JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 1 Continuous Probability Distributions Many...

JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 1

Continuous Probability Distributions

• Many continuous probability distributions, including: UniformNormalGammaExponentialChi-SquaredLognormalWeibull


Standard Normal Distribution

Standard Normal Distribution

-5 -4 -3 -2 -1 0 1 2 3 4 5

Z

• Table A.3: “Areas Under the Normal Curve”


Applications of the Normal Distribution• A certain machine makes electrical resistors having a mean

resistance of 40 ohms and a standard deviation of 2 ohms. What percentage of the resistors will have a resistance less than 44 ohms?

• Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44

P(X<44) = P(Z< +2.0) = 0.9772

Therefore, we conclude that 97.72% will have a resistance less than 44 ohms. What percentage will have a resistance greater than 44 ohms?

-5 0 5

2

4044

Z

XZ


The Normal Distribution “In Reverse”• Example:

Given a normal distribution with = 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.

• Solution: If P(Z < k) = 0.45, what is the value of k?

45% of area under curve less than kk = -0.125 from table in back of book

Z = (x- μ) / σ

Z = -0.125 = (x-40) / 6

X = 39.25

-5 0 5

-5 0 5

39.25 40

X values

Z values


Normal Approximation to the Binomial• If n is large and p is not close to 0 or 1,

orif n is smaller but p is close to 0.5, then

the binomial distribution can be approximated by the normal distribution using the transformation:

• NOTE: When we apply the theorem, we apply a continuity correction: add or subtract 0.5 from X to be sure the value of interest is included (drawing a picture helps you determine whether to add or subtract

• To find the area under the normal curve to the left of x+.0.5, use:

npq

npXZ

npq

npxZP

05.


Look at example 6.15, pg. 191-192

The probability that a patient recovers is p = 0.4

If 100 people have the disease, what is the probability that less then 30 survive?

n = 100 μ = np =____ σ = sqrt(npq)= ____

if x = 30, then z = _____________ (Don’t forget the continuity correction.)

and, P(X < 30) = P (Z < _________) = _________(Less than 30 for a discrete distribution is 29 or less. Adding .5 to 29 insures we “cover” the maximum value.)


The Normal Approximation: Your Turn

Using μ and σ from the previous example,

1. What is the probability that more than 50 survive?

More than 50 for a discrete distribution is 51 or greater. How do we include 51 and not 50 ?

-5 0 5

DRAW THE PICTURE!!


The Normal Approximation: Your Turn 2

Using μ and σ from the previous example,

1. What is the probability that exactly 45 survive if we use the normal approximation to the binomial?

Hint: Find the area under the curve between two values.

2. What is the probability that exactly 45 survive if we use the binomial distribution? b(45;100,0.4)

-5 0 5

DRAW THE PICTURE!!


Gamma & Exponential Distributions

• Related to the Poisson Process (discrete)Number of occurrences in a given interval or region “Memoryless” process

• Sometimes we’re interested in the number of events that occur in an area (eg flaws in a square yard of cotton).

• Sometimes we’re interested in the time until a certain number of events occur.

• Area and time are variables that are measured.• Typical problem statement: The length of time in days

between student complaints about too much homework follows a gamma distribution with α = 3 and β = 7. What is the probability that up to two weeks will elapse before 6 complaints are heard?


Gamma Distribution• The density function of the random variable X with

gamma distribution having parameters α (number of occurrences) and β (time or region).

x > 0.

μ = αβ

σ2 = αβ2

)!1()(

)(

1)( 1

nn

exxfx

Gamma Distribution

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8

x

f(x)


Exponential Distribution

0 5 10 15 20 25 30

• Special case of the gamma distribution with α = 1.

x > 0.

Describes the time until or time between Poisson events.

μ = β

σ2 = β2

x

exf

1

)(


Is It a Poisson Process?

• For homework and exams in the introductory statistics course, you will be told that the process is Poisson. 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?

An average of 3.5 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. How long before the next call?


Poisson Example Problem 1An 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 1 minute will elapse before 2 calls arrive?

β = 1 / λ = 1 / 2.7 = 0.3704 α = 2 X=1

What is the value of P(X ≤ 1)?

Can we use a table? Must we integrate?

Can we use Excel?

)!1()()(

1)( 1

nnwhereexxf

x


Poisson Example Solution Based on the Gamma Distribution

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 1 minute will elapse before 2 calls

arrive? β = 1/ 2.7 = 0.3704 α = 2

= 0.7513

Excel = GAMMADIST (1, 2, 0.3704, TRUE)

1

0

7.27.22

7.21

0

2

7.2

7.2)1(

xx

x

eex

dxexXP


Poisson Example Problem 2

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 expected time before the next call arrives?

α = next call =1 λ = 1/ β = 2.7

Expected value of time (Gamma) = μ = α β

Since α =1

μ = β = 1 / 2.7 = 0.3407 min.

Note that when α = 1 the gamma distribution is known as the exponential distribution.

JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 1 Continuous Probability Distributions Many...

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