Transcript chapter 1(part 2)

CHAPTER 1
EQT 271
(PART 2)
BASIC STATISTICS
1.4 PROBABILITY DISTRIBUTION

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
CONTINUOUS • NORMAL DISTRIBUTION
DISTRIBUTIONS
THE 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.
Example of Binomial Distribution
1.
2.
3.
A university found that 20% of its students withdraw
without completing the introductory statistics course.
Assume that 20 students registered for the course. Find
the probability….
In Kuala Lumpur, 30% of workers take public
transportation daily. In a sample of 10 workers,….
Suppose 20% of the marbles packed in a box are
red in color. Suppose 4 marbles are chosen at
random. Find the probability of …
1) Calculate probability using formula
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 )
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 1
Given
that X~B(12, 0.4), find
a ) P( X  2)
b) P(2  X  5)
c) E ( X )
d ) Var ( X )
Try to find the probabilities
using table. Do you get the
same answer???
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 1
• 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
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Solution
X=number of workers take public transportation
daily.
X ~ B (10,0.3)
i ) P( X  3)
10
3
7
 C3 (0.3) (0.7)
 0.2668
ii ) P( X  3)
 1  P( X  0)  P( X  1)  P( X  2)
 110C0 (0.3) 0 (0.7)10 10C1 (0.3)1 (0.7)9 10C2 (0.3) 2 (0.7)8
 1  0.02825  0.12106  0.2335
 0.61719
iii )Standard deviation  npq
 10(0.3)(0.7)
 1.449
Extra Exercise
• In a large shipment of automobile tires, 5%
have a certain flaw. A sample of four tires
was chosen to be installed on a car. Let X
be the random variable of tires have flaw.
a) What is probability that at least one of
the tires has a flaw?
b) What is the probability that exactly three
of the tires have no flaw?
Powerpoint Templates
Page 15
The Poisson Distribution
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,...


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
REMEMBER!!!!!!!!!!!!!!!!!!!!
Average
Rate
Mean

EXAMPLE 2
Given that X ~ Po (4.8), find
a) P( X  0)
b) P( X  9)
c) P( X  1)
Try to find the probabilities
using table. Do you get the
same answer???
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
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 2
• 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 23
Exercise 3
• 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
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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
NORMAL DISTRIBUTION
INTRODUCTION
NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION
NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
INTRODUCTION
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
Normal Distribution
The Normal Distribution has:
 mean = median = mode
 symmetry about the center
 50% of values less than the mean
and 50% greater than the mean
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.
Why Standardize ... ?
It can help you make decisions about your
data.
 It also makes life easier because we only need
one table (the Standard Normal Distribution
Table), rather than doing calculations
individually for each value of mean and
standard deviation.

Example: Professor Willoughby is marking a
test.
Here are the students results (out of 60 points):
20, 15, 26, 32, 18, 28, 35, 14, 26, 22, 17
 Most students didn't even get 30 out of 60, and most will fail.The
test must have been really hard, so the Prof decides to
Standardize all the scores and only fail people 1 standard
deviation below the mean.
 The Mean is 23, and the Standard Deviation is 6.6, and these
are the Standard Scores:
 -0.45, -1.21, 0.45, 1.36, -0.76, 0.76, 1.82, -1.36, 0.45, -0.15, 0.91
 Only 2 students will fail (the ones who scored 15 and 14 on the
test)

  1010
  20
X
1030  1010
Z
1
20
Z
X 

Z
Standard normal distribution
Total area =1
0.5
0.5
Z
 0
EXAMPLE 4
Determine the probability or area for the portions
of the Normal distribution described. (using the
normal 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 (0  Z  z )   ( z ) ( Appendix 2)
P( Z  3)
 0.5  P(0  Z  3)
 0.5  (3)
 0.5  0.4987  0.0013
P( Z  2.53)
P( Z  2.53)
2.P( Z  2.53)
Z  2.53
P( Z  2.53)
 0.5  P(0  Z  2.53)
 0.5  0.4943
 0.0057
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)
CALCULATE THE PROBABILITIES USING
CALCULATOR
-
-
Mode: SD
1
Shift 3
P(1) for P( Z  z )
Q (2) for P(0  Z  z )
R (3) for P( Z  z )
EXAMPLE 4
Determine the probability or area for the portions
of the Normal distribution described. (using the
calculator)
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)
1. P( Z  3)  0.0135
2.P( Z  2.53)  0.057
3.P ( Z  1.51)  0.93448
4.P(1  Z  2.93)
5.P (0.74  Z  3.1)
6.P (2  Z  1)
EXAMPLE 5
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)
 0.13567
Z
x

EXERCISE 3
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 4
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 5
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 value of Z
P ( Z  Z )  

z
Example 5
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 6
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
NORMAL DISTRIBUTION
INTRODUCTION
NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION
NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
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 )
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 7
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
NORMAL DISTRIBUTION
INTRODUCTION
NORMAL APPROXIMATION OF THE BINOMIAL
DISTRIBUTION
NORMAL APPROXIMATION OF THE POISSON
DISTRIBUTION
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 

X ~ P0 ( )
Rule
1.  10
X ~ N ( ,  )
Example 8
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
Poisson Approximation of the Binomial
Distribution
When the number of observations or trials n in
a binomial experiment is relatively large
(n≥30) and probability of success is very small
(np<5) , the Poisson probability distribution
can be used to approximate binomial
probabilities.
 A convenient rule is that such approximation is
acceptable when

n  30 and np  5
Example 9
Suppose a life insurance company insures
the lives of 5000 men aged 42.
If actual studies show that 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 more than 10 claims
during a given year.
Solutions
Let X = number of claims/number of any 42-yearold man will die in a given year
where X~B(4000,0.001).
Using binomial distribution;
P(X>10)  1  P ( X  10)
Not available
in binomial
tables!!!!
use Poisson approximation since
n  4000  30 and
np  4000(0.001)  4  5
  mean  np  4000(0.001)  4
X ~ P0 (4)
P( X  10)  1  P( X  10)
 1 0.9972
 0.0028
Look at
Poisson table!
EXERCISE 6
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 7
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 8
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 9
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?
EXERCISE 10
 The probability that a person will develop
an infection even after taking a vaccine
that was supposed to prevent the infection
is 0.003.
 In a random sample of 200 people in a
community who got the vaccine, what is
the probability that two or fewer people will
be infected?