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transcript
Chapter 4
Probability Distributions
Lesson 4-1/4-2
Random Variable
Probability Distributions
This chapter will deal the construction of probability distribution.
By combining the methods of descriptive statistics in Chapter 2 and those of probability presented in Chapter 3.
Probability Distributions will describe what will probably happen instead of what actually did happen.
Combining Descriptive Methods
and Probabilities
In this chapter we will construct probability distributions by
presenting possible outcomes along with the relative
frequencies we expect.
Definitions
A random variable is a variable (typically
represented by X) that has a single
numerical value, determined by chance, for
each outcome of a procedure.
A probability distribution is a graph, table,
or formula that gives the probability for each
value of the random variable.
2 Types of Random Variables
Discrete Random Variable has either a
finite number of values or countable number
of values, where “countable” refers to the fact
that there might be infinitely many values, but
they result from a counting process.
Continuous Random Variable has infinitely
many values, and those values can be
associated with measurements on a
continuous scale in such way that there are
no gaps or interruptions.
Example – Page 192, #2
Identify the given variable as being discrete or continuous.
A. The cost of making a randomly selected movie.
B. The number of movies currently being shown in U.S.
theaters.
C. The exact running time of a randomly selected movie.
Discrete
Discrete
Continuous
Example – Page 192, #2
D. The number of actors appearing in a randomly selected
movie.
E. The weight of the lead actor in a randomly selected
movie.
Discrete
Continuous
Graphs
The probability histogram is very similar to a relative
frequency histogram, but the vertical scale shows probabilities.
Requirements for a Discrete
Probability Distribution
1.
2.
P x( ) 1
P x0 ( ) 1
where x assumes all possible values
for every individual value of x
Describing a Distributions
Center
Value that indicates where the middle of the data set is
located.
Variation
Measures the amount that the values vary among
themselves
Distribution
Shape of the distribution of data
Outliers
Sample values that lie very far away from the vast
majority of the other sample values
Mean, Variance and Standard
Deviation of a Probability
Distribution
[ ( )] x P x
22 ( )
x P x
2 2 2( ) x P x
2 2( ) x P x
Mean
Variance
Variance (shortcut)
Standard Deviation
Roundoff Rule for μ, σ, and σ²
Round results by carrying one more decimal
place than the number of decimal places used
for the random variable x.
If the values of x are integers, round μ, σ, and
σ² to one decimal place.
Example – Page 192, #4
Determine whether a probability distribution is given. If probability
distribution is described, find its mean and standard deviation.
When manufacturing DVDs for Sony, batches of DVD’s are
randomly selected and the number of defects x is found for
each batch.
x P x( )
0 0.502
1 0.365
2 0.098
3 0.011
4 0.001
P x( ) 0.977 1
Not a probability distribution
Example – Page 192, #6
Determine whether a probability distribution is given. If probability
distribution is described, find its mean and standard deviation.
The Telektronic Company provides life insurance policies
for its top four executives, and the random variable x is the
number of those employees who live through next year.
x P x( )
0 0.0000
1 0.0001
2 0.0006
3 0.0387
4 0.9606
P x( ) 1
A probability distribution
Example – Page 192, #6
0 0.0000 0 0
1 0.0001 0.0001 0.0001
2 0.0006 0.0012 0.0024
3 0.0387 0.1161 0.3483
4 0.9606 3.8424 15.37
x P x( ) x P x( )
3.9598
x P x2 ( )
15.7204
Example – Page 192, #6
3.9598 4 x P xμ ( )
x P x2 2 2σ ( ) μ 2
15.7204 3.9598 .04038
2σ σ 0.04038 0.2009 0.2
Example – Page 192, #6
Use the TI to find the mean, variance, and standard deviations.
x P(x) StatCALC
Identifying Unusual Results
According to the range rule of thumb, most
values should lie within 2 standard deviation
of the mean.
We can therefore identify “unusual”
values by determining if they lie outside
these limits:
Maximum usual value = + 2
Minimum usual value = – 2
Identifying Unusual Results
Probabilities
Rare Event Rule
Given the assumption that boys and girls are equally likely the probability of a particular observed event (such as 13 girls in 14 births) is extremely small, we conclude that the assumption is probably not correct.
Unusually high
x successes among n trials is an unusually highnumber of successes if P(x or more) is very small (such as 0.05 or less).
Unusually low
x successes among n trials is an unusually lownumber of successes if P(x or fewer) is very small (such as 0.05 or less).
Example – Page 194, #18
x (girls) P(x)
0 0.000
1 0.001
2 0.006
3 0.022
4 0.061
5 0.122
6 0.183
7 0.209
8 0.183
9 0.122
10 0.061
11 0.022
12 0.006
13 0.001
14 0.000
Assume that in a test of gender-selections
technique, in clinical trial results in 12 girls and
14 births. Refer to Table 4-1 and find the
indicated probabilities.
A. Find the probability of exactly 12 girls
in 14 births.
P x( 12) 0.006
Example – Page 194, #18
x (girls) P(x)
0 0.000
1 0.001
2 0.006
3 0.022
4 0.061
5 0.122
6 0.183
7 0.209
8 0.183
9 0.122
10 0.061
11 0.022
12 0.006
13 0.001
14 0.000
P x( 12)
P x P x P x( 12) ( 13) ( 14)
B. Find the probability of 12 or more
girls in 14 births.
0.006 0.001 0.000
0.007
Example – Page 194, #18
C. Which probability is relevant for determining whether 12
girls in 14 births is unusually high: the results from part
(A) or part (B)
Part (B), the occurrence of 12 girls among 14 would be
unusually high if the probability of 12 or more
girls is very small 0.007 ≤ 0.05
Example – Page 194, #18
D.Does 12 girls in 14 births suggest that the gender-
selection technique is effective? Why or why not?
Yes, because the probability of 12 or more girls is
very small (0.007). The result of 12 or more girls could
not easily occur by chance.
Definition
The expected value of a discrete
random variable is denoted by E, and
it represents the average value of the
outcomes.
It is obtained by find the value of E
E x P x( )
The expected value of a discrete
random variable is denoted by E, and
it represents the average value of the
outcomes.
It is obtained by find the value of E
E x P x( )
Example – Page 192, #12
When you give the casino $5 for a bet on the number 7 in a
roulette, you have 1/38 probability of winning $175 and 37/38
probability of losing $5. If you bet $5 that outcome is an odd
number, the probability of winning $5 is 18/38, and the
probability of losing $5 is 20/38.
Example – Page 192, #12
When you give the casino $5 for a bet on the number 7 in a roulette, you
have 1/38 probability of winning $175 and 37/38 probability of losing $5. If
you bet $5 that outcome is an odd number, the probability of winning $5 is
18/38, and the probability of losing $5 is 20/38.
A. If you bet $5 on the number 7, what is your expected
value?x P x x P x( ) ( )
1 175175
38 38
37 1855
38 38
10
38
E10
0.263138
The expected loss is $0.263
for a $5 bet.
Example – Page 192, #12
When you give the casino $5 for a bet on the number 7 in a roulette, you
have 1/38 probability of winning $175 and 37/38 probability of losing $5. If
you bet $5 that outcome is an odd number, the probability of winning $5 is
18/38, and the probability of losing $5 is 20/38.
B. If you bet $5 that the outcome is an odd number, what is
your expected value?x P x x P x( ) ( )
18 905
38 38
20 1005
38 38
10
38
E10
0.263138
The expected loss is $0.263
for a $5 bet.
Lesson 4-3
Binomial Probability Distribution
Binomial Probability
Distribution
Specific type of discrete probability
distribution
The outcomes belong to two
categories
pass or fail
acceptable or defective
success or failure
Example of a Binomial
Distribution
Suppose a cereal manufacturer puts pictures of famous
athletes on cards in boxes of cereal, in the hope of
increasing sales. The manufacture announces that 20%
of the boxes contain a picture of Tiger Woods, 30% a
picture of Lance Armstrong, and the rest a picture of
Serena Williams.
You buy 5 boxes of cereal. What’s the probability you
get exactly 2 pictures of Tigers Woods?
Requirements for a Binomial
Probability Distribution
1) The procedure has a fixed number of trials
2) Trials are must be independent. (The
outcome of any individual trial doesn’t affect
the probabilities in the other trials.)
3) Each trial must have all outcomes classified
into two categories
4) Probabilities must remain constant for each
trial.
Example – Page 203, #4
Determine whether the given procedure results in a
binomial distribution.
Rolling a loaded dice 50 times and find the number
of times that 5 occurs.
Binomial
Notation for a Binomial
Distribution
There are n number of fixed trials
Let p denote the probability of success
Let q denote the probability of failure
q = 1 – p
Let x denote the number of success in n
independent trials:
Let P(x) denotes the probability of getting exactly
x successes among the n trials
x n0
Important Hints
Be sure that x and p both refer to the same category being called a success.
When sampling without replacement, the events can be treated as if they were independent if the sample size is no more than 5% of the population size.
Where n N0.05
Mathematical Symbols
Phrase Math Symbols
“at least” or “no less than” ≥
“more than” or “greater than” >
“fewer than” or “less than” <
“no more than” or “at most” ≤
“exactly” =
Methods for Finding Probabilities
of a Binomial Distribution
Using the Binomial Probability Formula
Using Table A-1 in Appendix A
Using the TI
Binomial Probability Formula
x n xn
n x xP x p q
!
! !( )
Number of outcomes
with exactly x
successes among
n trials
Probability of x
successes among
n trials for any
particular order
x n x
n xP x C p q( )
Binomial Probability Formula
x n x
n xP x C p q( )
for x = 0, 1, 2, ……,n
where
n = number of trials
x = number of success among n trials
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 – p)
Example 1 – Cereal
Suppose you buy 5 boxes of cereal. Where n = 5 and p =
0.2. What’s the probability you get exactly 2 pictures of Tiger
Woods?
5 2
5 5!10
2 2! 5 2 !C
First find the total number of outcomes.
There are 10 ways to get 2
Tiger pictures in 5 boxes.
Example 1 – Cereal
Suppose you buy 5 boxes of cereal. Where n = 5 and p = 0.2.
What’s the probability you get exactly 2 pictures of Tiger Woods?
2 3( 2) 10(0.20) (0.80) 0.2048P X
There are 10 ways to get 2 Tiger pictures in 5 boxes.
2nd Vars
Example – Cereal
The following table show the probability distribution function
(p.d.f) for the binomial random variable, X.
X = Tiger P(X)
0 0.32768
1 0.4096
2 0.2048
3 0.0512
4 0.0064
5 0.00032
Sum 1
(5,0.20,0) 0.32768
(5,0.20,1) 0.4096
(5,0.20,2) 0.2048
(5,0.20,3) 0.0512
(5,0.20,4) 0.0064
(5,0.20,5) 0.00032
Binompdf
Binompdf
Binompdf
Binompdf
Binompdf
Binompdf
Example – Cereal
The following table show the cumulative distribution function
(c.d.f) for the binomial random variable, X.
0 1 2 3 4 5
( ) 0.32768 0.4096 0.2048 .0512 .0064 .00032
( 0) ( 1) ( 2) ( 3) ( 4) ( 5)( )
0.32768 0.73728 0.94208 0.99328 0.99968 1
cdf
X
P X
P X P X P X P X P X P XP X
(5,0.20,0) 0.32768
(5,0.20,1) 0.73728
(5,0.20,2) 0.94208
(5,0.20,3) 0.99328
Binomcdf
Binomcdf
Binomcdf
Binomcdf
(5,0.20,4) 0.99968
(5,0.20,5) 1
Binomcdf
Binomcdf
Example – Cereal
0 1 2 3 4 5
( ) 0.32768 0.4096 0.2048 .0512 .0064 .00032
( 0) ( 1) ( 2) ( 3) ( 4) ( 5)( )
0.32768 0.73728 0.94208 0.99328 0.99968 1
cdf
X
P X
P X P X P X P X P X P XP X
Construct a histogram of the pdf and cdf using X[0, 6]1 and Y[0, 1]0.01
pdf cdf
TI – Binomial Probability
Computing exact probabilities
2nd/Vars/binompdf
• binompdf(n, p, x)
pdf: probability distribution function
Computing less than or equal to probabilities
2nd/Vars/binomcdf
• binomcdf(n, p, x)
cdf: cumulative distribution function
Example – Page 204, #18
x n xP x C p q6 2( )
Use the Binomial Probability Formula. Assume that a
procedure yields a binomial distribution.
n x p6, 2, 0.45
P x6 22( 2) 15(0.45) 0.55
0.2779 0.278
Binomial Probability Table
Example – Page 204, #12
Using Table A-1. Assume that a procedure yields a
binomial distribution.
n x p7, 2, 0.01
P x( 2) 0.002
Example – Page 204, #14
Using Table A-1. Assume that a procedure yields a
binomial distribution.
n x p6, 5, 0.99
P x( 5) 0.057
Example - Nielson Media
According to Nielson Media research, 75% of US households
have cable TV.
A). In a random sample of 15 households, what is the
probability that 10 have cable?
P x( 10)
binompdf
0.2361
(15,0.75,10)
x = households, p = 0.75, n = 15
Example - Nielson Media
The probability of getting exactly 10 households out 15
with cable is 0.1651. In 100 trials of this experiment we would
expect (100)(0.1651) = 17 trials to result in 10 households
with cable.
Example - Nielson Media
According to Nielson Media research, 75% of US households
have cable TV.
B). In a random sample of 15 households, what is the
probability that fewer than 13 have cable?
P x P x P x P x( 13) ( 0) ( 1) ... ( 12)
binomcdf(15, 0.75, 12) 0.7639P x( 12)
Example - Nielson Media
There is 0.7639 probability that in a random sample
of 15 households, less than 13 will have cable.
In 100 trials of this experiment, we would expect
(0.7639)(100) = 76 trials to result in fewer than 13
households that have cable.
Example – Nielson Media
According to Nielson Media research, 75% of US households
have cable TV.
C). In a random sample of 15 households, what is the
probability that at least 13 have cable?
P x P x P x P x( 13) ( 13) ( 14) ( 15)
binomcdf
0.2361
1 (15,0.75,12)
P x1 ( 12)
Example – Nielson Media
There is 0.2361 probability that in a random sample of 15
households, at least 13 will have cable. In 100 trials at
of this experiment we would expect about (100)(0.2361) = 24
trials to result in at least 13 households have cable
Example – Page 205, #28
An article in USA Today stated that “Internal surveys
paid by the directory assistance providers show that
even the most accurate companies give out wrong
numbers 15% of the time.” Assume that you are
testing such a provider by making 10 requests and also
assume that the provider gives the wrong telephone
number 15% of the time.
A). Find the probability of getting one wrong number.
P x( 1)
binompdf
0.347
(10,0.15,1)
x = the number of wrong answers, n = 10, p = 0.15
Example – Page 205, #28
B). Find the probability of getting at most one wrong number.
P x P x P x( 1) ( 0) ( 1)
0.544 binomcdf (10,0.15,1)
C). If you do get at most one wrong number, does it appear
that the rate of wrong numbers is not 15%, as claimed?
No, since 0.544 > 0.05 “at most one wrong number” is
not an unusual occurrence when the error rate is 15%
Example – Page 206, #32A study was conducted to determine whether there were significant
difference between medical students admitted through special programs
(such as affirmative action) and medical students admitted through the
regular admission criteria. It was found that the graduation rate was
94% for the medical students admitted through special programs.
A). If 10 students from the special programs are randomly
selected, find the probability that at least nine of them
graduated.
P x P x P x( 9) ( 9) ( 10) 1 ( 8)P x
binomcdf1 (10,0.94,8)
0.8824
x = the number of special program student who graduate
n = 10, p = 0.94
Example – Page 206, #32
B). Would it be unusual to randomly select 10 students from
the special programs and get only seven that graduate?
Why or why not?
P x( 7) binompdf (10,0.94,7)
0.01681
Yes, since 0.017 < 0.05 getting only 7 graduates would
be unusual result.
Lesson 4-4
Binomial Distribution: Mean,
Variance and Standard Devation
Any Discrete
Probability Distribution Formulas
Std. Dev = [ x 2 • P(x) ] – µ 2
Mean µ = [x • P(x)]
Variance 2= [ x 2 • P(x) ] – µ 2
Binomial Distribution
Std. Dev. = n • p • q
Where
n = number of fixed trials
p = probability of success in one of the n trials
q = probability of failure in one of the n trials
Mean µ = n • p
Variance 2 = n • p • q
Interpretation of Results
Maximum usual values = µ + 2
Minimum usual values = µ – 2
The range rule of thumb suggests that values
are unusual if they lie outside of these limits:
Example – Page 210, #2
Finding , , and Unusual Values. Assume that a procedure
yields a binomial distributions with n trials and the probability
of success for one trial is p.
n = 250, p = 0.45
250(0.45)
112.5
np
(250)(0.45)(1 0.45)
7.86
npq
min 2 112.5 2(7.86) 96.8
max 2 112.5 2(7.86) 128.2
Example – Page 210, #6
Several students are unprepared for a multiple-choice quiz
with 10 questions, and all of their answers are guesses.
Each question has five possible answers, and only one
of them is correct.
A). Find the mean and standard deviation for the number
of correct answers for such students.
n 10
p1
0.205
q 1 0.20 0.80
npμ
10(0.20)
2
npqσ
(10)(0.20)(0.80)
1.264
1.3
Example – Page 210, #6
B). Would it be unusual for a student to pass by guessing
and getting at least 7 correct answers? Why or why not?
Yes, since 7 is greater than or equal to the maximum
usual value, it would be unusual for a student to
pass by getting at least 7 correct answers.
max 2 2 2(1.3) 4.6
Example – Page 210, #10
The Central Intelligence Agency has specialists who
analyze the frequencies of letters of the alphabet in an
attempt to decipher intercepted messages. In standard
English text, the letter r is used at a rate of 7.7%
A). Find the mean and standard deviation for the number
of times the letter r will be found on a typical page of
2600 characters.
n 2600
p 0.077
q 1 0.077
0.923
npμ
2600(0.077)
200.2
npqσ
(2600)(0.077)(0.923)
13.6
Example – Page 210, #10
B). In the intercepted message sent to Iraq, a page of
2600 characters is found to a have the letter r occurring
175 times. Is this unusual?
μ 2σ unusual values are outside of these
boundaries.
200.2 2(13.6) 173 and 227.4
No, since 175 is within the above limits it would not
be considered an unusual result.
Example – Page 212, #16
Car Crashes: For drivers in the 20 – 24 age bracket, there
is a 34% rate of car accidents in one year (based on the data
from the National Safety Council). An insurance investigator
finds that in a group of 500 randomly selected drivers age
20 – 24 living in New York City, 42% had accidents in the last
year.
A) How many drivers in the New York City group of 500
had accidents in the last year?
(500) (0.42) 210
Example – Page 212, #16
B) Assuming that the same 34% rate applies to New York
City, find the mean and standard deviation for the number
of people in groups of 500 that can expected to have
accidents
500(0.34)
170.00
np
500(0.34)(1 0.34)
10.6
npq
Example – Page 212, #16
C) Based on the preceding results, does 42% result for the
New York City drivers appear to be unusually high when
compared to the 34% rate for the general population?
Does it appear that higher insurance rates for New York
City drivers justified?
max 2
170 2(10.6) 191.2
Yes, the result of 210 accidents for NYC drivers are
unusually high if the 34% rate for the general population
applies. Yes, it appears that the higher rates for NYC
drivers are justified.
Lesson 4-5
The Poisson Distribution
Examples of Poisson
Distribution
Planes arriving at an airport
Arrivals of people in a line
Cars pulling into a gas station
DefinitionThe Poisson distribution is a discrete probability distribution that applies to occurrences of some event over a specified interval. The random variable x is the number of occurrences of the event in an interval. The interval can be time, distance, area, volume, or some similar unit.
P(x) = where e 2.71828µ x • e -µ
x!
Formula
Requirements for a
Poisson Distribution
Random variable (x) is the number of occurrences of an
event over some interval.
Occurrences must be random
Occurrences must be independent of each other
Occurrences must be uniformly distributed over the
interval being used.
mean is μ
Standard deviation is σ μ
Difference from a Binomial
Distribution
The binomial distribution is affected by the
sample size n and the probability p, whereas
the Poisson distribution is affected only by
the mean .
In a binomial distribution the possible values
of the random variable x are 0, 1, … n, but a
Poisson distribution has possible x values of
0, 1, …., with no upper limit.
Example – Page 216, #2
Use a Poisson Distribution to find probability.
If μ = 0.5, find P(2).
x eP x
x
μμ( )
!
eP
2 0.50.5(2)
2!
0.0758
Example – Page 216, #6
Currently, 11 babies are born in the village of Westport
(population 760) each year.
A). Find the mean number of births per day.
11μ 0.0301
365
Example – Page 216, #6
B). Find the probability that on a given day, there is no births.
P x poissonpdf
P x
( 0) (11/ 365,0)
( 0) 0.970312
2ndvars/C:poissonpdf(μ,x)
11
365 x = births
Example – Page 216, #6
C). Find the probability that on a given day, there is at least
one birth.
P x P x P x P x( 1) ( 1) ( 2) ... ( 11)
P x
poissonpdf
1 ( 0)
1 (11/ 365,0)
0.0297
Example – Page 216, #6
D). Based on the preceding results, should medical birthing
personnel be on permanent standby, or should they be
called in as needed? Does this mean that Westport
mothers might not get the immediate medical attention
they would be likely to get in a more populated area?
The personnel should be called as needed. Yes; this
does mean that women giving birth might not get
the immediate attention available in more populated
areas.
Example – Page 216, #8
Homicide Deaths: In one year, there were 116 homicide deaths
in Richmond, Virginia (Based on “A Classroom Note on the
Poisson Distribution: A Model for Homicidal Deaths in
Richmond, VA for 1991,” in Mathematics and Computer
Education, by Winston A. Richards). For a randomly selected
day, find the probability that the number of homicide deaths is
a. 0 b. 1 c. 2 d. 3 e. 4
Compare the calculated probabilities to these actual results:
268 days (no homicides); 79 days (1 homicides); 17 days
(2 homicides); 1 day (3 homicides); no days with more than
3 homicides.
Example – Page 216, #8
Let x = the number of homicides per day
1160.3178
365
a. P(x = 0) = poissonpdf(0.3178, 0) = 0.7277
b. P(x = 1) = poissonpdf(0.3178, 1) = 0.2313
c. P(x = 2) = poissonpdf(0.3178, 2) = 0.0368
d. P(x = 3) = poissonpdf(0.3178, 3) = 0.0038
e. P(x = 4) = poissonpdf(0.3178, 4) = 0.0003
Example – Page 216, #8
x frequency Relative frequency P(x)
0 268 0.7342 = 268/365 0.7277
1 79 0.2164 0.2313
2 17 0.0466 0.0368
3 1 0.0027 0.0038
4 or more 0 0.0000 0.0004 (by subtraction)
365 1.000 1.0000
The agreement between the observed relative frequencies
and the probabilities predicted by the Poisson formula
is very good.
Poisson as Approximation
to Binomial
The Poisson distribution is sometimes used to
approximate the binomial distribution when n is
large and p is small.
Rule of thumb
n 100
np 10
If the two conditions are satisfied we can use
the Poisson distribution as an approximation to
the binomial distribution where
= np