Ch 5

Report
Ch 5 Probability: The
Mathematics of Randomness
5.1.1 Random Variables and Their
Distributions
• A random variable is a quantity that (prior to
observation) can be thought of as dependent
on chance phenomena.
Toss a coin 10 times
X=# of heads
Toss a coin until a head
X=# of tosses needed
More random variables
Toss a die
X=points showing
Plant 100 seeds of pumpkins
X=% germinating
Test a light bulb
X=lifetime of bulb
Test 20 light bulbs
X=average lifetime of bulbs
Types of Random Variable
• A discrete random variable is one that has
isolated or separated possible values.
(Counts, finite-possible values)
• A continuous random variable is one that can be
idealized as having an entire (continuous) interval
of numbers as its set of possible values.
(Lifetimes, time, compression strength 0infinity)
Probability Distribution
• To specify a probability distribution for a random
variable is to give its set of possible values and (in one
way or another) consistently assign numbers between
0 and 1 –called probabilities– as measures of the
likelihood that be various numerical values will occur.
• It is basically a rule of assigning probabilities to random
events.
Roll a die, X=# showing
x
f(x)
1 2 3 4 5 6
1/6 1/6 1/6 1/6 1/6 1/6
Probability Distribution
The values of a probability distribution must
be numbers on the interval from 0 to 1.
The sum of all the values of a probability
distribution must be equal to 1.
Example
Inspect 3 parts. Let X be the # of parts with
defect.
Find the probability distribution of X.
First, what are possible values of X?
(0, 1,2,3)
When does X take on each value?
Outcome
{NNN}
{NND, NDN, DNN}
{NDD, DND, DDN}
{DDD}
X
0
1
2
3
• If P(D)=0. 5, then P(N)=0.5 We have
x
f(x)=P(X=x)
0
1/8
1
3/8
2
3/8
3
1/8
Cumulative Distribution Function
• Cumulative distribution F(x) of a random
variable X is
F ( x )  P( X  x )
• Since (for discrete distributions) probabilities
are calculated by summing values of f(x), for a
discrete distribution
F ( x)   f ( z )
zx
“Defect” Example continued
We had
f(0)=1/8, f(1)=f(2)=3/8, f(3)=1/8.
Therefore
•
•
•
•
F(0)=f(0)=1/8;
F(1)=f(0)+f(1)=1/8+3/8=1/2;
F(2)=f(0)+f(1)+f(2)=7/8;
F(3)=f(0)+f(1)+f(2)+f(3)=1.
0,
1/8,

F(x)  1/2,
7/8,

1,
x  0;
0  x  1;
1  x  2;
2  x  3;
x  3.
Summaries of Discrete Probability Distributions
Given a set of numbers, for example,
S= {1, 1, 1, 1, 3, 3, 4, 5, 5 , 6 }
To calculate the average of these 10 numbers, you can use
1  1  1  1  3  3  4  5  5  6 30

3
10
10
Or you can get it this way
(1)(4) (3)(2) (4)(1) (5)(2) (6)(1)
10
4
2
1
2
1
 (1)( )  (3)( )  (4)( )  (5)( )  (6)( )
10
10
10
10
10
 3.
Mean or Expectation
The above result is the weighted sum of x values:
x
Weights
f(x)
1
4/10
0.4
3
2/10
0.2
4
1/10
0.1
5
2/10
0.2
6
1/10
0.1
The weights are actually the probability distribution for a random
variable X .
The weighted sum, 3, is called the mean or mathematical
expectation of random variable X.
Definition
Let X be a discrete random variable with probability
distribution f(x). The mean or expected value of X is
μ  E(X)   xf(x)
x
The expectation or mean of a random variable X is
the long run average of the observations from the
random variable X.
The mean of X is not necessarily a possible value for X.
Back to “defect” example
• What is the average number of defective
items we expect to see?
X
0
1
2
3
f(x)
1/8
3/8
3/8
1/8
1
3
3
1 3
  E ( X )   x  f ( x )  0   1  2   3  
8
8
8
8 2
Variance
• The mean is a measure of the center of a
random variable or distribution
X:
x= -1
f(x) ¼
0
½
1
¼
Y:
y= -2 0
2
g(y) ¼ ½ ¼
Both variables have the same center: 0
Which has a larger variability?
Variance
• Deviation from the center (mean)
XX
The average of XX is zero!
• Consider the weighted average of (XX)2
E(X0)2=(1)(1/4)+(0)(1/2)+(1)(1/4)=1/2
E(Y0)2=(4)(1/4)+(0)(1/2)+(4)(1/4)=2
Y is more variable!
Definition
• Let X be a discrete random variable with probability
distribution f(x) and mean 
• The variance of X is
σ  E[(X μ) ]   (x  μ) f(x)
2
2
2
x
The positive square root of the 2 is called standard
deviation of X, denoted by .
Back to “defect” example
X
0
1
2
3
f(x)
1/8
3/8
3/8
1/8
 2  Var ( X )  E ( x   )2   ( x   )2  f ( x)
1
3
3
1
2
2
2
 (0  1.5)   (1  1.5)   (2  1.5)   (3  1.5) 
8
8
8
8
2
1 2 3 2 3 2 1
E( X )  0   1   2   3   3
8
8
8
8
 2  Var ( X )  E ( X 2 )   2  3  1.52  0.75
2
2
   2  0.75
Binomial Distribution
In many applied problems, we are
interested in the probability that an event
will occur x times out of n.
For example
Inspect 3 items. X=# of defects.
D=a defect, N=not a defect, Suppose P(N)=5/6
No defect: (x=0)
NNN  (5/6)(5/6)(5/6)
One defect: (x=1)
NND  (5/6)(5/6)(1/6)
NDN  same
DNN  same
Two defect: (x=2)
NDD  (5/6)(1/6)(1/6)
DND  same
DDN  same
Three defect: (x=3)
DDD (1/6)(1/6)(1/6)
Binomial distribution
• x f(x)
0 (5/6)3
1 3(1/6)(5/6)2
2 3(1/6)2(5/6)
3 (1/6)3
How many ways to
choose x of 3 places for D.
__ __ __
 3  1   5 
f ( x)      
 x  6   6 
x
[P(D)]# of D
3 x
[1-P(D)]3- # of D
In general:
n independent trials
p probability of a success
x=# of successes
n
  ways to choose x places for sucesses,
 x
n x
f ( x)    p (1  p ) n  x
 x
• Roll a die 20 times. X=# of 6’s,
n=20, p=1/6
x
20  x
20
  1   5 
f ( x)       
 x  6   6 
 20   1   5 
p( x  4)       
 4  6   6 
4
16
• Flip a fair coin 10 times. X=# of heads
10  x
10 1   1 
f ( x)      
 x  2   2 
x
10 1 
   
 x  2 
10
More example
• Pumpkin seeds germinate with probability
0.93. Plant n=50 seeds
• X= # of seeds germinating
 50
x
50  x





f ( x)  
0
.
93
0
.
07
x 
 
 50
48
2





P( X  48)  
0
.
93
0
.
07
 48
 
Mean of Binomial Distribution
• X=# of defects in 3 items
x f(x)
0 (5/6)3
1 3(1/6)(5/6)2
2 3(1/6)2(5/6)
3 (1/6)3
Expected long run average of X?
The mean of a binomial distribution
•
Binomial distribution
n= # of trials,
p=probability of success on each trial
X=# of successes
n x
E ( x )     x   p (1  p)n x  np
 x
• Let’s check this.
x
3 x
3
  1   5 
E ( X )   X   x  f ( x)   x     
x 0  x   6   6 
3
3
2
2
3
5
 1  5 
1 5
1
 0*    1*3     2*3      3*    1/ 2
6
 6  6 
6 6
6
• Use binomial mean formula
E ( x)    np  3*1/ 6  1/ 2
More examples
• Toss a die n=60 times, X=# of 6’s
known that p=1/6
μ=μX =E(X)=np=(60)(1/6)=10
We expect to get 10 6’s.
Variance for Binomial distribution
• σ2=np(1-p)
where n is # of trials and p is probability of
a success.
• From the previous example, n=3, p=1/6
Then
σ2=np(1-p)=3*1/6*(1-1/6)=5/12
5.1.5 Poisson Distribution
• It is often useful to define a random variable
that counts the number of “events” that occur
within certain boundaries.
• # of defects in an item
• # of radioactive particles emitted by a
substance
• # of telephone calls received by an operator
within a certain time limit.
• A random variable X with a Poisson
distribution takes the values x=0, 1, 2, 3,…
with a probability distribution function
e 
f ( x)  P( X  x) 
, x  0,1, 2,...
x!

x
Exercise
• The number of monthly breakdowns of the kind of
computer used by an office is a random variable
having the Poisson distribution with λ=1.6. Find the
probabilities that this kind of computer will function
for a month
– Without a breakdown;
– With one breakdown;
– With two breakdowns.
Glass Sheet Flaws Example
• A quality inspector at a glass manufacturing
company inspects sheets of glass to check for
any slight imperfections. Suppose that the
number of these flaws in a glass sheet has a
Poisson distribution with parameter λ=0.5.
(This implies that the expected number of
flaws per sheet is only 0.5.)
1) What is the probability that there are no flaws
in a sheet? (percentage of the glass sheets in
“perfect condition”
e0.5 0.50
P( X  0) 
 e0.5  0.607
0!
So about 61% of the glass sheets are in
“perfect” condition.
2) Sheets with 2 or more flaws are scrapped by
the company. Find this probability.
P( X  2)  1  P( X  0)  P( X  1)
 1
e
0.5
0
0.5
1
0.5 e 0.5

 0.090
0!
1!
About 9% of the glass sheets need to be
scrapped and recycled through the company’s
manufacturing process.
Poisson Cumulative Probabilities with lambda=0.5
0.8
0.6
0.0
0.1
0.7
0.2
y
0.3
0.4
0.9
0.5
0.6
1.0
Poisson Probabilities with lambda=0.5
0
2
4
6
x
8
10
0
2
4
6
x
8
10
3) Looking at 3 sheets of glass, what is the
probability that we can find a total of 4 flaws?
e1.51.54
P(Y  4) 
4!
• This is a new Poisson random variable. λ=1.5
Something to think about:
• How many sheets of glass we need to inspect
to have a 90% chance of finding at least one
flaws?
Exercise
• In a certain city, medical expenses are given as
the reason for 75% of all personal
bankruptcies. Calculate the probability that
medical expenses will be given as the reason
for two of the next three personal
bankruptcies filed in that city.
To get a driver’s license, you must first pass a written
examination. If you fail in an exam, you can take it the
second time until you pass it. Suppose you fail an exam with
probability 0.3, and all the tests are independent.
1) What is the probability that you have to take the test twice in order to
pass the written exam?
2) What is the probability that you have to take the test more than 2 times
in order to pass the written exam?

similar documents