5.2 Continuous Random Variable
Recall Discrete Distribution
• For a discrete distribution, for example
Binomial distribution with n=5, and p=0.4, the
probability distribution is
x
0
1
2
3
4
5
f(x) 0.07776 0.2592 0.3456 0.2304 0.0768 0.01024
A probability histogram
0.3
0.2
P(x)
0.1
0.0
0
1
2
3
x
4
5
How to describe the distribution of a
continuous random variable?
• For continuous random variable, we also represent probabilities
by areas—not by areas of rectangles, but by areas under
continuous curves.
• For continuous random variables, the place of histograms will be
taken by continuous curves.
• Imagine a histogram with narrower and narrower classes. Then
we can get a curve by joining the top of the rectangles. This
continuous curve is called a probability density (or probability
distribution).
Continuous distributions: Density
Function
• For any x, P(X=x)=0. (For a continuous
distribution, the area under a point is 0.)
• Can’t use P(X=x) to describe the probability
distribution of X
• Instead, consider P(a≤X≤b)
Density Function
• A probability density function for a continuous
random variable X is a nonnegative function
f(x) with



f ( x)dx  1
• And such that for all a≤b, one is willing to
assign P[a≤X≤b] according to
b
P[a  X  b]   f ( x)dx
a
Density function
0.20
0.15
0.00
• P(a≤X≤b) is the area
between a and b
0.05
0.10
y
• The area under the
curve is 1
0.25
• A curve f(x):
f(x) ≥ 0
0
2
4
6
x
8
10
0.00
0.05
0.10
y
0.15
0.20
0.25
P(2≤X≤4)= P(2≤X<4)= P(2<X<4)
0
2
4
6
x
8
10
Cumulative Probability function
• For X continuous with probability density f(x)
x
F ( x)  P[ X  x]   f (t )dt

• We can get the density function f(x) from F(x)
by differentiation
d
F ( x)  f ( x)
dx
x
f ( x)  e , x  0
x
F ( x)   e dt  1  e
t
x
0
conversely
d
x
x
f ( x)  (1  e )  e
dx
The normal distribution
• A normal curve: Bell shaped
• Density is given by
2

1
(x  ) 
f ( x) 
exp  

2
2
 2


• μand σ2 are two parameters: mean and variance
of a normal population
(σ is the standard deviation)
0.06
0.04
0.02
0.00
fx
0.08
0.10
0.12
The normal—Bell shaped curve:
μ=100, σ2=10
90
95
100
x
105
110
0.2
0.1
0.0
fx1
0.3
0.4
Normal curves:
(μ=0, σ2=1) and (μ=5, σ 2=1)
-2
0
2
4
x
6
8
Normal curves:
0.2
0.1
0.0
y
0.3
0.4
(μ=0, σ2=1) and (μ=0, σ2=2)
-3
-2
-1
0
x
1
2
3
Normal curves:
0.0
0.2
0.4
fx1
0.6
0.8
1.0
(μ=0, σ2=1) and (μ=2, σ2=0.25)
-2
0
2
4
6
8
0.2
0.1
0.0
y
0.3
0.4
The standard normal curve:
μ=0, and σ2=1
-3
-2
-1
0
x
1
2
3
Table B.3 gives probabilities for standard normal (Numerical Integration – No
formula for
From Table B.3
= 0.9332
If
, then
Example:
Example (p323 #7)
• In a grinding operation, there is an upper
specification of 3.15 in. on a dimension of a
certain part after grinding. Suppose that the
standard deviation of this normally distributed
dimension for parts of this type ground to any
particular mean dimension μ is σ=.002 in.
Suppose further that you desire to have no more
than 3% of the parts fail to meet specifications.
What is the maximum μ (minimum machining
cost) that can be used if this 3% requirement is to
be met?
200
200
y2
50
100
150
150
0
y
100
50
0
3.140
3.142
3.144
3.146
x
3.148
3.150
3.152
3.142
3.144
3.146
3.148
x2
3.150
3.152
3.15
200
150
100
50
y
P( X  3.15)  0.03
0
So 3.15 is 1.88 σ above the mean.
3.15-1.88*0.002=3.146
3.140
3.142
3.144
3.146
3.148
3.150
3.152
0.4
x
0.1
0.2
P( Z  1.88)  0.03
0.0
fz
0.3
P( Z  _____)  0.03
-3
-2
-1
0
z
1
2
3
Exponential distribution
• The exponential distribution is a continuous
probability distribution with
f ( x) 
1

e
 x /
,x 0
• Exponential distributions are often used to
describe waiting times until occurrence of
events.
Density curves
1 x/2
f ( x)  e , x  0
2
x
1.0
0.0
0.6
0.4
0.0
0.2
0.2
0.4
y
dexp(x, 0.5)
0.6
0.8
0.8
1.0
f ( x)  e , x  0
0
1
2
3
x
alpha=1
4
5
0
1
2
3
x
alpha=2
4
5
Mean and Variance
• Mean of an exponential distribution is

1
0

  E( X )   x
e  x /  dx  
• Variance of the exponential distribution is

  Var ( X )   ( x   )
2
0
2
1

e  x /  dx   2
Cumulative Probability Function
0.8
0.6
0.2
0.4
pexp(x, 0.5)
0.6
0.4
0.2
0.0
0.0
pexp(x, 1)
0.8
1.0
1.0
F ( x)  1  e  x / 
0
1
2
3
x
alpha=1
4
5
0
1
2
3
x
alpha=2
4
5
2
P( X  2)  F (2)  1  e  0.8647
P(X<2) on density curve f(x)
When alpha=1
0.0
0.6
0.4
0.0
0.2
0.2
0.4
y
pexp(x, 1)
0.6
0.8
0.8
1.0
1.0
P(X<2) on CDF F(x)
0
1
2
3
x
4
5
0
1
2
3
x
alpha=1
4
5
Relationship between Exponential Distribution and Poisson
Ships arrive 2/ hour. The number of ships arriving in 1 hour is a Poisson
random variable with λ=2
Let’s define a new random variable X to be the waiting time until the first ship.
P(X≥x)=P(0 ships by time x)
F(x)=P(X≤x) =1-P(0 ships by time x)
An exponential distribution with mean
f ( x) 
has
1

e x /
As above with
Memoryless property: If we have already waited H hours and haven’t seen a ship,
our expected waiting time is
. Our expected waiting time is like starting all
over again. The probability of a ship showing up in the next 5 minutes is the same
as when we started.
Force of Mortality Function
• The force of Mortality Function is (p.760):
h(t ) 
f (t )
f (t )

P(T  t ) 1  F (t )
• H(t)dt is the probability of dying in time t to
t+dt if we are still living in t.
• For exponential distribution
1
e x / 
f (t )
f (t )
1

h(t ) 



 x /
P(T  t ) 1  F (t ) 1  (1  e ) 
• So the exponential distribution has a constant
force-of-mortality.
• If the lifetime of an engineering component is
described using a constant force of mortality,
there is no (mathematical) reason to replace
such a component before it fails.
• The distribution of its remaining life from any
point in time is the same as the distribution of
the time till failure of a new component of the
same type.
The geometric distribution is also memoryless. X = time to
success. The
expected tosses to next head doesn’t depend on how long we have been tossing
without getting a head.
Lifetimes of glasses in a restaurant might be exponential. Motor lifetimes or
people’s lifetimes are not. Sometimes lifetimes can be modeled with a lognormal
distribution where
Section 5.2.3 Weibull Distribution
Very commonly lifetimes of motors, etc. are
modeled with Weibull distributions. A Weibull
distribution is a generalization of an
exponential distribution and provides more
flexibility in terms of distributional shape.
For Weibull distribution
F ( x)  1  e
 ( x /  )
d
  1  ( x /  )
f ( x) 
F ( x)   x e
dx

1
  E ( X )  (1  )

Gamma function
 For
the force-of-mortality is a decreasing function, for example, a
product break in a period.

Constant force-of-mortality. Exponential distribution

Increasing force-of-mortality, for example, a product that wears out.
• For component with increasing force-ofmortality (IFM) distribution, such components
are retired from service once they reach a
particular age, even if they have not failed.
Exercise
The lifetime of a certain type of battery has an
exponential distribution with average lifetime
100 hours. 5 batteries are installed at the
same time and suppose that the operations of
the batteries are independent. Find the
probability that only 2 batteries are still
working after 50 hours.