2   2
(e 2  1)
ETM 620 - 09U
Example
The average rate of water usage (in thousands of gallons per hour)
by a certain community is know to involve the lognormal
distribution with parameters μ = 5 and σ =2. What is the
probability that, for any given hour, more than 50,000 gallons are
used?
P(X > 50,000) = __________________________
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ETM 620 - 09U
Bivariate Normal distribution
 Random variables [X, Y] that follow a bivariate normal
distribution have a joint density function,

1
1

f ( x, y ) 
exp 
2
2
1


2 x y 1   2



 x  
x

  x



2
 x   x  y   y

 2 
  x   y
  y y

  y
 




2

 


and a joint probability,
b1 b2
P(a1  X  b1,a2  Y  b2 ) 
 f ( x, y )dxdy
a1 a2
 This would be pretty nasty to calculate, except …
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X ∼ N (μX, σ2X )
and
Y ∼ N (μY , σ2Y )
ETM 620 - 09U
So…
 It can be shown (see pp. 161-162) that,

 y
E(Y x )   y   
 x

x   x 

Var (Y x    2y 1   2

E X y  x   x
 y


 y
y


Var (X y 
 


2
x

1   
2
 This allows us to use the standard normal approach to solve several
important problems.
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ETM 620 - 09U
Example
 Example7-14, pg. 163
Given the information regarding shear strength and weld
diameter in the example, what is the probability that the
strength specification (1080 lbs.) can be met if the weld
diameter is 0.18 in.?
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ETM 620 - 09U