Download Familiar Continuous Distributions.

Introduction to the
Continuous Distributions
The Uniform Distribution
Equally Likely
• If Y takes on values in an interval (a, b) such that
any of these values is equally likely, then
c, for a  y  b
f ( y)  
 0, otherwise
• To be a valid density function, it follows that
1
c
ba
Uniform Distribution
• A continuous random variable has a uniform
distribution if its probability density function is
given by
 1
, for a  y  b

f ( y)   b  a
 0, otherwise
Uniform Mean, Variance
• Upon deriving the expected value and variance for
a uniformly distributed random variable, we find
ab
E (Y ) 
2
is the midpoint of the interval
(b  a ) 2
and V (Y ) 
12
Example
• Suppose the round-trip times for deliveries from a
store to a particular site are uniformly distributed
over the interval 30 to 45 minutes.
• Find the probability the delivery time exceeds 40
minutes.
• Find the probability the delivery time exceeds 40
minutes, given it exceeds 35 minutes.
• Determine the mean and variance for these
delivery times.
The Normal Distribution
Bell-shaped Density
• The normal random variable has the famous
bell-shaped distribution. The most commonly
used continuous distribution.
• The normal distribution is used to approximate
other distributions (see Central Limit Theorem).
• For a normal distribution with E(Y) = m and
V(Y) = s2.
1
 ( y  m )2 /(2s 2 )
f ( y) 
e
s 2
Standard Normal Curve
• For the standard normal distribution
E(Y) = 0 and V(Y) = 1.
1  y2 / 2
f ( y) 
e
2
Normal Probabilities
• For finding probabilities, we compute
1
P ( a  Y  b) 
s 2

b
a
e
 ( y  m )2 /(2s 2 )
dy
• Or, at least approximate the value numerically
using an algorithm like Simpson’s Rule (Calc. 1?)
Width of Interval
• Find the percentage (or probability) for the
interval 24 < X < 26.4
P(24 < X < 26.4) = 0.4772
Example
• For a normal distribution with m = 30 and s = 2
• find percentage of data which falls in the interval
between 30 and 33.4.
• First, sketch a "bell-curve", centered at 30, and
shade the region of interest.
About 45.5%
Rest of the Half
• Exactly one-half (an area of 0.5) lies below the
mean and half lies above the mean.
• Find the percentage of data which falls in the
interval greater than 33.4.
Example
• Suppose the hours spent studying per week for
students in normally distributed with a mean of 18.4
hours, standard deviation of 2.5.
• What percentage of students study more than 20
hours per week?
The Rest of the Half
If we determine the area above the
mean and less than 20.
The "area in the tail" is
the rest of the half.
50% - 23.89%
= 26.11%
Continuing...
• This time determine the percentage of students
who study less than 22 hours per week?
Backwards?
• For a standard normal distribution, find z
such that P( Z < z ) = 0.8686
• For a normal distribution with m = 5 and
s = 1.5, find b such that P( Y < b ) = 0.8686
• If a soft drink machine fills 16-ounce cups
with an average of 15.5 ounces, what is the
standard deviation given that the cup
overflows 1.5% of the time?
Exponential Distribution
A special case of the
Gamma Distribution
Time till arrival?
• Consider W, the time until the first arrival.
Number of
customers
t
T
• W is a continuous random variable. What
can we say about its probability distribution?
Inter-arrival times
• If the average arrivals per unit time equals l, the
probability that zero arrivals have occurred in the
interval (0, w) is given by the Poisson distribution
F(w) = P(W < w) = 1 – P(W > w)
(l w)0 el w
 1  p(0)  1 
 1  e l w
0!
Sometimes written F ( w)  1  e w / b
where b = 1/ l is the average inter-arrival time
(e.g., “minutes per arrival”).
Exponential Distribution
• A continuous random variable W whose distribution
and density functions are given by
1  e w / b , 0  w  
F ( w)  
otherwise
 0,
and
 1 w/ b
, w0
 e
f ( w)   b
 0, otherwise

is said to have an exponential distribution with
parameter (“average”) b .
Exponential Random Variables
Typical exponential random variables may include:
• Time between arrivals (inter-arrival times)
• Service time at a server (e.g., CPU, I/O device, or
a communication channel) in a queueing network.
• Time to failure (“lifetime”) of a component.
0.4
0.2 arrivals per minute
dgamma( x 2.5) 0.2
0
0
2
Distributions
for W,
time4 till first arrival:
x
cumulative distribution
lambda = 0.2
0.2
1
0.1
0.5
0
5
10
15
0
5
10
15

E (W )   w(0.2e0.2 w )dw  5 minutes
0
( using integration-by-parts )
As expected, since average time is 1/0.2 = 5 minutes/arrival.
Exponential mean, variance
• If W is an exponential random variable with
parameter b the expected value and variance
for W are given by
E (W )  b and V (W )  b
Also, note that
E (W 2 )  2b 2 , and in general,
E (W n )  (n!) b n
2
CO concentrations
• Air samples in a city have CO concentrations that
are exponentially distributed with mean 3.6 ppm.
• For a randomly selected sample, find the
probability the concentration exceeds 9 ppm.
• If the city manages its traffic such that the mean
CO concentration is reduced to 2.5 ppm, then what
is the probability a sample exceeds 9 ppm?
Memoryless
• Note P(W > w) = 1 – P(W < w)
= 1 – (1 – e-lw) = e-lw
• Consider the conditional probability
P(W > a + b | W > a ) = P(W > a + b)/P(W > a)
• We find that
P(W  a  b | W  a )
The only
continuous
memoryless
random variable.
e  l ( a b )
 lb
 la  e
e
 P(W  b)
Gamma Distribution
• The exponential distribution is a special case of the
more general gamma distribution:
 y 1e y / b
, 0 y
 
f ( y )   b ( )
 0,
otherwise

where the gamma function is

( )   y 1e y dy
0
For the exponential, choose  = 1 and note (1) = 1.
1
dexp ( x 0.2)
Gamma Density Curves
dexp ( x 0.5) 0.5
dexp ( x 1)
the shape parameter, 
0
5
10
Gamma functionx facts:
1
(1)  1;
( )  (  1)(  1),   1;
dgamma( x 1)
dgamma( x 2) 0.5
dgamma( x 3)
(n)  (n  1)!, n  Z .
0
15
0
2
4
x
1
0.5
Exponential mean, variance
• If Y has a gamma distribution with parameters
 and b the expected value and variance for
Y are given by
E (Y )  b and V (Y )  b
2
In the case of  = 1, the values for the
exponential distribution result.
Recognize the distribution
• Find E(Y) and V(Y) by inspection
given that
4 y 2e 2 y , 0  y  
f ( y)  
otherwise
 0,
Chi-Square Distribution
• As another special case of the gamma distribution,
consider letting  = v/2 and b = 2, for some positive
integer v.
v / 21  y / 2
 y
e
, 0 y
 v/2
f ( y )   2 (v / 2)
 0,
otherwise

This defines the Chi-square distribution. Note the
mean and variance are given by
E (Y )  (v / 2)(2)  v, V (Y )  b  2v
2