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Probability Distributions





A probability function is a function which assigns probabilities
to the values of a random variable.
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:
1. The random experiment consists of n identical trials.
2. Each trial can result in one of two outcomes, which we denote
by success, S or failure, F.
3. The trials are independent.
4. 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.
Examples:
1.
No. of getting a head in tossing a coin 10 times.
2.
No. of getting a six in tossing 7 dice.
3.
A firm bidding for contracts will either get a contract or not
EXAMPLE


Check whether the distribution is a probability
distribution.
Solution
X
0
1
2
3
4
P(X=x)
0.125
0.375
0.025
0.375
0.125
4
 P( X  x)  P( X  0)  P( X  1)  P( X  2)  P( X  3)  P( X  4)
0
= 0.125+0.375+0.025+0.375+0.125
= 1.025
1

# so the distribution is not a probability distribution.
A binomial experiment consist of n identical trial with probability
of success, p in each trial. The probability of x success in n trials
is given by
P( X  x)  nCx p x q n  x ;
x  0,1, 2....n
The Mean and Variance of X if X ~ B(n,p) are
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
a) P ( X  2)
b(12, 0.4), find
b) P ( X  3)
c) P ( X  4)
d) P (2  X  5)
e) E( X )
f) Var( X )
SOLUTIONS:
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
Bin. table
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  k  the following guidelines can
be used:
EXAMPLE

In a Binomial Distribution, n =12 and p = 0.3. Find the
following probabilities.
a) P( X  5)  P( X  4)  0.7237
b) P( X  5)  0.8822
c) P( X  9)  1  P( X  8)  1  0.9983  0.0017
d) P(5  X  9)  P( X  9)  P( X  5)
 0.9998  0.8822  0.1176
e) P(3  X  5)  P( X  5)  P( X  2)
 0.8822  0.2528 
Bin. table
EXAMPLE
In August 2009, David and Maria conducted a survey for
Fortune magazine to examine CEO`s attitudes toward
employee`s personal problems. 30% of the CEOs interviewed
felt that personal problems were none of the company`s
business. Assume that this result is true for the current
population of CEOs. Using the Binomial distribution tables, in
a random samples of 15, find the probability that
(i)
(ii)
(iii)
(iv)
The number of CEOs who hold this view is 10.
The number of CEOs who hold this view is between 9 to
12.
The number of CEOs who hold this view is at most 7.
Find the mean and standard deviation of binomial
distribution.
THE POISSON DISTRIBUTION


Poisson distribution is the probability distribution of the
number of successes in a given space*.
*space can be dimensions, place or time or combination of
them
Examples:
1.
No. of cars passing a toll booth in one hour.
2.
No. defects in a square meter of fabric
3.
No. of network error experienced in a day.
A random variable X has a Poisson distribution and it is referred
to as a Poisson random variable if and only if its probability
distribution is given by

e 
P( X  x) 
x!
x
for x  0,1, 2,3,...
A random variable X having a Poisson distribution can also be
written as
X
Po ( )
with E ( X )   and Var ( X )  
EXAMPLE :
Given that X
a) P( X  0);
, fin
Po (4.8)
b) P( X  9);
c) P( X  1)
Solution:
e 4.8 4.80
a) P( X  0)  P( X  0) 
 0.0082
0!
e 4.8 4.89
b) P( X  9) 
 0.0307
9!
or using cumulative Possion distribution table
0.0307
P( X  9)  P
( X  9)  P( X  8)  0.9749  0.9442  0.0307
c) P( X  1)  1  P( X  0)  1  0.0082  0.9918
EXAMPLE :
16
15
POISSON APPROXIMATION OF BINOMIAL
PROBABILITIES
The Poisson distribution is suitable as an approximation
of Binomial probabilities when n is large and p is small.
Approximation can be made when n  30 , and either
np  5 or nq  5
Example:

0.9786
EXERCISE 1
1.
Given that X ~ B(2, 0.4)
Find P( X  0), P( X  2), P( X  2), P( X  1), E ( X ),Var ( X ).
(ans: 0.36, 0.16, 1.0, 0.64, 0.8, 0.48).
2.
In Kuala Lumpur, 30% of workers take public transportation. In
a sample of 10 workers,
i) what is the probability that exactly three workers take public
transportation daily? (ans: 0.2668)
ii) what is the probability that at least three workers take
public transportation daily? (ans: 0.6172)
3. Let X ~ P0 (12). Using Poisson distribution table, find
i) P( X  8) and P( X  8) (ans: 0.1550, 0.0655)
ii) P( X  4) and P( X  4)(ans: 0.9977, 0.9924)
iii) P(4  X  14)
(ans: 0.7697)
4. Last month ABC company sold 1000 new watches. Past
experience indicates that the probability that a new watch will
need repair during its warranty period is 0.002. Compute the
probability that:
i) At least 5 watches will need to warranty work. (ans: 0.0527)
ii) At most than 3 watches will need warranty work. (ans: 0.8571)
iii) Less than 7 watches will need warranty work. (ans: 0.9955)
THE NORMAL DISTRIBUTION
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
Numerous continuous variables have distribution closely
resemble the normal distribution.
The normal distribution can be used to approximate various
discrete probability distribution.
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
CHARACTERISTICS OF NORMAL
DISTRIBUTION
‘Bell Shaped’


Symmetrical
Mean, Median and Mode
are Equal
f(X)
σ
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an infinite
theoretical range:
+  to  
X
μ
Mean = Median = Mode
Many Normal Distributions
By varying the parameters μ and σ, we obtain different
normal distributions
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,   0
and a standard deviation of 1,  2  1

Z is denoted by Z ~ N (0,1)

Thus, its density function becomes
PATTERNS FOR FINDING AREAS UNDER THE STANDARD NORMAL CURVE
EXAMPLE
a)
Find the area under the standard normal curve of P(0  Z  1)
a)
Find the area under the standard normal curve of P(2.34  Z  0)
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)
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
Z table
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
SOLUTIONS
a) P( Z  Z )  0.25;
Z  0.675
b) P( Z  Z )  0.36;
Z  0.3585
c) P( Z  Z )  0.983;
Z  2.12
d) P( Z  Z )  0.89;
Z  1.2265
Z table
EXAMPLE (TRANSFORM X INTO Z)
Suppose X is a normal distribution N(25,25). Find
a) P(24  X  35)
b) P( X  20)
Solutions
35  25 
 24  25
a) P(24  X  35)  P 
Z
  P(0.2  Z  2)
5 
 5
 P( Z  2)  P( Z  0.2)
=P(0<Z  2)+P(0<Z<0.2)=0.4472+0.0793=0.5565
20  25 

b) P( X  20)  P  Z 

5


 P( Z  1)
 P( Z  1)  0.84134
0.5+0.3413 = 0.8413
EXERCISES 2:
1. Suppose X is a normal distribution, N(70,4). Find
a) P (67  X  75)
b)
P( X  74)
2. Suppose the test scores of 600 students are normally distributed
with a mean of 76 and standard deviation of 8. The number of
scoring is from 70 to 82.
EXAMPLE
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)
with   np and  2  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
Therefore, X N (135, 74.25)
154.5  135 

PZ 
  P( Z  2.26)  0.5  0.4881  0.0119
74.25 

EXERCISES
Suppose that 5% of the population over 70 years
old has disease A. Suppose a random sample of
9600 people over 70 is taken. What is the
probability that fewer than 500 of them have
disease A?
Answer:
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
then X
N ( ,  )
Po ( ), if   10,
EXAMPLE & SOLUTION:
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?
X is the number of customers 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)  0.5  0.4846  0.9846
45 

EXERCISE 3 :
1.
Reported that the mean weekly income of a
shift foreman in the glass industry is normally
distributed with a mean of $1000 and
standard deviation of $100. What is the
probability of selecting a shift foreman in the
glass industry whose income is
a)
Between $1000 and $1100.
Between $790 and $1000.
Between $840 and $1200.
b)
c)
2. A study by Great Southern Home Insurance
revealed that none of the stolen goods were
recovered by the homeowners in 80 percent of
reported thefts.
a)
b)
During a period in which 200 thefts occurred,
what is the probability that no stolen goods
were recovered in 170 and more of the
robberies?
During a period in which 200 thefts occurred,
what is the probability that no stolen goods
were recovered in at least 150 of the
robberies?