Transcript Chapter 1 Probability

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
Example 1.19


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.
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
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 )  nC x 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 1.20
Given that X
a) P( X  2)
b) P( X  3)
c) P( X  4)
d) P(2  X  5)
e) E( X )
f) Var( X )
b(12,0.4), find
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
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 1.21

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 1.22
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)
The number of CEOs who hold this view is 10.
(ii) The number of CEOs who hold this view is between 9 and
12.
(iii) The number of CEOs who hold this view is at most 7.
(iv) Find the mean and standard deviation of binomial
distribution.
Solution:
i) P( X  10) 
ii) P (9  X  12) 
iii) P( X  7) 
iv)   np 
  npq 
Poisson Distribution
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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 1.23
Consider a Poisson random variable with   3 . Calculate the
following probabilities :
a) Write the distribution of Poisson
b) P( X  0) 
c) P( X  1) 
d) P( X  1) 
Solution:
a) Write the distribution of Poisson
b)
c)
d)
Example 1.24
The average number of traffic accidents on a certain section of
highway is two per week. Assume that the number of accidents
follows a Poisson distribution with mean is 2.
i)
ii)
Find the probability of no accidents on this section of highway
during a 1-week period
Find the probability of a most three accidents on this section of
highway during a 2-week period.
Solution:
i) P( X  0) 
ii) P ( X  3) 
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 1.25

0.9786
Example 1.25
Suppose a life insurance company insures the lives of 4000 men
aged 42. If actuarial studies show the probability that any 42 year
old man will die in a given year to be 0.001, find the exact
probability that the company will have to pay x = 4 claims during a
given year.
Solution:
Step:
1.
Write the distribution of Binomial X ~ B(4000,0.001)
2.
The value for n is large and value of p is too small, check
whether np  5
3.
If yes, proceed to solve using Poisson Approximation. Use
formula e    x
x!
Exercise 1.4
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)
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
CHARACTERISTICS OF NORMAL
DISTRIBUTION
‘Bell Shaped’


Symmetrical
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
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 ,
and a standard deviation of 1,  2  1

Z is denoted by Z ~ N (0,1)

Thus, its density function becomes
 0
Calculating Probabilities for a General Normal Random Variable





Mostly, the probabilities involved x, a normal random variable
with mean, and standard deviation,
Then, you have to standardized the interval of interest, writing it
in terms of z, the standard normal random variable.
Once this is done, the probability of interest is the area that you
find using the standard normal probability distribution.
Normal probability distribution, X ~ N (  , 2 )
Need to transform x to z using
X 
Z

PATTERNS FOR FINDING AREAS UNDER THE STANDARD NORMAL CURVE
Example 1.25
Z table
Example 1.26
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)
Exercise 1.5
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)
Answer : a) 0.1736, b) 0.4783, c) 0.8078, d) 0.9812, e) 0.0614
Example 1.27
Z table
Exercise 1.6
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 1.28
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
Example 1.29
An assessment test is used to measure a person’s readiness for
college. The mathematics scores in the test are scaled to have a
normal distribution with mean 500 and standard deviation 100.
i.
ii.
What is the probability that the people taking the test will
score below 350?
Remedial assistance will be given to students in the bottom
10%. What is the maximum score of this group of students?
Solution:
i. P( X  350) 
ii. P( Z  max  500 ) 
100
Exercise 1.7
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 is:
Answer : 1. a) 0.927 b) 0.0228 2. 328 students
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
How do calculate Binomial Probabilities Using the Normal
Approximation?





Find the necessary values of n and p. Calculate   np and   npq
Write the probability you need in terms of x.
Correct the value of x with appropriate continuous correction
factor (ccf).
Convert the necessary x-values to z-values using
x  0.5  np
z
npq
Use Standard Normal Table to calculate the approximate
probability.
Example 1.30
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 

Exercise 1.8
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 less than 500 of them have disease
A?
Answer: 0.8186
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 1.31
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 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 1.9
The average number of accidental drowning in United States per
year is 3.0 per 100000 population. Find the probability that in a city
of population 400000 there will be less than 10 accidental drowning
per year.
Answer : 0.2358
Exercise 1.10
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)
b)
Between $1000 and $1100.
Between $790 and $1000.
c)
Between $840 and $1200.
Answer : a) 0.3413, b) 0.4821, c) 0.9224