4 Continuous Probabilities

DSC1007 Lecture 5

Continuous Probability Distributions

Continuous Random Variables

Probability Density Function

Uniform Distribution

Cumulative Distribution Function

Exponential Distribution

Normal Distribution

Computing Probabilities for Normal Distribution

Sum of Normally Distributed R.V.s

Central Limit Theorem

Normal Approximation to Binomial

Continuous Probability Distributions

In a class of 50 students, Mike said his height was 1.70m.

You randomly select a

student from the

class. What is the

probability that his or

her height is exactly

1.70m?


A continuous random variable can take any value in

some interval.

Examples : Temperature, Weight, Time


X P(X)

0 0.343

1 0.441

2 0.189

3 0.027

Discrete Probability Distribution

How to count the

probabilities for continuous

random variables?


student from the

class. What is the


her height is exactly

1.70m?

Concept 1: For a continuous random variable x,

the probability for x to be a particular value is

nearly zero.


student from the

class. What is the


her height is between

1.69m and 1.71m?

Concept 2: It makes sense to measure the

probability of a random variable over a range of

values.

You randomly select

an adult. Is he or she

more likely to have a

height of exactly 2m

or have a height of

exactly 1.7m?

Concept 3: It is still possible to compare the

likelihood of the probabilities of a random

variables taking different values.

PDF and CDF

Histogram

We can describe the probability distribution of a

continuous r.v. X by its probability density function

(pdf), usually denoted by f (x ) .


-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

x

f (x)

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0


a b

area =

The probability density function (pdf) of a r.v. X has the

following characteristics:

Area under the pdf curve is equal to 1

Probability that X lies between values a and b is equal to the

area under the curve between a and b

x

P(a X b)

f (x)

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0


a b

Q1 : Given the pdf f(x), how do we compute the area

under the curve between a and b?

x

f (x)

In general, we will have to evaluate the integral

area = P(a X b)

PDF for a womans height in US

PDF for a mans height in US

http://www.johndcook.com/blog/2008/07/20/why-heights-are-normally-distributed/

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0


x

f (x)

Q2 : Given the pdf f(x), how do we compute the

mean and variance of X ?

Mean = =

Variance = = ( )2

Discrete vs Continuous

Discrete Continuous

Probability distribution

x1 p1 x2 p2

xn pn

p d f f (x)

= 1

=1

= 1

Mean

=1

Variance 2 ( )

2

=1 ( )2

For a given t, the cumulative distribution function

(cdf) F(t) of a continuous r.v. X is defined by

= =


-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0t

f (t) 0 F(t) 1

By definition :

F(t) is a non-decreasingfunction of t

P(X t)

F(t) is the probability that X does not exceed t.


It is also easy to see that :

P(X > t) = 1 F(t)

P(c X d) = F(d) F(c)

For a given t, the cumulative distribution function

(cdf) F(t) of a continuous r.v. X is defined by

= =

Note: P(X t) = P(X < t)


X is uniform on [a ,b] if X is equally likely to take any value in the range from a to b.


a b

f (t)

t

Example

Suppose that travel time (by bus) between Clementi and NUS is uniformly distributed between 12 and 20 minutes.

1. What is the mean travel time?

2. What is the probability that the travel time exceeds 14 minutes?

3. What is the probability that the travel time is between 14 and 18 minutes?

If the r.v. X is equally likely to take any value in the range

from a to b (where b > a), then we say that X obeys a

uniform distribution over [a ,b].


a b

1/(ba)

f (t)

PDF of X

t

=

1

0 otherwise

Equation of f(t) ?

We write : X U[a ,b]

area = 1

CDF of Uniform Distribution

a b

1/(ba)

f (t)

PDF of X

t

=

1

0 otherwise

CDF of X ?

F(t) = P(X t)

CDF of Uniform Distribution

a b

1

F (t)

t

=

0 <

1 >

Equation of F(t) ?CDF of X

F(t) = P(X t)

Mean & Variance of Uniform Distribution

a b

1/(ba)

f (t)PDF

t

X U[a,b]

=

1

0 otherwise

E (X) = ?

Var (X) = ?

(a+b)/2

( )2

12

Example

The mens 100 meter sprint at the 1996 Olympic Games in Atlanta was a hotly contested event between three athletes.

Donovan Bailey

(Canada)

Frankie Fredericks

(Namibia)Ato Boldon

(Trinidad)

Example

Assume that the probability distribution of the time to run the race is the same for the three, and that this time obeys a uniform distribution between 9.75 seconds and

9.95 seconds.

What is the probability that Donovans time will beat the previous record of 9.86 seconds?

What is the probability that the winning time will beat the previous records of 9.86 seconds?


Often arises as the distribution of amount of time until an event occurs.

Useful for waiting line problems.

A random variable X is said to be an exponential r.v. with rate

parameter ( > 0) if it has the pdf

f (x ) = e x for x > 0


It can be shown that

Mean E ( X) =

Variance Va r ( X ) =

1

1

2

= 1

It can be shown that

X has the cdf

Poisson and Exponential

There is a close relationship between the two distributions.

= =

! for = 0, 1, 2, . . .

N is Poisson r.v. with parameter (>0):

X is exponential r.v. with parameter ( > 0) :

f(x) = ex for x > 0

Poisson and Exponential

Parameter Poisson Exponential

Mean 1

Example:Suppose the usage duration of a particular ATM is an exponential r.v. with a mean of 3 mins. If someone arrives at the ATM immediately ahead of you, find the probability that you have to wait (a) more than 2 minutes, and (b) between 2 and 3 minutes

X is exponential r.v. with parameter ( > 0)

PDF f(x) = ex for x> 0

CDF


= 1

X = usage duration (in mins) of ATM is exponential r.v. with parameter = 1/3.

(a) P ( X > 2 ) = 1 P ( X < 2 ) = 1 F ( 2 ) = e 2 0.51342

(b) P ( 2 < X < 3 ) = P ( X < 3 ) P ( X < 2 ) = ( 1 e 3 ) ( 1 e 2 )

X = usage duration (in mins) of ATM is exponential r.v. with parameter = 1/3.

(a) P ( X > 2 ) = 1 P ( X < 2 ) = 1 F ( 2 ) = e 2 0.51342

(b) P ( 2 < X < 3 ) = P ( X < 3 ) P ( X < 2 ) = ( 1 e 3 ) ( 1 e 2 )


Example:Suppose the usage duration of a particular ATM is an exponential r.v. with a mean of 3 mins. If someone arrives at the ATM immediately ahead of you, find the probability that you have to wait (a) more than 2 minutes, and (b) between 2 and 3 minutes



CDF = 1


Excel Function : EXPONDIST (x, , cumulative)

= 1

cumulative = 0 f (x )

1 F (x )



CDF = 1


Excel Function : EXPONDIST (x, , cumulative)

= 1/3

Example

P (X > 2) = 1 - EXPONDIST (2, 1/3, 1)

P (2 < X < 3) = EXPONDIST (3, 1/3, 1) - EXPONDIST (2, 1/3, 1)

Suppose the usage duration of a particular ATM is an exponential r.v. with a mean of 3 mins. If someone arrives at the ATM immediately ahead of you,

find the probability that you have to wait (a) more than 2 minutes, and

(b) between 2 and 3 minutes

Normal Distribution

Density function is the familiar bell-shaped curve

Normal Distribution

t

f(t)

A r.v. X that obeys a Normal distribution is completely described by

two parameters : mean and standard deviation .

We write : X N( ,).

= 1

22

()2

22

Note: f(t) is symmetric around the mean

f(t) is highest at its mean

Normal Distribution

PDFs of 4 Normally distributed r.vs

Many phenomena obey the Normal distribution:

PSLE scores

Height of a group of people (e.g., DSC1007 students)

Stock return over short periods of time

Width of steel plate from a production process

Consumer demand for a product on a given day

But ... some phenomena are not Normally distributed:

Stock returns over longer periods of time

Income distributions

Normal Distribution

Suppose : X N( ,)

How do we compute probabilities such as : P ( a X b ) ?

Computing Probabilities for Normal Distributions

Lets first look at the standard Normal case: Z N(0 , 1)

How do we compute : P ( a Z b ) ?

= 1

22

()2

22

Consider the standard Normal random variable Z ~ N(0,1)

Determine the following:

P(Z 1.55) = ?

P(0.55 Z 0.50) = ?

Computing Probabilities for Standard Normal Distributions

We have seen how to use a standard Normal table to

compute probabilities for a r.v.

Z N(0,1).

Next :

We can use the same table to compute probabilities for any

r.v. that obeys a Normal distribution :

X N( ,).


If X N( ,), then the r.v. Z defined by

obeys a standard Normal distribution.


=

There is a special relationship between N(0,1) and N( ,).

Because :

In other words :

~ ,

~ (0,1)

Application of Normal Distribution

x 3

2 +

+ 2 + 3

68.26%

95.44%

99.72%

Normal Distribution

Are you normal?

Mean = 100 & SD = 15

The IQ of famous people

Example

The personnel department of Ztel, a large communications company, is reconsidering its hiring policy. Each applicant for a job must take a standard exam, and the hire or no-hire

decision depends at least in part on the result of the exam. The scores of all applicants have

been examined closely. They are approximately normally distributed with mean 525 and

standard deviation 55.

The current hiring policy occurs in two phases. The first phase separates all applicants into three categories:

automatic accepts test scores 600 or above

automatic rejects test scores 425 or below

maybe the rest

All the maybes are passed on to a second phase where their previous job experience, special talents, and other factors are used as hiring criteria.

Questions

What is the percentage of applicants who are automatic accepts or rejects?

If ZTel wants to automatically accept 15% of applicants and automatically reject 10% of them,

how should the standards be changed?

Power of Collaboration

A Retail Example

Two retail chains (lets call them COURTS and Challenger) are planning to

sell the Apple iPad. Demands at the two retail chains are random variables.

Let

X = daily demand for the iPad at COURTS

Y = daily demand for the iPad at Challenger

Suppose that X and Y are Normally distributed with

X = 800 Y = 160

X = 500 Y = 100

Corr(X,Y) = 0.23

A Retail Example

Q1: What is probability that the demand for iPads

in COURTS exceed 1000?

A Retail Example

Q2: Challenger would like to stock enough iPad so that the

probability that all customers will be able to purchase and

carry home an iPad is 98%.

How many iPad does Challenger need to stock per day?

A Retail Example

t

f(t)

160Stock

Q3: COURTS would like to stock enough iPad so that the

probability that all customers will be able to purchase and

carry home an iPad is 98%.

How many iPads does COURTS need to stock per day?

A Retail Example

Lets summarize:To ensure that the probability that all customers will be able to

purchase and carry home an iPad is 98% ,

COURTS would need to stock : 1827

Challenger would need to stock : 365

Total :

2192

A Retail Example

A weighted sum of jointly Normal r.v.s

is a Normal r.v.

(d) What is the distribution of W?

Suppose that X and Y are jointly Normal random variables, and

W = aX + bY

(a) What is E(W)?

(b) What is Var(W)?

(c) What is SD(W)?

= a E(X) + b E(Y) = a X + b Y

= a2 Var (X ) + b2 Var (Y ) +2 ab X Y Corr(X ,Y )

Sums of Normal RVs

Suppose that COURTS and Challenger are considering a merger

(or just a merger of their warehousing operations).

Q4 : What is the distribution of the demand for iPads from the combined company?

Q5 : How many iPads would the combined warehouse need to

stock per day in order to achieve the 98% customer service requirement?

Sums of Normal RVs

Q4 : What is the distribution of the demand for iPads from the combined company?

We know that W = X + Y ( total demand for iPads at Courts &

Challenger ) is a Normal r.v.

Mean W = X + Y = 800 + 160 = 960

Variance W2 = X

2 + Y2 + 2X Y Corr(X,Y)

=5002 + 1002 + 2(500)(100)(0.23) = 283,000

Std Deviation W2 531.98

Corr(X,Y) = 0.23

A Retail Example

X and Y are Normally distributed with

X =800 Y =160

X = 500 Y=100

Q5 : How many iPads would the combined warehouse need tostock per day in order to achieve the 98% customer service requirement?

Corr(X,Y) = 0.23

A Retail Example


X =800 Y =160

X = 500 Y=100

Q5 : How many iPads would the combined warehouse need tostock per day in order to achieve the 98% customer service requirement?

Corr(X,Y) = 0.23

P( Z [(b960)/531.98] ) = 0.98

[(b960)/531.98] ) ) 2.054

b 2052.7

Need to find b such that : P(W b) = 0.98

A Retail Example


X =800 Y =160

X = 500 Y=100

Normal Distribution Approximation

Suppose X1, X2, . . ., Xn are independent and identically distributed (i.i.d.)

random variables with

E(Xi) = and SD(Xi) =

SD(Sn) =

E(Sn) = n

Var(Sn) = n2

Let Sn = X1 + X2 + . . . + Xn

n

Sum of Random Variables

Suppose X1, X2, . . ., Xn are independent and identically distributed (i.i.d.)

random variables with

E(Xi) = and SD(Xi) =

SD(Sn) =

E(Sn) = n

Var(Sn) = n2

The Central Limit Theorem (for the sum). If n is large (say, at

least 30), then Sn is approximately Normally distributed with

mean n and standard deviation n .

Let Sn = X1 + X2 + . . . + Xn

n


Let An = 1+2++

E(An) =

Var(An) = 2 /n

0.000

0.050

0.100

0.150

0.200

0.250

1 3 5 7 9

11

13

15

17

0.000

0.050

0.100

0.150

0.200

0.250

1 3 5 7 9

11

13

15

17

0.000

0.050

0.100

0.150

0.200

0.250

1 3 5 7 9

11

13

15

17

Example: cast a die n times

n = 3

n = 2


n = 1

n does not have to be very large (30 is often good enough), especially if the probability distribution of Xi is nice (i.e., it is not too skewed to the left or to the right, and that its tails are not too wide.

The Central Limit Theorem (CLT) makes use of only two pieces of information: the mean and standard deviation of Xi .

(This goes a long way in justifying the use of the standard deviation as a

measure of spread.)

The Normal distribution can arise in at least three ways :(i) as a natural model for many physical processes,

(ii) as a sum of several Normal random variables, or

(iii) as an approximation of a sum (or average) of several iid r.v.s

Some Comments on CLT

Example: An electrical component is guaranteed by its suppliers to have

2% defective components. To check a shipment, you test a

random sample of 500 components.

If the suppliers claim is true, what is the probability of finding 15 or more

defective components?

Let X be the number of defective components found during the test.

Then X is Binomial (500, 0.02).

P(X15) = 1 P(X14)

Approximating Binomial with Normal

Probability of finding 15 or more defective components

0.08136 (using BINOMDIST)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 5

10

15

20

25

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 5

10

15

20

25

Examples: p = 0.8

n = 20

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 5

10

15

20

25

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 5

10

15

20

25

n = 15

n = 25

n = 10

Then Y would be a good approximation of X if n is large.

Suppose that X is Binomial (n, p).

Then :

E(X) = np

Var(X) = np(1-p)

Let Y be a Normal r.v. with :

mean np

variance np(1p)

Approximating Binomial with Normal

Good rule of thumb : use the Normal distribution as an

approximation of the Binomial distribution if

np 5 and n(1 p) 5

Back to Example:

Let X be the number of defective components found during the test. X is Binomial (500, 0.02).

np = 500 * 0.02 = 10 > 5

n (1p) = 500 * 0.98 = 490 > 5

X ~ N (10, 3.130)

P(X 15) 1 P(X

4 Continuous Probabilities

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