This document discusses probability distributions for discrete variables. It begins by defining a probability distribution as a relative frequency distribution of all possible outcomes of an experiment. It provides examples of probability distributions for discrete variables like the binomial distribution. It discusses key aspects of probability distributions like the mean, standard deviation, and different types of distributions like binomial. It provides examples of calculating probabilities, means, and standard deviations for binomial distributions. It discusses the basic characteristics of the binomial distribution and provides an example of constructing a binomial distribution and calculating related probabilities.
Probability Distribution (Discrete Random Variable)Cess011697
Learning Competencies:
- to find the possible values of a random variable.
illustrates a probability distribution for a discrete random variable and its properties.
- to compute probabilities corresponding to a given random variable.
There are some exercises for you to answer.
Determining the Mean, Variance, and Standard Deviation of a Discrete Random Variable
Visit the website for more services: https://cristinamontenegro92.wixsite.com/onevs
Chapter 4 part3- Means and Variances of Random Variablesnszakir
Statistics, study of probability, The Mean of a Random Variable, The Variance of a Random Variable, Rules for Means and Variances, The Law of Large Numbers,
Probability Distribution (Discrete Random Variable)Cess011697
Learning Competencies:
- to find the possible values of a random variable.
illustrates a probability distribution for a discrete random variable and its properties.
- to compute probabilities corresponding to a given random variable.
There are some exercises for you to answer.
Determining the Mean, Variance, and Standard Deviation of a Discrete Random Variable
Visit the website for more services: https://cristinamontenegro92.wixsite.com/onevs
Chapter 4 part3- Means and Variances of Random Variablesnszakir
Statistics, study of probability, The Mean of a Random Variable, The Variance of a Random Variable, Rules for Means and Variances, The Law of Large Numbers,
Chapter 5 part1- The Sampling Distribution of a Sample Meannszakir
Mathematics, Statistics, Population Distribution vs. Sampling Distribution, The Mean and Standard Deviation of the Sample Mean, Sampling Distribution of a Sample Mean, Central Limit Theorem
: Random Variable, Discrete Random variable, Continuous random variable, Probability Distribution of Discrete Random variable, Mathematical Expectations and variance of a discrete random variable.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 5: Discrete Probability Distribution
5.1: Probability Distribution
This presentation contains the topic as follows, probability distribution, random variable, continuous variable, discrete variable, probability mass function, expected value and variance and examples
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 6: Normal Probability Distribution
6.4: The Central Limit Theorem
Detail Description about Probability Distribution for Dummies. The contents are about random variables, its types(Discrete and Continuous) , it's distribution (Discrete probability distribution and probability density function), Expected value, Binomial, Poisson and Normal Distribution usage and solved example for each topic.
I am Ben R. I am a Statistics Assignment Expert at statisticshomeworkhelper.com. I hold a Ph.D. in Statistics, from University of Denver, USA. I have been helping students with their homework for the past 5 years. I solve assignments related to Statistics.
Visit statisticshomeworkhelper.com or email info@statisticshomeworkhelper.com.
You can also call on +1 678 648 4277 for any assistance with Statistics Assignments.
Chapter 5 part1- The Sampling Distribution of a Sample Meannszakir
Mathematics, Statistics, Population Distribution vs. Sampling Distribution, The Mean and Standard Deviation of the Sample Mean, Sampling Distribution of a Sample Mean, Central Limit Theorem
: Random Variable, Discrete Random variable, Continuous random variable, Probability Distribution of Discrete Random variable, Mathematical Expectations and variance of a discrete random variable.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 5: Discrete Probability Distribution
5.1: Probability Distribution
This presentation contains the topic as follows, probability distribution, random variable, continuous variable, discrete variable, probability mass function, expected value and variance and examples
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 6: Normal Probability Distribution
6.4: The Central Limit Theorem
Detail Description about Probability Distribution for Dummies. The contents are about random variables, its types(Discrete and Continuous) , it's distribution (Discrete probability distribution and probability density function), Expected value, Binomial, Poisson and Normal Distribution usage and solved example for each topic.
I am Ben R. I am a Statistics Assignment Expert at statisticshomeworkhelper.com. I hold a Ph.D. in Statistics, from University of Denver, USA. I have been helping students with their homework for the past 5 years. I solve assignments related to Statistics.
Visit statisticshomeworkhelper.com or email info@statisticshomeworkhelper.com.
You can also call on +1 678 648 4277 for any assistance with Statistics Assignments.
A chi-squared test is a statistical hypothesis test that is valid to perform when the test statistic is chi-squared distributed under the null hypothesis, specifically Pearson's chi-squared test and variants
differences between the observed values
Please put answers below the boxes1) A politician claims that .docxLeilaniPoolsy
Please put answers below the boxes
1)
A politician claims that he is supported by a clear majority of voters. In a recent survey, 35 out of 51 randomly selected voters indicated that they would vote for the politician. Use a 5% significance level for the test. Use Table 1.
a.
Select the null and the alternative hypotheses.
H0: p = 0.50; HA: p ≠ 0.50
H0: p ≤ 0.50; HA: p > 0.50
H0: p ≥ 0.50; HA: p < 0.50
b.
Calculate the sample proportion. (Round your answer to 3 decimal places.)
Sample proportion
c.
Calculate the value of test statistic. (Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)
Test statistic
d.
Calculate the p-value of the test statistic. (Round intermediate calculations to 4 decimal places. Round "z" value to 2 decimal places and final answer to 4 decimal places.)
p-value
e.
What is the conclusion?
Do not reject H0; the politician is not supported by a clear majority
Do not reject H0; the politician is supported by a clear majority
Reject H0; the politician is not supported by a clear majority
Reject H0; the politician is supported by a clear majority
2)
Consider the following contingency table.
B
Bc
A
22
24
Ac
28
26
a.
Convert the contingency table into a joint probability table. (Round your intermediate calculations and final answers to 4 decimal places.)
B
Bc
Total
A
Ac
Total
b.
What is the probability that A occurs? (Round your intermediate calculations and final answer to 4 decimal places.)
Probability
c.
What is the probability that A and B occur? (Round your intermediate calculations and final answer to 4 decimal places.)
Probability
d.
Given that B has occurred, what is the probability that A occurs? (Round your intermediate calculations and final answer to 4 decimal places.)
Probability
e.
Given that Ac has occurred, what is the probability that B occurs? (Round your intermediate calculations and final answer to 4 decimal places.)
Probability
f.
Are A and B mutually exclusive events?
Yes because P(A | B) ≠ P(A).
Yes because P(A ∩ B) ≠ 0.
No because P(A | B) ≠ P(A).
No because P(A ∩ B) ≠ 0.
g.
Are A and B independent events?
Yes because P(A | B) ≠ P(A).
Yes because P(A ∩ B) ≠ 0.
No because P(A | B) ≠ P(A).
No because P(A ∩ B) ≠ 0.
3)
A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited the salon were 72, 55, 49, 35, 39, 23, and 77. It can be assumed that weekday customer visits follow a normal distribution. Use Table 2.
a.
Construct a 90% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places and "t" value to 3 decimal places, and final answers to 2 decimal places.)
Confidence interval
to
b.
Construct a 99% confidence interval for the average number of customers who visit the .
1. Consider the following partially completed computer printout fo.docxjackiewalcutt
1. Consider the following partially completed computer printout for a regression analysis where the dependent variable is the price of a personal computer and the independent variable is the size of the hard drive.
Based on the information provided, what is the F statistic?
About 8 .33
Just over 2.35
About 4.76
About 69.5
4 points
QUESTION 2
1. The standard error of the estimate is a measure of
total variation of the Y variable.
the variation around the sample regression line.
explained variation.
the variation of the X variable.
4 points
QUESTION 3
3.Nintendo Sony would like to test the hypothesis that a difference exists in the average age of users of a Wii, a PlayStation, or an Xbox console game. The following data represent the age of a random sample of Wii, PlayStation, and Xbox users.
Wii
PlayStation
Xbox
37
26
31
31
21
20
47
24
38
29
24
31
36
25
30
Using α = 0.05, the conclusion for this hypothesis test would be that because the test statistic is
more than the critical value, we cannot conclude that there is a difference in the average age of users of a Wii, a PlayStation, or an Xbox console game.
less than the critical value, we cannot conclude that there is a difference in the average age of users of a Wii, a PlayStation, or an Xbox console game.
more than the critical value, we can conclude that there is a difference in the average age of users of a Wii, a PlayStation, or an Xbox console game.
less than the critical value, we can conclude that there is a difference in the average age of users of a Wii, a PlayStation, or an Xbox console game.
4 points
QUESTION 4
1. The relationship of Y to four other variables was established as Y = 12 + 3X1 - 5X2 + 7X3 + 2X4. When X1 increases 5 units and X2 In a sample of n = 23, the Student's t test statistic for a correlation of r = .500 would be:
2.559
2.819
2.646
can’t say without knowing α (alpha)
4 points
QUESTION 5
1. Given the following ANOVA table (some information is missing), find the F statistic.
3.71
0.99
0.497
4.02
4 points
QUESTION 6
1. Examine the following two-factor analysis of variance table:
Complete the analysis of variance table.
MSA = 40.928, F Factor A =3.35, SSB = 85.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 66, MSE = 12.143
MSA = 40.928, F Factor A = 3.35, SSB = 85.35, Factor B df = 4, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1 SSE = 789.29, SSE df = 66, MSE = 12.143
MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 5, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1, SSE = 789.29, SSE df = 65, MSE = 12.143
MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 65, MSE = 12.143
4 points
QUESTION 7
1. The critical value for a two-tailed test of H0: ß1 = 0 at a (alpha) = .05 in a simple regression with 22 observations is:
+ or - 1.725 ...
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
2. What is a Probability Distribution?
• Precisely, a probability distribution is
•a relative frequency distribution of all
•possible outcomes of an experiment
0
0.05
0.1
0.15
0.2
0.25
0.3
Under
$20,000
$20,000
to <
$40,000
$40,000
to <
$60,000
$60,000
to <
$80,000
$80,000
to <
$100,000
Above
$100,000
ProbabilityP(x)
Income
Income Relative frequency P(x)
Under $20,000 0.2380
$20,000 to < $40,000 0.2607
$40,000 to < $60,000 0.1941
$60,000 to < $80,000 0.1408
$80,000 to < $100,000 0.0907
Above $100,000 0.0757
1
Results =
Positive
(PO)
Results =
Negative
(NE) Total
Woman is pregnant (Y) 0.85 0.05 0.9
Woman is not pregnant (N) 0.03 0.07 0.1
Total 0.88 0.12 1
(a)
(b)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Y & PO Y & NE N & PO N & NE
ProbabilityP(I&j)
Joint Events (i & j)
Examples of Probability Distribution
2
3. Different types of Probability Distribution
Probability distributions of
Continuous Variables
Probability distributions of
Discrete Variables
x
P(x)
80 80.5 81 81.5 82.0 82.5
Examples:
- Uniform distribution
- Normal distribution
(b)
x
P(x)
0 1 2 3 4
Examples:
- Binomial distribution
- Poisson distribution
(a)
3
4. Random variable:
A variable is a characteristic that varies from one component of a population to another. It
becomes a random variable when its values vary randomly or by chance
Discrete random variable:
A variable that can take only integer values (e.g. 0, 1, and 2)
Continuous random variable:
A variable that can take any value in a specified interval falling within its plausible range
(e.g. 5, 5.6, 8.8, 10, and 12.4)
4
5. Working problem 5.1:
Decide whether the following random variables x are discrete or continuous.
Explain your answer
1. Human height
2. Human weight
3. Temperature
4. Number of graduating students
5. Number of student in a stat class
6. House area in square feet
7. The number of eggs that a hen lays in a day
8. The number of emergency room patients in one day in a hospital
5
6. The probability distribution of discrete variables
It is a relative frequency distribution, which lists each possible value of a discrete
random variable and its associated probability. It should also satisfy the criteria:
6
7. Example: Does the Table below describe a probability distribution?
The probability distribution of discrete variables
probabilities p(x) corresponding to a random variable x
x P(x)
0 0.15
1 0.2
2 0.05
3 0.25
4 0.25
5 0.3
6 0.1
Solution:
•The probability values here are less than one, which satisfies the condition:
• However, the sum of probabilities
which is not equal one.
Because the sum requirement is not satisfied, we conclude that Table 5.1 does not describe
a probability distribution. 7
8. Example: In an experiment in which two coins are tossed, what are the possible
outcomes? If the outcome of interest is ‘head’, what are the probabilities of zero head,
one head, and two heads? Construct a probability distribution illustrating this
experiment.
Solution:
Outcomes resulting from tossing two coins
Trial number Possible outcomes x = number of heads P(x)
1 T, T 0 1/4
2 T, H 1 1/4
3 H, T 1 1/4
4 H, H 2 1/4
x = number of
heads
P(x)
0 1/4
1 1/2
2 1/4
Probability values associated with
the number of heads
8
10. Working problem 5.3:
(a) Does P(x) = x/6, where x can be 0, 1, or 5 determine a probability distribution? Explain why?
(b) Does P(x) = x/5, where x can be 0 or 1 , 2, or 3 determine a probability distribution? Explain why?
(c) Does P(x) = x/3, where x can be 0 or 1 , or 2 determine a probability distribution? Explain why?
Working problem 5.4: In a survey conducted by a Real-Estate broker, houses in a certain city
were classified by the number of bathrooms available. The results of this survey are shown
below. The first column represents the number of bathrooms, and the second column
represents the number of houses. Construct a probability distribution representing this survey.
Number of bathrooms per
house
Number of houses
1 400
1.5 840
2 330
2.5 920
3 360
3.5 420
10
11. The mean and standard deviation of the probability distribution of discrete variables
The mean of a discrete random variable x (also called the expected value of x) is
expressed by:
The standard deviation of a discrete variable x is expressed by:
11
12. Example: For the data of the Table below, calculate the mean of the discrete variable
representing the number of bedrooms per house.
The mean and standard deviation of the probability distribution of discrete variables
Calculations of mean of a discrete variable
Bedrooms per house Number of houses P(x) xP(x)
1 200 0.1 0.1
2 310 0.155 0.31
3 330 0.165 0.495
4 600 0.3 1.2
5 450 0.225 1.125
6 110 0.055 0.33
2000
1 3.56
12
13. Example: For the data of the Table below, calculate the standard deviation of the discrete
variable representing the number of bedrooms per house.
Number of
rooms/house
Number of
houses
P(x) xP(x) x2 x2P(x)
1 200 0.1 0.1 1 0.1
2 310 0.155 0.31 4 0.62
3 330 0.165 0.495 9 1.485
4 600 0.3 1.2 16 4.8
5 450 0.225 1.125 25 5.625
6 110 0.055 0.33 36 1.98
2000 1 3.56
14.61
13
14. Working Problem: 5.5
In a survey conducted by a real-estate broker, houses in a certain city were classified by
the number of full bathrooms available. The results of this survey are shown in the table
below. If a house is picked from this area randomly:
- What is the chance that it will have 3 bathrooms?
- What is the chance that it will have less than 3 bathrooms?
Frequency of houses by the number of full bathrooms
Number of full bathrooms per house Number of houses
1 45
2 410
3 320
4 38
5 12
14
15. Working Problem: 5.6
In a survey conducted by a real-estate broker, houses in a certain city were classified by the
number of full bathrooms available. The results of this survey are shown in the table below.
- Calculate the mean number of bathrooms per house
- Calculate the standard deviation of the number of bathrooms per house.
Frequency of houses by the number of full bathrooms
Number of full bathrooms per house Number of houses
1 45
2 410
3 320
4 38
5 12
15
16. Working problem 5.7:
A person plans to invest a $100,000. An investment consultant offered this person five investment packages the return
of each will depend on whether the economy next year will be fair or good. The table below lists the anticipated return
on each investment.
According to the investment consultant, next year’s economy has a 45% chance of being good and a 55% chance of
being fair.
a. Determine the probability distribution of each random variable M, S, C, and R
b. Determine the expected value of each random variable.
c. Which investment has the best expected payoff? Which has the worst?
16
17. The binomial probability distribution
Or
Parameters of the
Binomial Distribution, n & P
The mean and variance
of a binomial distribution
are:
17
18. Basic characteristics of the binomial distribution are as follows:
•The binomial distribution is the most appropriate probability distribution for sampling from an
infinitely large population, where p represents the fraction of defective or nonconforming items in
the population, x represents the number of nonconforming items found in a random sample of n
items. Examples of variables that are best approximated by a binomial distribution include success
and failure in business, passing or failing a course, product defects, rejected items, and head or tail
in tossing a coin.
•In a binomial experiment, there are n identical trials. In other words, the given experiment is
repeated n times, where n is a positive integer.
•Each repetition of a binomial experiment is called a trial or a Bernoulli trial (after Jacob Bernoulli).
•Each trial has two and only two outcomes (success or failure, pass or fail, head or tail, etc).
•The probability of success is denoted by p and that of failure by q or (1-p).
•Binomial trials are independent. This means that the outcome of one trial does not affect the
outcome of the next trial or any other trial.
18
20. Example: There are four flights daily from Atlantic City Airport, New Jersey, to Atlanta Airport,
Georgia, via Air Tran Airline. Suppose the probability that any flight arrives on time is 0.70.
• Construct a binomial distribution of the random variable “number of on-time arrivals”.
• What is the probability that none of the flights will arrive on time tomorrow?
• What is the probability that exactly two of the flights will be on time?
Solution:
If we are willing to assume that the variable arrival time follows a binomial
distribution, we can construct this distribution using the binomial equation,
and the two parameters n and p. In symbols, the x variable is described as
x ~ b(n, p) or x ~ b(4, 0.7).
x= 0,1,2, …
x P(x)
0 0.00810
1 0.07560
2 0.26460
3 0.41160
4 0.24010
20
21. x P(x)
0 0.00810
1 0.07560
2 0.26460
3 0.41160
4 0.24010
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0 0 0 0 0
P(x)
Number of flights on time (x)
Binomial distribution (n = 4, p = 0.7)
Binomial Distributions of flight on-time arrivals
The probability that none of the flights will arrive on time tomorrow = P(x= 0) = 0.0081
The probability that exactly two of the flights will be on time = P(x=2) = 0.26460 (about 26% chance
that exactly two flights will be on time)
21
22. Example: Using the results of the previous Example, what is the expected value of arrival on time,
and what is the extent of variation (variance)?
Solution:
n = 4, p = 0.7
22
23. Working problem 5.8:
There are three flights daily from Charlotte, NC , to New York, NY (JFK), via U.S. Air .
Suppose the probability that any flight arrives late is 0.10.
- Construct a binomial probability distribution
- What is the expected value and the variance?
- What is the probability that none of the flights are late today?
-What is the probability that exactly one of the flights is late today?
- What is the probability that exactly two of the flights are late today?
23
24. Working problem 5.9:
For the experiment of tossing a coin 6 times, construct a binomial probability
distribution.
- What is the expected value and the variance?
- What is the probability that 2 heads will turn up?
-What is the probability that five heads will turn up?
- What is the probability that no heads will turn up?
24
25. Working problem 5.10: A manufacturer of electromagnetic combined clutch-brake units
used for automobiles finds that on average 1.2% of the units fails at high-speed
accelerated test. This finding represents historical value based on routine inspection of
100 units daily. If we let x be the random variable representing the number of failing
brake units,
- Construct a binomial distribution of this variable
- Determine the expected value and the variance of failure of brakes
25
26. Working problem 5.11: A manufacturer of electromagnetic combined clutch-brake units
used for automobiles finds that on average 1.2% of the units fails at high-speed
accelerated test. This finding represents historical value based on routine inspection of
100 units daily. If we let x be the random variable representing the number of failing
brake units:
- What is the probability that exactly one brake unit will fail?
- What is the probability that less than two brake units will fail?
- What is the probability that more than 4 brake units will fail?
26
27. Working Problem 5.12:
There are five bus trips from JFK Airport to Newark Airport.
Suppose the probability that any bus arrives late is 0.20.
- Construct a binomial distribution describing the random event of bus late
arrivals.
-Calculate the mean and the variance
27
28. Working Problem 5.13:
There are five bus trips from JFK Airport to Newark Airport.
Suppose the probability that any bus arrives late is 0.20.
What is the probability that none of the bus trips will be late today?
What is the probability that one of the bus trips will be late today?
28
29. Example: You enter the class to find out that the teacher is given a previously unannounced
quiz of 10 multiple-choice questions, each with one correct answer out of possible 4 answer
choices (A, B, C, and D). Since you were not prepared for the quiz, you decided to make
a random guess.
• Does this situation reflect a binomial experiment?
If the answer is ‘yes’, formulate the probability function
• What are the values of the two parameters, n and p of the binomial distribution?
• What is your chance to answer exactly 3 questions correctly?
Solution:
This experiment fully satisfies the binomial model. The reasons are:
• A fixed number of trials = n = 10
• The trials are independent (the answer of one question is independent of the answer of another
question)
• Each trial is associated with two outcomes, correct or wrong
• Each trial is associated with fixed probability p = 1/4, since the chance of a correct answer is one
out of four. Thus, the probability remains constant
n = 10, p = 0.25
Your chance to answer exactly 3 questions correctly:
29
30. Example: In the previous example, find the probabilities of answering one, two,
three,…., or ten questions correctly. Construct the Binomial Distribution.
Binomial distribution of multiple-choice quiz
x P(x) Cumulative probability
0 0.05631 0.05631
1 0.18771 0.24403
2 0.28157 0.52559
3 0.25028 0.77588
4 0.14600 0.92187
5 0.05840 0.98027
6 0.01622 0.99649
7 0.00309 0.99958
8 0.00039 0.99997
9 0.00003 1.00000
10 0.00000 1.00000
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0 1 2 3 4 5 6 7 8 9 10
P(x)
x
Binomial Distributions at p = 0.25 and n = 10
2.500 expected value
1.875 variance
1.369 standard deviation
30
31. 30+ 60+ 70+
Working problem 5.14
Based on a worldwide mortality statistics, there is about 60% chance that a person aged 30
will be alive at age 70. Suppose that five people aged 30 are selected at random.
- Find the probability that the number alive at age 70 will be exactly two
- Find the probability that the number alive at age 70 will be at most two
- Find the probability that the number alive at age 70 will be at least three
- Find the probability that the number alive at age 70 will be none
- Determine the probability distribution of the number alive at age 70
- Determine the expected value and the variance of the number alive at age 70
31
32. Working problem 5.15:
A charity for deaf and blind offered 1000 tickets each of $5 to collect money. Only the first 4
draws win prices. The prices for the first 4 tickets are $200, $150, $100, and $50. If you buy
one ticket, what would be your expected value of gain?
Working problem 5.16:
In a multiple-choice exam of 20 questions, 5 possible answers were assigned for each
question and only one answer is correct. If a student is answering these questions
randomly, would this represent a binomial experiment? If the answer is ‘yes’, formulate the
probability function
What are the values of the two parameters, n and p of the binomial distribution? Calculate
the expected value and the variance.
- What is your chance to answer exactly 2 questions correctly?
- What is your chance to answer less than 3 questions correctly?
-What is your chance to answer more than 15 questions correctly? 32
33. Working problem 5.17:
In a multiple-choice exam of 100 questions, 4 possible answers were assigned for each
question and only one answer is correct. If a student is answering these questions randomly,
how many questions would he/she be expected to answer?
Working problem 5.18:
In a multiple-choice exam of 60 questions, 4 possible answers were assigned for each question
and only one answer is correct. A student made a serious attempt at the first 40 questions. This
student typically answer multiple-choice tests with a rate of success of about 80%. How many
questions will this student be expected to answer correctly out of the first 40 questions?
If the same student then attempted to answer the remaining 20 questions randomly. Would you
consider his attempts as a binomial experiment? If the answer is ‘yes”, what is the expected
number of questions that this student will answer correctly out of these 20 questions? Estimate
the total number of questions that this student and predict his letter grade?
Note:
G ≥ 90 = A
80 ≤ G < 90 = B
70 ≤ G < 80 = C
60 ≤ G < 70 = D
33
34. Example: If 15% of the bolts produced by a machine are defective, what is the probability
that out of 3 bolts selected at random (a) 0, (b) 1, (c) less than 2, bolts will be defective?
Using Microsoft Excel® to Perform Binomial Distribution Analysis
Solution:
P = 0.15, 1- P = 0.85
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3
P(X)
X
Binomial distribution (n = 3, p = 0.15)
X P(X)
0 0.61413
1 0.32513
2 0.05738
3 0.00338
34
35. 1
2
3
4
Using Microsoft Excel® to Perform Binomial Distribution Analysis
• Go to Excel Spreadsheet
• Click fx on the button bar
• Select Statistical from the
“Or select a category” drop
down list box
• Select BINODIST from the
“Select a function” list
• Click OK
P = 0.15, q = 0.85
Example:
?)2(
?)1(
?)0(
xP
xP
xP
35
36. •To answer Part (a) of this Example,
type 0 in the Number_s window,
and 3 in the Trials Window, 0.15 in the
Probability_s window, and False in
the Cumulative Window.
• You should be able to see the value
of the P(x = 0) as illustrated by the
ellipse in the Figure, or you can
Click Ok and the value will be
presented
P = 0.15, q = 0.85Example: ?)2()(?)1()(?)0()( xPcxPbxPa
5
6
614.085.015.0
0
3
)0(
30
xP
Note:
Under Cumulative, we use
“FALSE” for getting the exact
probability value (i.e. P(x=0) or
P(x=1)), or “TRUE” for getting
cumulative probability
36
37. 939.0325.0614.0
)1()0()2(
xPxPxP
325.085.015.0
1
3
)1(
21
xP
Note:
Under Cumulative, we use
“FALSE” for getting the exact
probability value (i.e. P(x=0) or
P(x=1)), or “TRUE” for getting
cumulative probability
37
38. Working problem 5.19 (Use Excel to solve this problem):
A manufacturer of electromagnetic combined clutch-brake units used for automobiles finds that
on average 1.2% of the units fails at high-speed accelerated test. This finding represents
historical value based on routine inspection of 100 units daily. If we let x be the random variable
representing the number of failing brake units:
- What is the probability that exactly one brake unit will fail?
- What is the probability that less than two brake units will fail?
- What is the probability that more than 4 brake units will fail?
38
39. Working problem 5.20:
You enter the class to find out that the teacher is given a previously unannounced quiz of
eight multiple-choice questions, each with one correct answer out of possible 4 answer
choices (A, B, C, and D). Since you were not prepared for the quiz, you decided to make a
random guess.
- Use Excel function to determine your chance to answer exactly 3 questions correctly?
- Use Excel function to determine your chance to answer exactly 2 questions correctly?
- Use Excel function to determine your chance to answer no questions correctly?
39
40. What is a Poisson Probability Distribution?
• A probability distribution used for modeling discrete variables with the concern being with
the number of events occurring per a given time or space interval.
• The interval may represent any duration of time (e.g. per hour, per shift, per week, etc.) or
any specified region (e.g. per unit area, per box, per shipment, etc.).
• In most applications, the Poisson distribution is often used to model the events of largely
unlikely failure or rare product defects.
• The probability of an event described by a Poisson distribution is given by:
,....3,2,1,0
!
P(x)
x
x
x
e
where x is the number of target events,
P(x) is the probability of x events in a
given time or space interval, l is the
expected average number of events per
interval, and e = 2.7182818 is the base of
the natural logarithm system.
• The mean and variance of the Poisson distribution are given by:
np
np
2
40
41. 168.0
1234
0498.081
!4
)3(
!
4)P(x
34
e
x
ex
Example: The mean number of accidents on the job per month at a certain
company is 3. Model this variable by a Poisson distribution and determine the
probability that in any given month four accidents will occur at this company?
3
3
2
The mean and variance of this
distribution are:
Poisson Probability Distribution-Applications
0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
P(X)
X
Poisson distribution (= 3)
0216.0
!7
)3(
!
)7(7
...............................
224.0
!2
)3(
!
)2(2
1494.0
!1
)3(
!
)1(1
0498.0
!0
)3(
!
)0(0
37
32
31
30
e
x
e
Pxat
e
x
e
Pxat
e
x
e
Pxat
e
x
e
Pxat
x
x
x
x
X P(X)
0 0.04979
1 0.14936
2 0.22404
3 0.22404
4 0.16803
5 0.10082
6 0.05041
7 0.02160
8 0.00810
9 0.00270
10 0.00081
11 0.00022
12 0.00006
13 0.00001
14 0.00000
15 0.00000
16 0.00000
17 0.00000
41
42. Example: A fast-food restaurant was trying to extend breakfast service from 10 am to 10:30 am to
allow more customers, particularly those who wake up late, to be served breakfast. From past
experience, the restaurant found that on average, the number of people showing up for breakfast
service from 10:00 am to 10:30 am is 7.5 people. What advice would you give this restaurant?
To assist you in this task,
• The restaurant will allow the service to be extended if they can maintain more than 5 customers
coming for breakfast service in this period.
- What is the chance that the restaurant will maintain more than 5 customers during the added
period?
- What is the chance that the restaurant will maintain more than 10 customers during the added
period?
Poisson Probability Distribution-Applications
7585.0]1094.00729.00389.00156.000414.0000553.0[1
]
!5
5.7
!4
5.7
!3
5.7
!2
5.7
!1
5.7
!0
5.7
[1
)]5()4()3()1()0([1
!
-15)P(x15)P(x
....)8()7()6(
!
5)P(x
5.755.745.735.725.715.70
5
0
6
eeeeee
xPxPxPxPxP
x
e
or
xPxPxP
x
e
x
x
= 7.5
42
43. Example (Cont’d):
Poisson Probability Distribution-Applications
7585.05)P(x = 7.5
There is more than 75% chance that the restaurant will maintain more than 5
customers within the added half hour breakfast service.
The chance that the restaurant will maintain more than 10 customers during the
added period:
138.00.862241)01P(x1)01P(x
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0 2 4 6 8 10 12 14 16 18 20 22 24
P(X)
X
Poisson distribution (= 7.5)
There is a 75% chance the restaurant will maintain more than 5 customers and 14% chance it
may even have more than 10 customers. 43
44. Example: Suppose that a new knitting machine used by a mill only fails to operate one
time every 10 hours. This leads to a failure rate, , of 0.1failure per hour.
What is the probability of two failures occurring randomly during the 10 hour period?
Poisson Probability Distribution-Applications
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0 1 2 3 4 5
P(X)
X
Poisson distribution ( = 0.1)
X P(X)
Cumulative
probability
0 0.90484 0.90484
1 0.09048 0.99532
2 0.00452 0.99985
3 0.00015 1.00000
4 0.00000 1.00000
5 0.00000 1.00000
0045.0
!2
1.0
!
2)P(x
1.02
e
x
ex
44
45. Working problem 5.21: The mean number of on-site accidental
deaths per a period of five years in an electric-service company
is 1.0. Assuming that a Poisson distribution can best represent
this random variable, determine the probability that in any given
five-year period, three on-site accidental deaths will occur?
Construct the Poison distribution describing this variable.
What is the mean and variance of this
distribution?
Working problem 5.22: The mean number of car accidents in Peachtree street in Atlanta, Georgia
is 3 per month. Assuming that a Poisson distribution can best represent this random variable,
determine the probability that in any given month, 5 or more car accidents will occur? Construct the
Poison distribution describing this variable. What is the mean and variance of this distribution?
45
46. Working Problem 5.23:
Assume on-time flight per week is a rare situation for a particular airline. Suppose on average,
this number is about 2.
Construct a Poisson distribution of this random variable.
-What is the expected value and variance of the number of on-time flights per week?
- What is the chance that the number of on-time flights per week will exceed 5?
46
47. Example: Suppose that historically a machine producing bolts yields a rate
of 10% defective bolts. Find the probability that in a sample of 10 bolts
chosen at random exactly two will be defective? Calculate this probability
using Binomial and Poisson modeling of the variable.
Poisson & Binomial Probability Distribution-Applications
Using Binomial distribution:
n = 10, p = 0.1
1937.0)90.0(10.0
2
10
)2( 82
xnx
qp
x
n
xP
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0 1 2 3 4 5 6 7 8 9 10
P(X)
X
Binomial distribution (n = 10, p = 0.1)
Using Poisson distribution:
11.010 np
1839.0
!2
1
!
)2(
12
e
x
xP e
x
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0 1 2 3 4 5 6 7 8 9 10
P(X)
X
Poisson distribution ( = 1)
47
48. Using Microsoft Excel to Perform Poisson Distribution Analysis
Example: If the number of some type of metal defects is assumed to follow a
Poisson distribution with a mean value 1 per 1000 square yard.
•What is the probability of finding 2 defects?
•What is the probability of finding at most 2 defects?
•What is the probability of finding at least one defect?
184.0
!2
1
)2(
!
)2(
0
12
x
x
e
p
x
XP e
Manual Calculations:
920.0184.0368.0368.0
!2
1
!1
1
!0
1
)2()1()0(
!
)2(
0
121110
x x
eee
ppp
x
XP e
632.0368.01)0(1)1(1)1( xpxpxp
48
49. 1
2
3
4
Using Microsoft Excel to Perform Poisson Distribution Analysis ( =1)
•Go to Excel Spreadsheet
•Click fx on the button bar
•Select Statistical from the “Or
select a category” drop down list
box
•Select POISSON from the
“Select a function” list
•Click OK
49
50. 5
6184.0
!2
1
)2(
!
)2(
0
12
x
x
e
p
x
XP e
•Type 2 at the X window, 1 in Mean
window, and False in the
Cumulative Window.
•You should be able to see the
value of the P(x = 2) as illustrated
by the circle in Figure 5.16. or you
can Click Ok and the value will be
presented
Using Microsoft Excel to Perform Poisson Distribution Analysis ( =1)
50
52. Example: Suppose that a manufacturer typically finds 2% defects in hard
drive units produced for personal computers. Generate a Binomial
distribution of nonconforming drives using a sample size of 400 drives.
Using Microsoft Excel to Perform Discrete Probability Distribution Simulation
n = 400, and p = 0.02
Binomial Distribution
USING EXCEL®
52
56. Average Number of Defects ≈ 8
• This leads to average probability
= 8/400 = 0.02 (the p value)
• The Standard Deviation = 2.737
• This is approximately equal to
the theoretical :
8.2)98.0)(02.0)(400( npq
Note
56
59. To convert a histogram to a probability
Distribution:
1. Form a column labeled p(x)
2. Calculate p(x) for each value of x
By dividing the frequency of each x by
The total frequency (1000).
3. You can then make a graph using
Excel graph features
Note
Convert Histogram into a Probability Distribution
59
60. Example: If a company producing notebook computers historically finds
an average of 1.2 defects when inspecting random samples of constant
size of 60 disks. Using Excel Spreadsheet, simulate the probability
distribution of defects.
Using Microsoft Excel to Perform Discrete Probability Distribution Simulation
= 1.2
Poisson Distribution
USING EXCEL®
60
61. 2
1
3
1000 random
numbers
each representing
number of defects
per trial - Microsoft Excel data analysis
- Random Number Generator
- Generate, say 1000 numbers
- Select Poisson, and = 1.2.
- Press OK
61
In Chapter 4, we introduced key probability terms such as experiment, outcome, and event. We also indicated that any given statistical experiment can have several outcomes. Obviously, when an experiment is conducted, it is impossible to predict which of the many possible outcomes will actually occur. As a result, we have to make decisions under uncertain conditions. A probability distribution can help us making these decisions.
In general, a probability distribution is a relative frequency distribution of all possible outcomes of an experiment. Recall in chapter 4, we determined empirical probabilities from a relative frequency distribution similar to the one in Figure a, which shows probabilities associated with different income levels. We also used a contingency table to determine joint probabilities such as the ones illustrated in Figure b, which illustrates the joint probabilities in a pregnancy test experiment. These types of plots represent forms of probability distributions. We can also construct a probability distribution based on knowledge of probable patterns of the variable under consideration and the parameters controlling these patterns. These points will be demonstrated by examples later in this chapter.
The need for probability distributions stems from the fact that the classic definition of probability requires that the total number of independent trials of the experiment be finite. In practice, a finite number of trials are often not enough to obtain a stabilized probability. In addition, the problem with the classic definition of probability is that the words "equally likely" are often ambiguous. Since these words seem to be synonymous with "equally probable", the definition is circular because we are essentially defining probability in terms of itself.
Probability distributions can be classified into two main types: probability distributions of discrete variables, and probability distributions of continuous variables. The Figure here illustrates the difference between these two types of distribution. As can be seen in Figure a, a probability distribution of a discrete variable is represented by separated needles of heights indicating the values of probabilities, P(xi), corresponding to different values of x. Examples of probability distributions of discrete variables include the “binomial distribution” and the “Poisson distribution”. These distributions are discussed in the following sections. Figure b shows a probability distribution of a continuous variable, which is represented by adjacent bars of heights indicating the values of probabilities, P(xi), corresponding to different values of x. Examples of probability distributions of continuous variables include the “uniform distribution” and the “normal distribution”. These distributions will be discussed Chapter 6.
As discussed in Chapter 1, a discrete variable can take only integer values (e.g. 0, 1, and 2). Examples of situations involving discrete variables include: (a) tossing a coin (H or T), (b) inspection of a product (pass or fail), (c) the number of defects in a product (5 or 8, not 5.5), and (d) the number of students passing a course (20 or 25, not 23.6). A continuous variable can take any value in a specified interval falling within its plausible range. For example, the diameter of a fine metal rod may take a value of 40, 40.25, 40.75 or 41 millimeter and human weight may take values of 120 lb, 155 lb, or 165.8 lb. These data imply continuity of the variable values.
Human height (continuous)
Human weight (continuous)
Temperature (continuous)
Number of graduating students (discrete)
Number of student in a stat class (discrete)
House area in square feet (continuous)
The number of eggs that a hen lays in a day (discrete)
The number of emergency room patients in one day in a hospital (discrete)
The First Table here shows all possible outcomes associated with tossing two coins. The second Table illustrates the probability values associated with the number of heads. The discrete probability distribution resulted from this experiment is also shown here.
WP5.3
Solution:
Does P(x) = x/6, where x can be 0, 1, or 5 determine a probability distribution? Explain why?
P(x) = x/6
P(0) = 0
P(1) = 1/6
P(5) = 5/6
Sum = P(0) + P(1) + P(5) = 0 + (1/6) + (5/6) = 1
Then P(x) is a probability distribution
(b) Does P(x) = x/5, where x can be 0 or 1 , 2, or 3 determine a probability distribution? Explain why?
P(x) = x/5
P(0) = 0
P(1) = 1/5
P(2) = 2/5
P(3) = 3/5
Sum = P(0) + P(1) + P(2) + P(3) = 0 + (1/5) + (2/5) + (3/5) = 6/5 ≠ 1
Then P(x) is a not probability distribution
(c) Does P(x) = x/3, where x can be 0 or 1 , or 2 determine a probability distribution? Explain why?
P(x) = x/3
P(0) = 0
P(1) = 1/3
P(2) = 2/3
Sum = P(0) + P(1) + P(2) = 0 + (1/3) + (2/3) = 1
Then P(x) is a probability distribution
The reason the mean is called ‘the expected value’ is that the mean in an experiment represents the average value that we would expect to get if the experimental trials could continue indefinitely. The expected value represents a key parameter in many applications associated with decision theory (see illustration of stock market decision making at the end of this section).
The Binomial distribution is a probability distribution used to model discrete variables, or variables that can be described by observations placed in only one of two mutually exclusive categories, such as good or bad, fail or pass, stop or go, etc. The distribution associated with the example of tossing two coins, shown here, is essentially a binomial distribution described by two parameters: the number of trials n, and the probability p. In that case, n = 2 and p = 0.5.
In practice, the binomial distribution is widely used to determine the probability that an outcome will occur x times in n performances of an experiment. In this case, the variable x must be a discrete random variable and each repetition of the experiment must result in one of two possible outcomes. Statistically, if p is the probability that an event will happen in any single trial and q = 1 – p, is the probability that it will fail to happen, then the probability that the event will happen exactly x times in n trials (i.e. x successes and n-x failures) is given by the equation shown here.
Note that the main parameters of the binomial distribution are n and p. Knowledge of the values of these two parameters will allow us to completely simulate a variable following a Binomial distribution.
The shape of the binomial distribution depends strictly on the values of its two parameters, n and p. In general, as n increases, the binomial distribution will tend to be more symmetrical. In addition, when p = 0.5, the distribution also becomes symmetrical. Th Figure here illustrates these two points using binomial distributions of different values of n and p.
Note that the distribution is largely symmetrical as a result of using large sample size n.
Let us solve this example first without Excel
The Poisson distribution was named after the French mathematician Simeon D. Poisson. It is a probability distribution used for modeling discrete variables with the concern being with the number of events occurring during a given time or space interval. The interval may represent any duration of time (e.g. per hour, per shift, per week, etc.) or any specified region (e.g. per unit area, per box, per shipment, etc.). In most applications, the Poisson distribution is often used to model rare events. The probability of an event described by a Poisson distribution is given by the function in the box.