Introduction to the Continuous Distributions
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Transcript of Introduction to the Continuous Distributions
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Introduction to the Continuous Distributions
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The Uniform Distribution
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Equally Likely
• If Y takes on values in an interval (a, b) such that any of these values is equally likely, then
, for a( )
0, otherwisec y b
f y
• To be a valid density function, it follows that1c
b a
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Uniform Distribution
• A continuous random variable has a uniform distribution if its probability density function is given by
1 , for a( )
0, otherwise
y bf y b a
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Uniform Mean, Variance
• Upon deriving the expected value and variance for a uniformly distributed random variable, we find
( )2
a bE Y
is the midpoint of the interval
2( )( )12
b aV Y and
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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.
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The Normal Distribution
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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) = and V(Y) = .
2 2( ) /(2 )1( )2
yf y e
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Standard Normal Curve
• For the standard normal distribution E(Y) = 0 and V(Y) = 1.
2 / 21( )2
yf y e
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Normal Probabilities
• For finding probabilities, we compute2 2( ) /(2 )1( )
2
b y
aP a Y b e dy
• Or, at least approximate the value numerically using an algorithm like Simpson’s Rule (Calc. 1?)
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Width of Interval• Find the percentage (or probability) for the
interval 24 < X < 26.4
P(24 < X < 26.4) = 0.4772
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Example• For a normal distribution with = 30 and = 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%
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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.
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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?
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If we determine the area above the mean and less than 20.
The Rest of the Half
The "area in the tail" is
the rest of the half.
50% - 23.89% = 26.11%
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Continuing...• This time determine the percentage of students
who study less than 22 hours per week?
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Backwards?
• For a standard normal distribution, find z such that P( Z < z ) = 0.8686
• For a normal distribution with = 5 and = 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?
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Exponential Distribution
A special case of the Gamma Distribution
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Time till arrival?
• Consider W, the time until the first arrival.Number of customers
Tt
• W is a continuous random variable. What can we say about its probability distribution?
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Inter-arrival times• If the average arrivals per unit time equals , 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)0( )1 (0) 1 1
0!
www ep e
Sometimes writtenwhere = 1/ is the average inter-arrival time(e.g., “minutes per arrival”).
/( ) 1 wF w e
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Exponential Distribution• A continuous random variable W whose distribution
and density functions are given by
and
is said to have an exponential distribution with parameter (“average”)
/1 , 0( )
0, otherwise
we wF w
/1 , 0( )
0, otherwise
we wf w
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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.
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0.2 arrivals per minute
0 5 10 15
0.1
0.2
lambda = 0.2
0 2 40
0.2
0.4
dgamma x 2.5( )
xcumulative distribution
0 5 10 15
0.5
1
0.2
0( ) (0.2 )wE W w e dw
As expected, since average time is 1/0.2 = 5 minutes/arrival.
( using integration-by-parts )
Distributions for W, time till first arrival:
5 minutes
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Exponential mean, variance
• If W is an exponential random variable with parameter the expected value and variance for W are given by
2( ) and ( )E W V W
Also, note that2 2( ) 2 , and in general,
( ) ( !)n n
E W
E W n
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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?
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Memoryless
• Note P(W > w) = 1 – P(W < w) = 1 – (1 – e-w) = e-w
• Consider the conditional probability P(W > a + b | W > a ) = P(W > a + b)/P(W > a)
• We find that
( )
( | )
( )
a bb
a
P W a b W a
e ee
P W b
The only continuous memoryless random variable.
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Gamma Distribution• The exponential distribution is a special case of the more
general gamma distribution:
where the gamma function is
1 /
, 0( ) ( )
0, otherwise
yy e yf y
1
0( ) yy e dy
For the exponential, choose = 1 and note (1) = 1.
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Gamma Density Curves
0 5 10 15
0.5
1
dexp x 0.2( )
dexp x 0.5( )
dexp x 1( )
x
0 2 40
0.5
1
dgamma x 1( )
dgamma x 2( )
dgamma x 3( )
x
the shape parameter,
0 5 10 15
0.5
1
Gamma function facts:(1) 1;( ) ( 1) ( 1), 1;( ) ( 1)!, .n n n Z
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Exponential mean, variance
• If Y has a gamma distribution with parameters and the expected value and variance for Y are given by
2( ) and ( )E Y V Y
In the case of = 1, the values for the exponential distribution result.
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Recognize the distribution
• Find E(Y) and V(Y) by inspectiongiven that
2 24 , 0( )
0, otherwise
yy e yf y
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Chi-Square Distribution• As another special case of the gamma distribution, consider letting = v/2 and = 2, for some positive integer v.
This defines the Chi-square distribution. Note the mean and variance are given by / 2 1 / 2
/ 2 , 0( ) 2 ( / 2)
0, otherwise
v y
v
y e yf y v
2( ) ( / 2)(2) , ( ) 2E Y v v V Y v