Oct 06, 2015
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?
Continuous Random Variables
A continuous random variable can take any value in
some interval.
Examples : Temperature, Weight, Time
Continuous Random Variables
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?
You randomly select a
student from the
class. What is the
probability that his or
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.
You randomly select a
student from the
class. What is the
probability that his or
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 ) .
Probability Density Function
-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
Probability Density Function
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
Probability Density Function
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
Probability Density Function
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
= =
Cumulative Distribution Function
-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.
Cumulative Distribution Function
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)
Uniform Distribution
Uniform Distribution
X is uniform on [a ,b] if X is equally likely to take any value in the range from a to b.
Uniform Distribution
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].
Uniform Distribution
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?
Exponential Distribution
Exponential Distribution
Exponential Distribution
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
Exponential Distribution
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
Exponential Distribution
= 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 )
Exponential Distribution
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
Exponential Distribution
Excel Function : EXPONDIST (x, , cumulative)
= 1
cumulative = 0 f (x )
1 F (x )
X is exponential r.v. with parameter ( > 0)
PDF f(x) = ex for x> 0
CDF = 1
Exponential Distribution
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
Exponential Distribution
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( ,).
Computing Probabilities for Normal Distributions
If X N( ,), then the r.v. Z defined by
obeys a standard Normal distribution.
Computing Probabilities for Normal Distributions
=
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 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
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 and Y are Normally distributed with
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
Central Limit Theorem
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
Central Limit Theorem
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