Introduction to the Continuous Distributions. The Uniform Distribution

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

, for a( )

0, otherwise

c y bf y

• To be a valid density function, it follows that

1c

b a

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

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

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) = and V(Y) = .

2 2( ) /(2 )1( )

2yf y e

Standard Normal Curve

• For the standard normal distribution E(Y) = 0 and V(Y) = 1.

2 / 21( )

2yf y e

Normal Probabilities

• For finding probabilities, we compute

2 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?)

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 = 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%

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?

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%

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 = 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?

Exponential Distribution

A special case of the Gamma Distribution

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?

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 10!

www e

p e

Sometimes writtenwhere = 1/ is the average inter-arrival time(e.g., “minutes per arrival”).

/( ) 1 wF w e

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

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.

0.2 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

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 that

2 2( ) 2 , and in general,

( ) ( !)n n

E W

E W n

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-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

ee

eP W b

The only continuous memoryless random variable.

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 ey

f y

1

0( ) yy e dy

For the exponential, choose = 1 and note (1) = 1.

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

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.

Recognize the distribution

• Find E(Y) and V(Y) by inspectiongiven that

2 24 , 0( )

0, otherwise

yy e yf y

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 ey

f y v

2( ) ( / 2)(2) , ( ) 2E Y v v V Y v