Transcript Chapter 1- Lecture 2

Probability Distributions

A probability function - function when probability values are assigned to
all possible numerical values of a random variable (X).

Individual probability values may be denoted by the symbol P(X=x), in
the discrete case, which indicates that the random variable can have
various specific values.

All the probabilities must be between 0 and 1;
0≤ P(X=x)≤ 1

The sum of the probabilities of the outcomes must be 1.
∑ P(X=x)=1

It may also be denoted by the symbol f(x), in the continuous, which
indicates that a mathematical function is involved.
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Poisson
Continuous
Probability
Distributions
Normal
BINOMIAL DISTRIBUTION
An experiment in which satisfied the following characteristic is called a
binomial experiment:




The random experiment consists of n identical trials.
Each trial can result in one of two outcomes, which we denote by
success, S or failure, F.
The trials are independent.
The probability of success is constant from trial to trial, we denote
the probability of success by p and the probability of failure is equal
to (1 - p) = q.
Example:
No. of getting a head in tossing a coin 10 times.
o
Binomial distribution is written as X ~ B(n,p),
A binomial experiment consist of n identical trial with probability of success,
p in each trial.The probability of success in n trials is given by
P ( X  x )  nC x p x q n  x ;
Note:
n
x  0,1, 2....n
n
n!
Cx    
 x   n  x  ! x!
The Mean and Variance of X
If X ~ B(n, p), then
Mean
Variance
:  E ( X )  np
:2  V ( X )  np(1  p)  npq
Std Deviation :  npq
where n is the total number of trials, p is the probability of success and q is
the probability of failure.
EXAMPLE:
Given that X ~ B (12, 0.4), find
a)
b)
c)
d)
e)
f)
P ( X  2)
P ( X  3)
P ( X  4)
P (2  X  5)
E( X )
Var( X )
SOLUTION:
a) P ( X  2)  12C2 (0.4) 2 (0.6)10  0.0639
b) P ( X  3)  12C3 (0.4)3 (0.6)9  0.1419
c) P ( X  4)  12C4 (0.4) 4 (0.6)8  0.2128
d) P (2  X  5)  P( X  2)  P( X  3)  P( X  4)
 0.0639  0.1419  0.2128  0.4185
e) E ( X )  np  12(0.4)=4.8
f) Var ( X )  npq  12(0.4)(0.6)= 2.88
CUMULATIVE BINOMIAL DISTRIBUTION

When the sample is relatively large, tables of Binomial are often used. Since the
probabilities provided in the tables are in the cumulative form P  X  x 
the following guidelines can be used:
Bin.
table
EXAMPLE:
Bin.
table
EXAMPLE:
The probability that boards purchased by a cabinet manufacturer are unusable
for building cabinets is 0.10.The cabinet manufacturer bought eleven boards,
what is the probability that
a)
b)
c)
Four or more of the eleven boards are unusable for building cabinets?
At most two of the eleven boards are unusable for building cabinets?
None of the eleven boards are unusable for building cabinets?
Solution:
X : The number of unusable boards
X ~ B(11, 0.10)
a)
P  x  4  1  P  x  3  1  0.9815  0.0185
b) P  x  2  0.9104
c)
P  x  0  0.3138
Bin.
table
Exercise:
1. Given that X~B(20, 0.25) using tables of binomial probabilites, find
a) P  X  6 
b) P  X  4 
c) P  X  8 
d) P  X  9 
e) P  3  X  11
f) P  4  X  12 
g) P  5  X  10 
2. According to a survey, 45% of college students wear contact lense
a) What is the probability that exactly 3 of a random sample of 5 college
student wear contact lenses?
b) What is the probability that at least 7of a random sample of 15 college
student wear contact lenses?
c) What is the probability that not more than 7 of a random sample of 15
college student wear contact lenses?
Bin.
table
THE POISSON DISTRIBUTION

Poisson distribution is the probability of a given number of events occurring in
a fixed interval of time / space and other specified intervals such as
distance, area /measurement or volume .
Examples:
1.
The number of cars passing a toll booth in one hour (time).
2.
The number of defects in a square meter of fabric (area).

A random variable X has a Poisson distribution and it is written as
X ~ Po ( )
with

Mean, E ( X )  
Variance, V ( X )  
Probability distribution function of Poisson is given by
e   x
P( X  x) 
for x  0,1, 2,3,...
x!
EXAMPLE :
Given that X ~ Po (4.8) , find
a) P( X  0)
b) P( X  9)
c) P( X  1)
SOLUTION :
e 4.8 4.80
a) P ( X  0) 
 0.0082
0!
e 4.8 4.89
b) P ( X  9) 
 0.0307
9!
c) 1  P ( X  0)  1  0.0082
= 0.9918
EXAMPLE :
Poi.
table
Exercise:
1. Given that X ~ P0  7.9
, using tables of Poisson probabilities, find
a) P  X  7 
b) P  X  12 
c) P  X  8 
d) P  X  11
e) P  X  10 
f) P  4  X  14 
2. The numbers of cars sold by a new car dealer follows a Poisson distribution
with a mean of 13.5 cars sold in three days.
a) What is the probability that at least 6 cars a sold today?
b) Find the mean and standard deviation of Y, the number of cars sold in
two days. What is the probability that fewer than 10 cars sold in two
days?
c) Find the mean and standard deviation of W, the number of cars sold in
four days. What is the probability that at most 18 cars are sold in 4 days?
Poi.
table
POISSON APPROXIMATION OF BINOMIAL PROBABILITIES
Applicable to used Poisson distribution when the Binomial
experiment / trial has the following :
 n  30 and
 np  5 or
 nq  5
Example:
Given that X~B (1000, 0.004).
Find
a) P (X=7)
b) P (X<9)
Solution:
Since n = 1000 >30, np = 1000(0.004) = 4 < 5, thus Poisson
approximation is used with   np  4 . Therefore, X ~ P (4)
a)
b)
Poi.
table
Exercise:
1. Given that X ~ B  500,0.005 , find
a) P  X  5 
b) P  X  4 
c) P  X  8 
2. 0.4% of all 9 city’s voters are not in favour of a certain candidate for
mayor. Suppose a poll of 1000 voters in the city is taken. Find the
probability that 10 or more voters do not favour this candidates.
Poi.
table
THE NORMAL DISTRIBUTION


Numerous continuous random variables have distribution closely resemble
the normal distribution.
The normal distribution can be used to approximate various discrete prob.
dist.
A continuous random variable X is said to have a normal distribution
with parameters  and  2 , where       and  2  0,
if the pdf of X is
f ( x) 
1
e
 2
1  x 
 

2  
2
  x  
X is denoted by X ~ N (  ,  2 ) with E  X    and V  X    2
THE NORMAL DISTRIBUTION
 ‘Bell


Shaped’
Symmetric about the mean
Mean, Median and Mode are Equal
f(X)
- Location is determined by the mean, μ
- Spread is determined by the standard
deviation, σ
The random variable has an infinite
theoretical range:
+  to  
σ
μ
Mean = Median = Mode
X
The Standard Normal Distribution


Any normal distribution (with any mean and standard deviation
combination) can be transformed into the standard normal distribution (Z)
Need to transform X units into Z units using
Z
X 


The standardized normal distribution (Z) has a mean of 0,
standard deviation of 1,   1 .

Z is denoted by

Thus, its density function becomes
Z ~ N (0,1)
 0
and a
PATTERNS FOR FINDING AREAS UNDER THE STANDARD NORMAL CURVE
EXAMPLE:
Z table
EXERCISES:
Determine the probability or area for the portions of the Normal distribution
described.
a) P (0  Z  0.45)
b) P ( 2.02  Z  0)
c) P ( Z  0.87)
d) P (2.1  Z  3.11)
e) P (1.5  Z  2.55)
Z table
SOLUTIONS:
a) P(0  Z  0.45)  0.1736
b) P(2.02  Z  0)  0.47831
c) P(Z  0.87)  0.5  0.3078  0.8078
d)P( 2.1  Z  3.11)  0.4821  0.4991  0.9812
e) P(1.5  Z  2.55)  0.4946  0.4332  0.0614
EXAMPLE:
Z table
EXERCISES:
Determine Z such that
a) P ( Z  Z )  0.25
b) P ( Z  Z )  0.36
c) P ( Z  Z )  0.983
d) P ( Z  Z )  0.89
Z table
SOLUTIONS:
a) P( Z  Z )  0.25;
Z  0.675
b) P( Z  Z )  0.36;
Z  0.355
c) P( Z  Z )  0.983;
Z  2.12
d) P( Z  Z )  0.89;
Z  1.225
EXERCISES:
Z table
EXAMPLE:
to
Z table
Exercise:
At a certain community college, the time that is required by students to
complete the math competency examination is normally distributed with a
mean of 57.6 minutes and a standard deviation of 8 minutes. Find the
probability that a student takes
a) Longer than 1 hour to complete the examination
b) Between 56 minutes and I hour to complete the examination
Z table
NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION

When the number of observations or trials n in a binomial experiment is
relatively large, the normal probability distribution can be used to
approximate binomial probabilities. A convenient rule is that such
approximation is acceptable when
n  30, and both np  5 and nq  5
Given a random variable X ~ b(n, p), if n  30 and both np  5
and nq  5, then X ~ N (np, npq)
X  np
with Z 
npq
CONTINUOUS CORRECTION FACTOR
The continuous correction factor needs to be made when a
continuous curve is being used to approximate discrete probability
distributions.
 0.5 is added or subtracted as a continuous correction
factor according to the form of the probability statement as
follows:

c .c
a) P ( X  x ) 
 P ( x  0.5  X  x  0.5)
c .c
b) P ( X  x) 
 P ( X  x  0.5)
c .c
c) P ( X  x ) 
 P ( X  x  0.5)
c .c
d) P ( X  x ) 
 P ( X  x  0.5)
c .c
e) P ( X  x ) 
 P ( X  x  0.5)
c.c  continuous correction factor
Example:
In a certain country, 45% of registered voters are male. If 300
registered voters from that country are selected at random, find
the probability that at least 155 are males.
Solutions:
X is the number of male voters.
X ~ b(300, 0.45)
c .c
P( X  155) 
 P( X  155  0.5)  P( X  154.5)
np  300(0.45)  135  5
nq  300(0.55)  165  5

154.5  300(0.45) 
154.5  135 

PZ 
  P  Z 


300(0.45)(0.55)
74.25




 P (Z  2.26)
 0.01191
Z table
NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION

When the mean  of a Poisson distribution is relatively large,
the normal probability distribution can be used to approximate
Poisson probabilities. A convenient rule is that such
approximation is acceptable when
  10.
Given a random variable X ~ Po ( ), if   10, then X ~ N ( ,  )
with Z 
X 

EXAMPLE:
A grocery store has an ATM machine inside. An average of
5 customers per hour comes to use the machine. What is
the probability that more than 30 customers come to use
the machine between 8.00 am and 5.00 pm?
Solution:
X is the number of customers come to use the ATM machine in 9 hours.
X ~ Po (45)
  45  10
X ~ N (45, 45)
c .c
P ( X  30) 
 P ( X  30  0.5)  P ( X  30.5)
30.5  45 

PZ 
  P ( Z  2.16)
45 

 0.98461
Z table
Exercise:
An average of 10 patients are admitted per day to the
emergency room of a big hospital. What is the probability that
less 75 than patients are admitted to the emergency room in 7
days?
Z table
Continue
Lecture 3