JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 1

Continuous Probability Distributions

• Many continuous probability distributions, including: UniformNormalGammaExponentialChi-SquaredLognormalWeibull

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 2

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”

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 3

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

We conclude that 97.72% will have a resistance less than 44 ohms.

What percentage will have a resistance greater than 44 ohms? 2.28%

-5 0 5

2

4044

Z

XZ

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 4

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

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 5

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

5.0

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 6

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 than 30 survive?

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

if x = 30, then z =((30-0.5) – 40)/4.899 = -2.14 (Don’t forget the continuity correction.)

and, P(X < 30) = P (Z < _________) = _________(Subtract 0.5 from 30 because we are looking for P (X< x).

Less than 30 for a discrete distribution is 29 or less. How would the equation change if we wanted the probability that 30 or less survived?

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 7

The Normal Approximation: Your Turn

Using μ and σ from the previous example,

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 ?

Since we are looking for P (X>x) we add 0.5 to x.

z = ((50 + 0.5) – 40)/4.899 = 2.14

Answer: _____________

-5 0 5

DRAW THE PICTURE!!

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 8

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.

Calculate P (X>x) by adding 0.5 to x.

Calculate P (X<x) by subtracting 0.5 from x.

Determine P(X=x) by calculating the difference.

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

Note that n=100 is too large for the tables. Determine the value using the binomial equation.

0 …. 44 45 46 ….. 100

-5 0 5

DRAW THE PICTURE!!

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 9

Gamma & Exponential Distributions

• Sometimes we’re interested in the number of events that occur in a certain time period.Related to the Poisson ProcessNumber of occurrences in a given interval or region “Memoryless” processDiscrete

• If we are interested in the time until a certain number of events occur, we will use continuous distributions (exponential and gamma). The time until a number of Poisson events uses gamma

distribution with alpha = number of events and beta = mean time. See Examples 6.17 and 6.18

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 10

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)

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 11

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

)(

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 12

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?

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 13

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

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 14

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

JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 15

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.