Transcript Chapter 3 - Portal UniMAP

ROHANA
ROHANABINTI
BINTIABDUL
ABDULHAMID
HAMID
INSTITUT
INSTITUTE FOR
E FORENGINEERING
ENGINEERINGMATHEMATICS
MATHEMATICS(IMK)
(IMK)
UNIVERSITIMALAYSIA
MALAYSIAPERLIS
PERLIS
UNIVERSITI
Free Powerpoint Templates
CHAPTER 3
PROBABILITY DISTRIBUTION
(PART 1)
PROBABILITY DISTRIBUTION
• 3.1
Introduction
Binomial
distribution
• 3.2
• 3.3
Poisson
distribution
Normal
distribution
• 3.4
3.1 INTRODUCTION

A probability distribution is obtained
when probability values are assigned to
all possible numerical values of a
random variable.

Probability distribution can be classified
either discrete or continuous.
• BINOMIAL
DISTRIBUTION
DISCRETE
DISTRIBUTIONS • POISSON
DISTRIBUTION
CONTINUOS • NORMAL DISTRIBUTION
DISTRIBUTIONS
3.2 THE BINOMIAL DISTRIBUTION
Definition 3.1 :
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.
1) Calculate probability using formula
Definition 3.2 :
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)  Cx p q
n

x = 0, 1, 2, ......, n
x
n x
2) Calculate probability using table

Table of binomial can be used to find the probabilities
using the following rules as the guidelines.
i. P( X  x)  P( X  x)  P( X  x  1)
ii. P( X  x)  P( X  x  1)
iii . P( X  x)  P( X  x  1)  1  P( X  x)
iv. P( X  x)  1  P( X  x  1)
v. P( x1  X  x2 )  P( X  x2 )  P( X  x1 1)
vi. P( x1  X  x2 )  P( X  x2 1)  P( X  x1 )
Definition 3.3 :The Mean and Variance of X
If X ~ B(n,p), then
Mean
Variance
where
 n is the total number of trials,
 p is the probability of success and
 q is the probability of failure.
Standard
deviation
EXAMPLE 3.1
Given
that X~B(12, 0.4), find
a ) P ( X  2)
b) P (2  X  5)
c) E ( X )
d ) Var ( X )
SOLUTIONS
a) P( X  2)
 C2 (0.4) (0.6)  0.064
12
2
10
b) P(2  X  5)
 P( X  2)  P( X  3)  P( X  4)
 C2 (0.4) (0.6)  C3 (0.4) (0.6)  C4 (0.4) (0.6)
12
2
10
12
 0.064  0.142  0.213
 0.419
3
9
12
4
8
e) E ( X )  np
= 12(0.4)
=4.8
f) Var ( X )  npq
= 12(0.4)(0.6)
= 2.88
Exercise
• In Kuala Lumpur, 30% of workers take
public transportation daily. In a sample of
10 workers,
I. What is the probability that exactly three
workers take public transportation daily?
II.What is the probability that at least three
workers take public transportation daily?
III.Calculate the standard deviation of this
distribution.
Powerpoint Templates
Page 13
3.3 The Poisson Distribution
Definition 3.3
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   x
P( X  x) 
for x  0,1, 2,3,...
x!


λ (Greek lambda) is the long run mean
number of events for the specific time or space
dimension of interest.
A random variable X having
distribution can also be written as
X ~ Po ( )
with E ( X )   and Var ( X )  
a
Poisson
EXAMPLE 3.2
Given that X ~ Po (4.8), find
a) P( X  0)
b) P( X  9)
c) P( X  1)
SOLUTIONS
a) P ( X  0) 
b) P( X  9) 
e
4.8
0
4.8
9
e
4.8
 0.0082
0!
4.8
 0.0307
9!
c) 1  P ( X  0)  1  0.0082
= 0.9918
EXAMPLE 3.3
Suppose that the number of errors in a piece of
software has a Poisson distribution with
parameter   3 . Find
a) the probability that a piece of software has no
errors.
b) the probability that there are three or more
errors in piece of software .
c) the mean and variance in the number of
errors.
SOLUTIONS
e 3  30
a) P( X  0) 
0!
 e3  0.050
b)P( X  3)  1  P( X  0)  P( X  1)  P( X  2)
e 3  30 e 3  31 e 3  32
 1


0!
1!
2!
3 9
3  1
 1 e    
1 1 1 
 1  0.423  0.577
Exercise 1
• Phone calls arrive at the rate of 48 per
hour at the reservation desk for Regional
Airways
I. Find the probability of receiving three
calls in a 5-minutes interval time.
II.Find the probability of receiving more than
two calls in 15 minutes.
Powerpoint Templates
Page 20
Exercise 2
• An average of 15 aircraft accidents occurs
each year. Find
I. The mean, variance and standard
deviation of aircraft accident per month.
II.The probability of no accident during a
months.
Powerpoint Templates
Page 21
IMPORTANT!!!!
exactly two
=2
More than two/
Exceed two
2
Two or more/
At least two/
Two or more
2
less than two/
Fewer than two
At most two/
Two or fewer/
Not more than
two
2
2
DISCRETE
DISTRIBUTIONS
• BINOMIAL DISTRIBUTION
• POISSON DISTRIBUTION
CONTINUOUS • NORMAL DISTRIBUTION
DISTRIBUTIONS
3.4 NORMAL DISTRIBUTION
3.4.1 INTRODUCTION
3.4.2 NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION
3.4.3 NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
3.4 .1 INTRODUCTION
Definition 3.4
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
1
f ( x) 
e
 2
1  x 
 

2  
2
  x  
If X ~ N (  ,  2 ) then E  X    and V  X    2
Applications of normal distribution
Many naturally occurring random processes tend
to have a distribution that is approximately
normal. Examples can be found in any field, these
include:
 heights and weights of adults
 length and width of leaves
of the same species
 actual weights of rice in 5 kg bags sold in
supermarkets

The Standard Normal Distribution

The normal distribution with parameters
  0 and  2  1 is called a standard normal
distribution.

A random variable that has a standard normal
distribution is called a standard normal random
variable and will be denoted by
.
Z ~ N (0,1)
Standardizing A Normal Distribution
If X is a normal random variable with E ( X )   and V ( X )   2 ,
the random variable
X 
Z

is a normal random variable with E ( Z )  0 and V ( Z )  1.
That is Z is a standard normal random variable.
  1010
  20
X
1030  1010
Z
1
20
Z
X 

Z
Standard normal distribution
Total area =1
0.5
0.5
Z
 0
EXAMPLE 3.1
Determine the probability or area for the portions
of the Normal distribution described. (using the
table)
1. P ( Z  3)
2.P ( Z  2.53)
3.P ( Z  1.51)
4.P (1  Z  2.93)
5.P (0.74  Z  3.1)
6.P (2  Z  1)
SOLUTIONS
P( Z  3)
1. P( Z  3)
Z 3
Using
table P( Z  z )  ( z ) ( Appendix 1)
P( Z  3)  (3)
 0.0013
P( Z  2.53)
P( Z  2.53)
2.P( Z  2.53)
Z  2.53
Using
Z  2.53
table P( Z  z )  ( z ) ( Appendix 1)
P(Z  2.53)
 P(Z  2.53)
 (2.53)
 0.0057
3.P( Z  1.51)
 1  P ( Z  1.51)
 1 0.0655
 0.9345
4.P(1  Z  2.93)
 P( Z  1)  P( Z  2.93)
 0.1587  0.0017
 0.157
5.P(0.74  Z  3.1)
 1  P( Z  3.1)  P( Z  0.74)
 1  0.0010  0.2296
 0.7694
6.P(2  Z  1)
 P( Z  1)  P( Z  2)
 0.1587  0.0228
 0.1359
EXAMPLE 3.2
The masses of a well known brand of
breakfast
cereal
are
normally
distributed with mean of 250g and
standard deviation of 4g. Find the
probability of a packet containing more
than 254.4g.
SOLUTIONS
 Let
X be the r.v. “masses of cereal in grams”
where X~N(250, 16).
P( X  254.4)
254.4  250
 P( Z 
)
4
 P ( Z  1.1)
 0.1357
Z
x

EXERCISE 1
A battery has a lifetimes which are normally
distributed with a mean of 62 hours and a
standard deviation of 3 hours. What is
the probability of battery lasting less than
68 hours?
EXERCISE 2
A carton of orange juice has a volume which
is normally distributed with a mean of
120ml and a standard deviation of 1.8ml.
Find the probability that the volume is more
than 118ml.
EXERCISE 3
The pulse rate is a measure of the number
of heart beats per minute. Suppose that
the pulse rates for adults are assumed to
be normally distributed with a mean of 78
and a standard deviation of 12.
Find the probability that adults will have
the pulse rates between 60 and 100.
Find the area/probability
P ( Z  Z )  

z
Example 3.3
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
1. P( Z  Z )  0.25
  0.25
Z
Using
table P( Z  z )   ( Appendix III )
Z  0.6745
2. P(Z  Z )  0.36
  0.36
Z
Using
table P( Z  z )   ( Appendix III )
Z  0.3585
3. P( Z  Z )  0.983
1  0.983  0.017
Z  2.1201
4. P( Z  Z )  0.89
1  0.89  0.11
Z  1.2265
Example 3.4
In January 2003, the American worker spent
an average of 77 hours logged on to the
internet while at work. Assume that the
population mean is 77 hours, the times are
normally distributed, and the standard
deviation is 20 hours.
A person is classified as heavy user if he or
she is in the upper 20% of usage. How
many hours did a worker have to be logged
on to be considered a heavy user?
SOLUTIONS
 Let
X be the r.v. “hours of worker spent on
internet” where X~N(77, 202).
  0.2
P( Z  Z )  0.2
Z  0.8416
X  77
Z 
 0.8416
20
X  93.83 hours
Z
3.4 NORMAL DISTRIBUTION
3.4.1 INTRODUCTION
3.4.2 NORMAL APPROXIMATION OF THE
BINOMIAL DISTRIBUTION
3.4.3 NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
3.4.2 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.
X ~ B(n, p)
RULES
1. n  30
2. np  5 and nq  5
X ~ N (np, npq )
Definiton 3.5
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.
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 3.5
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
 Let
X be the r.v. “number of male voters” where
X~B(300, 0.45).
Rules :
1.n  300  30
2.np  300(0.45)  135  5
nq  300(0.55)  165  5
Hence, we use normal approximation
c .c
P( X  155)  P( X  155  0.5)  P( X  154.5)
P( X  154.5)
154.5  300(0.45)
 P( Z 
)
300(0.45)(0.55)
 P( Z  2.26)
 0.0119
2.26
3.4 NORMAL DISTRIBUTION
3.4.1 INTRODUCTION
3.4.2 NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION
3.4.3 NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
3.4.3 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.
Definition 3.6
Given a random variable X ~ Po ( ), if   10, then X ~ N ( ,  )
with Z 
X 

X ~ P0 ( )
Rule
1.  10
X ~ N ( ,  )
Example 3.6
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?
Solutions
Let X be the r.v. “number of customers per hour”
where X~P0 (5).
  5 for 1 hour
  45 for 9 hours (8am  5 pm)
Let X be the r.v. “number of customers for 9
hours” where X~P0 (45).
Rule
1.  45  10
use normal approximation
c .c
P( X  30)  P( X  30  0.5)  P( X  30.5)
P( X  30.5)
30.5  45
 P( Z 
)
45
 P( Z  2.16)
 1  P( Z  2.16)
 1  0.0154
 0.9846
EXERCISE 1
According to a survey by Duit magazine,
27% of women expect to support their
parents financially. Assume that this
percentage holds true for the current
population of all women. Suppose that a
random sample of 300 women is taken.
Find the probability that exactly 79 of the
women in this sample expect to support
their parents financially.
EXERCISE 2
Aonang Beach Resort Hotel has 120
rooms. In the spring months, hotel room
occupancy is approximately 75%.
I. What is the probability that 100 or more
rooms are occupied on a given day.
II. What is the probability that 80 or fewer
rooms are occupied on a given day?
EXERCISE 3
In a university, the average of the students
that come to the student health center is 5
students per hour. What is the probability
that at least 40 students will come to the
student health center from 9.00 am to 6.00
pm?
EXERCISE 4
Suppose that at a certain automobile plant
the average number of work stoppages
per day due to equipment problems during
the production process is 12.0. What is the
approximate probability of having 15 or
fewer work stoppages due to equipment
problems on any given day?