This document discusses probability distributions and provides examples of calculating probabilities using binomial distributions. It begins by defining a probability distribution as a table, graph or formula used to specify the possible values and probabilities of a discrete random variable. It then gives examples of probability distributions for number of assistance programs used by families and calculates related probabilities. The document introduces binomial distribution and provides two examples of calculating probabilities of outcomes for binomial processes, such as number of full term births out of total births. It describes key concepts like Bernoulli trials, processes and use of combinations and factorials to calculate probabilities for larger sample sizes.
A Probability Distribution is a way to shape the sample data to make predictions and draw conclusions about an entire population. It refers to the frequency at which some events or experiments occur. It helps finding all the possible values a random variable can take between the minimum and maximum statistically possible values.
A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution. Suppose we flip a coin two times and count the number of heads (successes).
A Probability Distribution is a way to shape the sample data to make predictions and draw conclusions about an entire population. It refers to the frequency at which some events or experiments occur. It helps finding all the possible values a random variable can take between the minimum and maximum statistically possible values.
A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution. Suppose we flip a coin two times and count the number of heads (successes).
Turning from discrete to continuous distributions, in this section we discuss the normal distribution. This is the most important continuous distribution because in applications many random variables are normal random variables (that is, they have a normal distribution) or they are approximately normal or can be transformed into normal random variables in a relatively simple fashion. Furthermore, the normal distribution is a useful approximation of more complicated distributions, and it also occurs in the proofs of various statistical tests.
Normal Distribution, also called Gaussian Distribution, is one of the widely used continuous distributions existing which is used to model a number of scenarios such as marks of students, heights of people, salaries of working people etc.
Each binomial distribution is defined by n, the number of trials and p, the probability of success in any one trial.
Each Poisson distribution is defined by its mean.
In the same way, each Normal distribution is identified by two defining characteristics or parameters: its mean and standard deviation.
The Normal distribution has three distinguishing features:
• It is unimodal, in other words there is a single peak.
• It is symmetrical, one side is the mirror image of the other.
• It is asymptotic, that is, it tails off very gradually on each side but the line representing the distribution never quite meets the horizontal axis
It includes various cases and practice problems related to Binomial, Poisson & Normal Distributions. Detailed information on where tp use which probability.
Today’s overwhelming number of techniques applicable to data analysis makes it extremely difficult to define the most beneficial approach while considering all the significant variables.
The analysis of variance has been studied from several approaches, the most common of which uses a linear model that relates the response to the treatments and blocks. Note that the model is linear in parameters but may be nonlinear across factor levels. Interpretation is easy when data is balanced across factors but much deeper understanding is needed for unbalanced data.
Analysis of variance (ANOVA) is a collection of statistical models and their associated estimation procedures (such as the "variation" among and between groups) used to analyze the differences among means. ANOVA was developed by the statistician Ronald Fisher. ANOVA is based on the law of total variance, where the observed variance in a particular variable is partitioned into components attributable to different sources of variation. In its simplest form, ANOVA provides a statistical test of whether two or more population means are equal, and therefore generalizes the t-test beyond two means. In other words, the ANOVA is used to test the difference between two or more means.Analysis of variance (ANOVA) is an analysis tool used in statistics that splits an observed aggregate variability found inside a data set into two parts: systematic factors and random factors. The systematic factors have a statistical influence on the given data set, while the random factors do not. Analysts use the ANOVA test to determine the influence that independent variables have on the dependent variable in a regression study.
Sir Ronald Fisher pioneered the development of ANOVA for analyzing results of agricultural experiments.1 Today, ANOVA is included in almost every statistical package, which makes it accessible to investigators in all experimental sciences. It is easy to input a data set and run a simple ANOVA, but it is challenging to choose the appropriate ANOVA for different experimental designs, to examine whether data adhere to the modeling assumptions, and to interpret the results correctly. The purpose of this report, together with the next 2 articles in the Statistical Primer for Cardiovascular Research series, is to enhance understanding of ANVOA and to promote its successful use in experimental cardiovascular research. My colleagues and I attempt to accomplish those goals through examples and explanation, while keeping within reason the burden of notation, technical jargon, and mathematical equations.
Turning from discrete to continuous distributions, in this section we discuss the normal distribution. This is the most important continuous distribution because in applications many random variables are normal random variables (that is, they have a normal distribution) or they are approximately normal or can be transformed into normal random variables in a relatively simple fashion. Furthermore, the normal distribution is a useful approximation of more complicated distributions, and it also occurs in the proofs of various statistical tests.
Normal Distribution, also called Gaussian Distribution, is one of the widely used continuous distributions existing which is used to model a number of scenarios such as marks of students, heights of people, salaries of working people etc.
Each binomial distribution is defined by n, the number of trials and p, the probability of success in any one trial.
Each Poisson distribution is defined by its mean.
In the same way, each Normal distribution is identified by two defining characteristics or parameters: its mean and standard deviation.
The Normal distribution has three distinguishing features:
• It is unimodal, in other words there is a single peak.
• It is symmetrical, one side is the mirror image of the other.
• It is asymptotic, that is, it tails off very gradually on each side but the line representing the distribution never quite meets the horizontal axis
It includes various cases and practice problems related to Binomial, Poisson & Normal Distributions. Detailed information on where tp use which probability.
Today’s overwhelming number of techniques applicable to data analysis makes it extremely difficult to define the most beneficial approach while considering all the significant variables.
The analysis of variance has been studied from several approaches, the most common of which uses a linear model that relates the response to the treatments and blocks. Note that the model is linear in parameters but may be nonlinear across factor levels. Interpretation is easy when data is balanced across factors but much deeper understanding is needed for unbalanced data.
Analysis of variance (ANOVA) is a collection of statistical models and their associated estimation procedures (such as the "variation" among and between groups) used to analyze the differences among means. ANOVA was developed by the statistician Ronald Fisher. ANOVA is based on the law of total variance, where the observed variance in a particular variable is partitioned into components attributable to different sources of variation. In its simplest form, ANOVA provides a statistical test of whether two or more population means are equal, and therefore generalizes the t-test beyond two means. In other words, the ANOVA is used to test the difference between two or more means.Analysis of variance (ANOVA) is an analysis tool used in statistics that splits an observed aggregate variability found inside a data set into two parts: systematic factors and random factors. The systematic factors have a statistical influence on the given data set, while the random factors do not. Analysts use the ANOVA test to determine the influence that independent variables have on the dependent variable in a regression study.
Sir Ronald Fisher pioneered the development of ANOVA for analyzing results of agricultural experiments.1 Today, ANOVA is included in almost every statistical package, which makes it accessible to investigators in all experimental sciences. It is easy to input a data set and run a simple ANOVA, but it is challenging to choose the appropriate ANOVA for different experimental designs, to examine whether data adhere to the modeling assumptions, and to interpret the results correctly. The purpose of this report, together with the next 2 articles in the Statistical Primer for Cardiovascular Research series, is to enhance understanding of ANVOA and to promote its successful use in experimental cardiovascular research. My colleagues and I attempt to accomplish those goals through examples and explanation, while keeping within reason the burden of notation, technical jargon, and mathematical equations.
Computational Pool-Testing with Retesting StrategyWaqas Tariq
Pool testing is a cost effective procedure for identifying defective items in a large population. It also improves the efficiency of the testing procedure when imperfect tests are employed. This study develops computational pool-testing strategy based on a proposed pool testing with re-testing strategy. Statistical moments based on this applied design have been generated. With advent of computers in 1980‘s, pool-testing with re-testing strategy under discussion is handled in the context of computational statistics. From this study, it has been established that re-testing reduces misclassifications significantly as compared to Dorfman procedure although re-testing comes with a cost i.e. increase in the number of tests. Re-testing considered improves the sensitivity and specificity of the testing scheme.
36086 Topic Discussion3Number of Pages 2 (Double Spaced).docxrhetttrevannion
36086 Topic: Discussion3
Number of Pages: 2 (Double Spaced)
Number of sources: 1
Writing Style: APA
Type of document: Essay
Academic Level:Master
Category: Psychology
Language Style: English (U.S.)
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Reference/Module
Learning Objectives
•Explain what the x2 goodness-of-fit test is and what it does.
•Calculate a x2 goodness-of-fit test.
•List the assumptions of the x2 goodness-of-fit test.
•Calculate the x2 test of independence.
•Interpret the x2 test of independence.
•Explain the assumptions of the x2 test of independence.
The Chi-Square (x2) Goodness-of-Fit test: What It Is and What It Does
The chi-square (x2) goodness-of-fit test is used for comparing categorical information against what we would expect based on previous knowledge. As such, it tests what are called observed frequencies (the frequency with which participants fall into a category) against expected frequencies (the frequency expected in a category if the sample data represent the population). It is a nondirectional test, meaning that the alternative hypothesis is neither one-tailed nor two-tailed. The alternative hypothesis for a x2 goodness-of-fit test is that the observed data do not fit the expected frequencies for the population, and the null hypothesis is that they do fit the expected frequencies for the population. There is no conventional way to write these hypotheses in symbols, as we have done with the previous statistical tests. To illustrate the x2 goodness-of-fit test, let's look at a situation in which its use would be appropriate.
chi-square (x2) goodness-of-fit test A nonparametric inferential procedure that determines how well an observed frequency distribution fits an expected distribution.
observed frequencies The frequency with which participants fall into a category.
expected frequencies The frequency expected in a category if the sample data represent the population.
Calculations for the x2 Goodness-of-Fit Test
Suppose that a researcher is interested in determining whether the teenage pregnancy rate at a particular high school is different from the rate statewide. Assume that the rate statewide is 17%. A random sample of 80 female students is selected from the target high school. Seven of the students are either pregnant now or have been pregnant previously. The χ2goodness-of-fit test measures the observed frequencies against the expected frequencies. The observed and expected frequencies are presented in Table 21.1.
TABLE 21.1Observed and expected frequencies for χ2 goodness-of-fit example
FREQUENCIES
PREGNANT
NOT PREGNANT
Observed
7
73
Expected
14
66
As can be seen in the table, the observed frequencies represent the number of high school females in the sample of 80 who were pregnant versus not pregnant. The expected frequencies represent what we would expect based on chance, given what is known about the population. In this case, we would expect 17% of the females to be pregnant .
Chapter 9: Two-Sample Inference
265
Chapter 9: Two-Sample Inference
Chapter 7 discussed methods of hypothesis testing about one-population parameters.
Chapter 8 discussed methods of estimating population parameters from one sample using
confidence intervals. This chapter will look at methods of confidence intervals and
hypothesis testing for two populations. Since there are two populations, there are two
random variables, two means or proportions, and two samples (though with paired
samples you usually consider there to be one sample with pairs collected). Examples of
where you would do this are:
Testing and estimating the difference in testosterone levels of men before and
after they had children (Gettler, McDade, Feranil & Kuzawa, 2011).
Testing the claim that a diet works by looking at the weight before and after
subjects are on the diet.
Estimating the difference in proportion of those who approve of President Obama
in the age group 18 to 26 year olds and the 55 and over age group.
All of these are examples of hypothesis tests or confidence intervals for two populations.
The methods to conduct these hypothesis tests and confidence intervals will be explored
in this method. As a reminder, all hypothesis tests are the same process. The only thing
that changes is the formula that you use. Confidence intervals are also the same process,
except that the formula is different.
Section 9.1 Two Proportions
There are times you want to test a claim about two population proportions or construct a
confidence interval estimate of the difference between two population proportions. As
with all other hypothesis tests and confidence intervals, the process is the same though
the formulas and assumptions are different.
Hypothesis Test for Two Population Proportion (2-Prop Test)
1. State the random variables and the parameters in words.
x1 = number of successes from group 1
x2 = number of successes from group 2
p1 = proportion of successes in group 1
p2 = proportion of successes in group 2
2. State the null and alternative hypotheses and the level of significance
Ho : p1 = p2 or Ho : p1 − p2 = 0
HA : p1 < p2
HA : p1 > p2
HA : p1 ≠ p2
HA : p1 − p2 < 0
HA : p1 − p2 > 0
HA : p1 − p2 ≠ 0
Also, state your α level here.
Chapter 9: Two-Sample Inference
266
3. State and check the assumptions for a hypothesis test
a. A simple random sample of size n1 is taken from population 1, and a simple
random sample of size n2 is taken from population 2.
b. The samples are independent.
c. The assumptions for the binomial distribution are satisfied for both
populations.
d. To determine the sampling distribution of p̂1 , you need to show that n1p1 ≥ 5
and n1q1 ≥ 5 , where q1 =1− p1 . If this requirement is true, then the sampling
distribution of p̂1 is well approximated by a normal curve. To determine the
sampling distribution of p̂2 , you need to show that n2p.
Defecation
Normal defecation begins with movement in the left colon, moving stool toward the anus. When stool reaches the rectum, the distention causes relaxation of the internal sphincter and an awareness of the need to defecate. At the time of defecation, the external sphincter relaxes, and abdominal muscles contract, increasing intrarectal pressure and forcing the stool out
The Valsalva maneuver exerts pressure to expel faeces through a voluntary contraction of the abdominal muscles while maintaining forced expiration against a closed airway. Patients with cardiovascular disease, glaucoma, increased intracranial pressure, or a new surgical wound are at greater risk for cardiac dysrhythmias and elevated blood pressure with the Valsalva maneuver and need to avoid straining to pass the stool.
Normal defecation is painless, resulting in passage of soft, formed stool
CONSTIPATION
Constipation is a symptom, not a disease. Improper diet, reduced fluid intake, lack of exercise, and certain medications can cause constipation. For example, patients receiving opiates for pain after surgery often require a stool softener or laxative to prevent constipation. The signs of constipation include infrequent bowel movements (less than every 3 days), difficulty passing stools, excessive straining, inability to defecate at will, and hard feaces
IMPACTION
Fecal impaction results from unrelieved constipation. It is a collection of hardened feces wedged in the rectum that a person cannot expel. In cases of severe impaction the mass extends up into the sigmoid colon.
DIARRHEA
Diarrhea is an increase in the number of stools and the passage of liquid, unformed feces. It is associated with disorders affecting digestion, absorption, and secretion in the GI tract. Intestinal contents pass through the small and large intestine too quickly to allow for the usual absorption of fluid and nutrients. Irritation within the colon results in increased mucus secretion. As a result, feces become watery, and the patient is unable to control the urge to defecate. Normally an anal bag is safe and effective in long-term treatment of patients with fecal incontinence at home, in hospice, or in the hospital. Fecal incontinence is expensive and a potentially dangerous condition in terms of contamination and risk of skin ulceration
HEMORRHOIDS
Hemorrhoids are dilated, engorged veins in the lining of the rectum. They are either external or internal.
FLATULENCE
As gas accumulates in the lumen of the intestines, the bowel wall stretches and distends (flatulence). It is a common cause of abdominal fullness, pain, and cramping. Normally intestinal gas escapes through the mouth (belching) or the anus (passing of flatus)
FECAL INCONTINENCE
Fecal incontinence is the inability to control passage of feces and gas from the anus. Incontinence harms a patient’s body image
PREPARATION AND GIVING OF LAXATIVESACCORDING TO POTTER AND PERRY,
An enema is the instillation of a solution into the rectum and sig
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One of the most developed cities of India, the city of Chennai is the capital of Tamilnadu and many people from different parts of India come here to earn their bread and butter. Being a metropolitan, the city is filled with towering building and beaches but the sad part as with almost every Indian city
How many patients does case series should have In comparison to case reports.pdfpubrica101
Pubrica’s team of researchers and writers create scientific and medical research articles, which may be important resources for authors and practitioners. Pubrica medical writers assist you in creating and revising the introduction by alerting the reader to gaps in the chosen study subject. Our professionals understand the order in which the hypothesis topic is followed by the broad subject, the issue, and the backdrop.
https://pubrica.com/academy/case-study-or-series/how-many-patients-does-case-series-should-have-in-comparison-to-case-reports/
Medical Technology Tackles New Health Care Demand - Research Report - March 2...pchutichetpong
M Capital Group (“MCG”) predicts that with, against, despite, and even without the global pandemic, the medical technology (MedTech) industry shows signs of continuous healthy growth, driven by smaller, faster, and cheaper devices, growing demand for home-based applications, technological innovation, strategic acquisitions, investments, and SPAC listings. MCG predicts that this should reflects itself in annual growth of over 6%, well beyond 2028.
According to Chris Mouchabhani, Managing Partner at M Capital Group, “Despite all economic scenarios that one may consider, beyond overall economic shocks, medical technology should remain one of the most promising and robust sectors over the short to medium term and well beyond 2028.”
There is a movement towards home-based care for the elderly, next generation scanning and MRI devices, wearable technology, artificial intelligence incorporation, and online connectivity. Experts also see a focus on predictive, preventive, personalized, participatory, and precision medicine, with rising levels of integration of home care and technological innovation.
The average cost of treatment has been rising across the board, creating additional financial burdens to governments, healthcare providers and insurance companies. According to MCG, cost-per-inpatient-stay in the United States alone rose on average annually by over 13% between 2014 to 2021, leading MedTech to focus research efforts on optimized medical equipment at lower price points, whilst emphasizing portability and ease of use. Namely, 46% of the 1,008 medical technology companies in the 2021 MedTech Innovator (“MTI”) database are focusing on prevention, wellness, detection, or diagnosis, signaling a clear push for preventive care to also tackle costs.
In addition, there has also been a lasting impact on consumer and medical demand for home care, supported by the pandemic. Lockdowns, closure of care facilities, and healthcare systems subjected to capacity pressure, accelerated demand away from traditional inpatient care. Now, outpatient care solutions are driving industry production, with nearly 70% of recent diagnostics start-up companies producing products in areas such as ambulatory clinics, at-home care, and self-administered diagnostics.
CRISPR-Cas9, a revolutionary gene-editing tool, holds immense potential to reshape medicine, agriculture, and our understanding of life. But like any powerful tool, it comes with ethical considerations.
Unveiling CRISPR: This naturally occurring bacterial defense system (crRNA & Cas9 protein) fights viruses. Scientists repurposed it for precise gene editing (correction, deletion, insertion) by targeting specific DNA sequences.
The Promise: CRISPR offers exciting possibilities:
Gene Therapy: Correcting genetic diseases like cystic fibrosis.
Agriculture: Engineering crops resistant to pests and harsh environments.
Research: Studying gene function to unlock new knowledge.
The Peril: Ethical concerns demand attention:
Off-target Effects: Unintended DNA edits can have unforeseen consequences.
Eugenics: Misusing CRISPR for designer babies raises social and ethical questions.
Equity: High costs could limit access to this potentially life-saving technology.
The Path Forward: Responsible development is crucial:
International Collaboration: Clear guidelines are needed for research and human trials.
Public Education: Open discussions ensure informed decisions about CRISPR.
Prioritize Safety and Ethics: Safety and ethical principles must be paramount.
CRISPR offers a powerful tool for a better future, but responsible development and addressing ethical concerns are essential. By prioritizing safety, fostering open dialogue, and ensuring equitable access, we can harness CRISPR's power for the benefit of all. (2998 characters)
CHAPTER 1 SEMESTER V - ROLE OF PEADIATRIC NURSE.pdfSachin Sharma
Pediatric nurses play a vital role in the health and well-being of children. Their responsibilities are wide-ranging, and their objectives can be categorized into several key areas:
1. Direct Patient Care:
Objective: Provide comprehensive and compassionate care to infants, children, and adolescents in various healthcare settings (hospitals, clinics, etc.).
This includes tasks like:
Monitoring vital signs and physical condition.
Administering medications and treatments.
Performing procedures as directed by doctors.
Assisting with daily living activities (bathing, feeding).
Providing emotional support and pain management.
2. Health Promotion and Education:
Objective: Promote healthy behaviors and educate children, families, and communities about preventive healthcare.
This includes tasks like:
Administering vaccinations.
Providing education on nutrition, hygiene, and development.
Offering breastfeeding and childbirth support.
Counseling families on safety and injury prevention.
3. Collaboration and Advocacy:
Objective: Collaborate effectively with doctors, social workers, therapists, and other healthcare professionals to ensure coordinated care for children.
Objective: Advocate for the rights and best interests of their patients, especially when children cannot speak for themselves.
This includes tasks like:
Communicating effectively with healthcare teams.
Identifying and addressing potential risks to child welfare.
Educating families about their child's condition and treatment options.
4. Professional Development and Research:
Objective: Stay up-to-date on the latest advancements in pediatric healthcare through continuing education and research.
Objective: Contribute to improving the quality of care for children by participating in research initiatives.
This includes tasks like:
Attending workshops and conferences on pediatric nursing.
Participating in clinical trials related to child health.
Implementing evidence-based practices into their daily routines.
By fulfilling these objectives, pediatric nurses play a crucial role in ensuring the optimal health and well-being of children throughout all stages of their development.
Artificial Intelligence to Optimize Cardiovascular Therapy
Probablity distribution
1. PROBABILITY DISTRIBUTIONPROBABILITY DISTRIBUTION
Dr Htin Zaw SoeDr Htin Zaw Soe
MBBS, DFT, MMedSc (P & TM), PhD, DipMedEdMBBS, DFT, MMedSc (P & TM), PhD, DipMedEd
Associate ProfessorAssociate Professor
Department of BiostatisticsDepartment of Biostatistics
University of Public HealthUniversity of Public Health
2. Probability distribution – expressed in table, graph, or formulaProbability distribution – expressed in table, graph, or formula
etc.etc.
- useful in data summarization and- useful in data summarization and
descriptiondescription
I. Probability distribution of discrete variablesI. Probability distribution of discrete variables
Definition: The probability distribution of a discrete randomDefinition: The probability distribution of a discrete random
variable is a table, graph, formula, or other device used tovariable is a table, graph, formula, or other device used to
specify all possible values of a discrete random variable alongspecify all possible values of a discrete random variable along
with their respective probabilitieswith their respective probabilities
3. Table 1.Table 1. NumberNumber of Assistance Programs Utilized byof Assistance Programs Utilized by
Families with Children in Head Start Programs inFamilies with Children in Head Start Programs in
Southern OhioSouthern Ohio
Number of programNumber of program FrequencyFrequency
11
22
33
44
55
66
77
88
6262
4747
3939
3939
5858
3737
44
1111
TotalTotal 297297
4. Table 2.Table 2. Probability DistributionProbability Distribution of Programs Utilized byof Programs Utilized by
Families with Children in Head Start Programs in SouthernFamilies with Children in Head Start Programs in Southern
OhioOhio
Number of programNumber of program (x)(x) P(X =x)P(X =x)
11
22
33
44
55
66
77
88
.2088.2088
.1582.1582
.1313.1313
.1313.1313
.1953.1953
.1246.1246
.0135.0135
.0370.0370
TotalTotal 1.00001.0000
5. GraphicalGraphical representation of the Probability Distribution ofrepresentation of the Probability Distribution of
Programs Utilized by Families with Children in Head StartPrograms Utilized by Families with Children in Head Start
Programs in Southern Ohio (See Text)Programs in Southern Ohio (See Text)
Two essential properties of a discrete random variableTwo essential properties of a discrete random variable
(1) 0(1) 0 ≤≤ P( X =x)P( X =x) ≤≤ 11
(2)(2) ΣΣ P(X =x)P(X =x) = 1= 1
Q1Q1: What is the probability that a randomly selected family: What is the probability that a randomly selected family
will be one who used three assistance programs?will be one who used three assistance programs?
A1A1:: P(X =P(X = 3) is 0.13133) is 0.1313
Q1Q1: What is the probability that a randomly selected family: What is the probability that a randomly selected family
used either oneused either one oror two programs?two programs?
A1A1:: PP(1(1UU 2) =2) = PP(1) +(1) + PP (2) = 0.2088 + 0.1582(2) = 0.2088 + 0.1582
= 0.3670= 0.3670
6. Table 3.Table 3. CumulativeCumulative Probability DistributionProbability Distribution of Number ofof Number of
Programs Utilized by Families with Children in Head StartPrograms Utilized by Families with Children in Head Start
Programs in Southern OhioPrograms in Southern Ohio
Number of programNumber of program (x)(x) CumulativeCumulative FrequencyFrequency
(PX(PX ≤≤ x)x)
11
22
33
44
55
66
77
88
.2088.2088
.3670.3670
.4983.4983
.6296.6296
.8249.8249
.9495.9495
.9630.9630
1.00001.0000
7. Cumulative Probability forCumulative Probability for xxii is written asis written as F (xF (xii)) == P(XP(X ≤ x≤ xii))
[[ XX is less than or equal to a specified value,is less than or equal to a specified value, xxii
eg.eg. FF (2) =(2) = PP (( XX ≤ 2) ]≤ 2) ]
The graph of a cumulativeThe graph of a cumulative probability distribution is called ‘probability distribution is called ‘OgiveOgive’’..
(See text)(See text)
By consulting the cumulativeBy consulting the cumulative probability distribution we canprobability distribution we can
answer the following questionsanswer the following questions
Q 3Q 3: What is the probability that a family picked at random will be: What is the probability that a family picked at random will be
one who used twoone who used two oror fewer programs?fewer programs?
Q 4Q 4: What is the probability that a family picked at random will be: What is the probability that a family picked at random will be
one who used fewer than four programs?one who used fewer than four programs?
Q 5Q 5: What is the probability that a family picked at random will be: What is the probability that a family picked at random will be
one who used fiveone who used five oror more programs?more programs?
Q 6Q 6: What is the probability that a family picked at random will be: What is the probability that a family picked at random will be
one who used between three and five programs, inclusive?one who used between three and five programs, inclusive?
8. A3: P (XA3: P (X ≤ 2) = 0.3670≤ 2) = 0.3670
A4: P ( X < 4A4: P ( X < 4) = P (X ≤ 3) = 0.4983) = P (X ≤ 3) = 0.4983
A5: P (X ≥ 5) = 1 – 0.6296 = 0.3704A5: P (X ≥ 5) = 1 – 0.6296 = 0.3704
A6: P ( 3 ≤ X ≤ 5) = 0.8249 – 0.3670 = 0.4579A6: P ( 3 ≤ X ≤ 5) = 0.8249 – 0.3670 = 0.4579
9. Three basic probability distributionsThree basic probability distributions
(1) The binomial distribution(1) The binomial distribution
(2) The Poisson distribution(2) The Poisson distribution
(3) The normal distribution(3) The normal distribution
10. II. The binomial distributionII. The binomial distribution
- Derived from a- Derived from a BernoulliBernoulli trial or process [Swiss Mathematiciantrial or process [Swiss Mathematician
James Bernoulli (1654-1705)]James Bernoulli (1654-1705)]
- When a random process or experiment or a trial can result in- When a random process or experiment or a trial can result in
only one of two mutually exclusive outcomes such as dead oronly one of two mutually exclusive outcomes such as dead or
alive, sick or well, full-term or premature, the trial is known as aalive, sick or well, full-term or premature, the trial is known as a
BernoulliBernoulli trial.trial.
11. TheThe BernoulliBernoulli processprocess
TheThe BernoulliBernoulli trials forms atrials forms a BernoulliBernoulli processprocess under theunder the
following conditions.following conditions.
(1) Each trial results in(1) Each trial results in one of two mutually exclusive outcomesone of two mutually exclusive outcomes..
One the success, the other failureOne the success, the other failure
(2) The probability of a(2) The probability of a success (success (pp)) remains constant from trial toremains constant from trial to
trial. The probability oftrial. The probability of failure (failure (1- p1- p) is denoted by () is denoted by (qq))
(3) The trials are(3) The trials are independentindependent; that is the outcome of any; that is the outcome of any
particular trial is not affected by the outcome of any other trial.particular trial is not affected by the outcome of any other trial.
12. Example 1: 85.8 % of pregnancies had full-term birth. What isExample 1: 85.8 % of pregnancies had full-term birth. What is
the probability thatthe probability that exactly three of fiveexactly three of five birth records randomlybirth records randomly
selected from that population will be full-term birth?selected from that population will be full-term birth?
FPFFP (F= Full-term, P=Premature)FPFFP (F= Full-term, P=Premature)
10110 (Coded form: F=1, P=0)10110 (Coded form: F=1, P=0)
According toAccording to multiplication rulemultiplication rule probability of above sequence is:probability of above sequence is:
P(1, 0, 1, 1, 0) =P(1, 0, 1, 1, 0) = pqppq = qpqppq = q22
pp33
NumberNumber SequenceSequence NumberNumber SequenceSequence
2 11100 7 011102 11100 7 01110
3 10011 8 001113 10011 8 00111
4 11010 9 010114 11010 9 01011
5 11001 10 011015 11001 10 01101
6 101016 10101
13. According toAccording to addition ruleaddition rule,,
the probability is sum of individual probabilitiesthe probability is sum of individual probabilities
ie. 10ie. 10 × q× q22
pp33
pp = 0.858,= 0.858, qq = 1 - 0.858 = 0.142= 1 - 0.858 = 0.142
10 × (0.142)10 × (0.142)22
(0.858)(0.858) 33
= 0.1276= 0.1276
14. Large sample procedure: Use of ‘combination’Large sample procedure: Use of ‘combination’
If a set consists ofIf a set consists of nn objects, and we wish to form a subset ofobjects, and we wish to form a subset of xx
objects from theseobjects from these nn objects, without regard to the order of theobjects, without regard to the order of the
objects in the subset, the result is called aobjects in the subset, the result is called a combinationcombination
Definition of combination: ADefinition of combination: A combinationcombination ofof nn objects takenobjects taken xx atat
a time is an unordered subset ofa time is an unordered subset of xx ofof nn objectsobjects
nn CC xx == nn ! /! / xx ! (! ( nn –– xx) ! ( ! is read factorial)) ! ( ! is read factorial)
55 CC 33 = 5= 5 ! / 3 ! ( 5 – 3) !! / 3 ! ( 5 – 3) !
= 5 · 4 ·3 · 2 · 1 / 3 · 2 · 1 · 2 · 1= 5 · 4 ·3 · 2 · 1 / 3 · 2 · 1 · 2 · 1
= 120 / 12= 120 / 12
= 10= 10
[Note 1: 0[Note 1: 0 ! = 1]! = 1]
15. Example 2: Smoking prevalence is 14% among pregnantExample 2: Smoking prevalence is 14% among pregnant
women in a county . If a random sample of size 10 selectedwomen in a county . If a random sample of size 10 selected
from that population, what is the probability that it will containsfrom that population, what is the probability that it will contains
exactly 4 mothers smoke during pregnancy?exactly 4 mothers smoke during pregnancy?
f (x) =f (x) = nn CC xx qq n-xn-x
pp xx
nn CC xx == nn ! /! / xx ! (! ( nn –– xx) !) !
1010 CC 44 = 10= 10 ! / 4 ! ( 10 – 4) !! / 4 ! ( 10 – 4) !
== 1010 ! / 4 ! 6 !! / 4 ! 6 !
= 210= 210
f (x) =f (x) = nn CC xx qq n-xn-x
pp xx
f (4) = 210f (4) = 210 (.86)(.86) 10-410-4
(.14)(.14) 44
= 210= 210 (.86)(.86)66
(.14)(.14) 44
16. Using Binomial Table (Table B)Using Binomial Table (Table B)
Example 3: Smoking prevalence is 14% among pregnantExample 3: Smoking prevalence is 14% among pregnant
women in a county . If a random sample of size 10 selectedwomen in a county . If a random sample of size 10 selected
from that population, what is the probability that it will containsfrom that population, what is the probability that it will contains
exactly 4exactly 4 mothers smoke during pregnancy?mothers smoke during pregnancy?
n = 10, p = .14, x = 4n = 10, p = .14, x = 4
Table B. Cumulative binomial probability distributionTable B. Cumulative binomial probability distribution
f (4) = P ( Xf (4) = P ( X≤ 4) -≤ 4) - P ( XP ( X≤ 3)≤ 3)
= .9927 - .9600= .9927 - .9600
= .0327= .0327
17. Using Binomial Table (Table B) when p > 0.5Using Binomial Table (Table B) when p > 0.5
Table B does not contain value ofTable B does not contain value of pp > 0.5. So> 0.5. So
P ( X = xP ( X = x || n , p > .5n , p > .5) =) = P ( X = n – xP ( X = n – x | n| n , 1- p, 1- p))
P ( XP ( X ≤≤ xx || n , p > .5n , p > .5) =) = P ( XP ( X ≥≥ n – xn – x | n| n , 1- p, 1- p))
When X is greater than or equal to some x whenWhen X is greater than or equal to some x when p > .5, usep > .5, use
P ( XP ( X ≥≥ xx || n , p > .5n , p > .5) =) = P ( XP ( X ≤≤ n – xn – x | n| n , 1- p, 1- p))
18. Example 4 usingExample 4 using Table BTable B: 55 % of residents said ‘childhood: 55 % of residents said ‘childhood
obesity is a serious problem’. If a random sample of 12obesity is a serious problem’. If a random sample of 12
residents is selected, what is the probability thatresidents is selected, what is the probability that exactly sevenexactly seven
of 12 residents (sample) said the statement is ‘a seriousof 12 residents (sample) said the statement is ‘a serious
problem’?problem’?
P ( X = xP ( X = x || n , p > .5n , p > .5) =) = P ( X = n – xP ( X = n – x | n| n , 1- p, 1- p))
P ( X = 7P ( X = 7 || 12 , p > .5512 , p > .55) =) = P ( X = 5P ( X = 5 | 12| 12 , p =.45, p =.45))
P ( X = 5P ( X = 5 | 12| 12 , p =.45, p =.45) = P (X ≤ 5) - P (X ≤ 4)) = P (X ≤ 5) - P (X ≤ 4)
= .5269 - .3044= .5269 - .3044 ←← [From Table B][From Table B]
== .2225.2225
19. Example 4 usingExample 4 using FormulaFormula : 55 % of residents said ‘childhood: 55 % of residents said ‘childhood
obesity is a serious problem’. If a random sample of 12obesity is a serious problem’. If a random sample of 12
residents is selected, what is the probability thatresidents is selected, what is the probability that exactly sevenexactly seven
of 12 residents (sample) said the statement is ‘a seriousof 12 residents (sample) said the statement is ‘a serious
problem’?problem’?
f (x) =f (x) = nn CC xx qq n-xn-x
pp xx
nn CC xx == nn ! /! / xx ! (! ( nn –– xx) !) !
1212 CC77 = 12= 12 ! / 7 ! ( 12 – 7) !! / 7 ! ( 12 – 7) !
== 1212 ! / 7 ! 5 !! / 7 ! 5 !
= 792= 792
f (x) =f (x) = nn CC xx qq n-xn-x
pp xx
f (7) =f (7) = 1212 CC77 qq12-712-7
pp77
= 792= 792(.45)(.45)55
(.55)(.55)77
== 0.22250.2225
20. [Note 2: How to calculate 12[Note 2: How to calculate 12 !! // 7 ! 5 ! by an easier method ?7 ! 5 ! by an easier method ?
1212 !! // 7 ! 5 ! = 12·11·10·9·8·7·6·5·4·3·2·17 ! 5 ! = 12·11·10·9·8·7·6·5·4·3·2·1// (7·6·5·4·2·3·2·1) 5·4·3·2·1(7·6·5·4·2·3·2·1) 5·4·3·2·1
= 12·11·10·9·8= 12·11·10·9·8 7 !7 ! // 7 !7 ! 5·4·3·2·15·4·3·2·1
= 12·11·10·9·8= 12·11·10·9·8 // 5·4·3·2·15·4·3·2·1
= 792 ]= 792 ]
[Note 3: mean of binomial distribution =[Note 3: mean of binomial distribution = μμ == npnp
variancevariance of binomial distribution =of binomial distribution = σσ 22
== np (1-p)np (1-p) ]]
21. III. The Poisson DistributionIII. The Poisson Distribution
[French Mathematician Denis Poisson (1781 – 1840)][French Mathematician Denis Poisson (1781 – 1840)]
IfIf xx is the number of occurrences of some random event in anis the number of occurrences of some random event in an
interval of time or space (or some volume of matter), theinterval of time or space (or some volume of matter), the
probability thatprobability that xx will occur is given by:will occur is given by:
f (x) = ef (x) = e--λλ
λλ xx
/ x/ x !!
(( λλ = a parameter of the distribution, the average number of= a parameter of the distribution, the average number of
occurrences of the random event in the interval or volume;occurrences of the random event in the interval or volume;
e = a constant , 2.7183)e = a constant , 2.7183)
22. The Poisson ProcessThe Poisson Process states that:states that:
1. The occurrences of the events are1. The occurrences of the events are independentindependent. The. The
occurrence of an event in an interval of space or time has nooccurrence of an event in an interval of space or time has no
effect on the probability of a second occurrence of the event ineffect on the probability of a second occurrence of the event in
the same, or any other, intervalthe same, or any other, interval
2. Theoretically2. Theoretically an infinite numberan infinite number of occurrences of the eventof occurrences of the event
must be possible in the intervalmust be possible in the interval
3. The probability of the3. The probability of the single occurrencesingle occurrence of the event in a givenof the event in a given
interval isinterval is proportional to the length of the intervalproportional to the length of the interval
4. In an4. In an infinitesimally smallinfinitesimally small portion of the interval, the probabilityportion of the interval, the probability
of more than one occurrence of the event isof more than one occurrence of the event is negligiblenegligible
Mean and variance are equal in Poisson distributionMean and variance are equal in Poisson distribution
23. When to use the Poisson Model?When to use the Poisson Model?
When counts are made of events or entities that are distributedWhen counts are made of events or entities that are distributed
at random in space or time the Poisson Model is usedat random in space or time the Poisson Model is used
Example 5: Accidental poisoning cases admitted to a generalExample 5: Accidental poisoning cases admitted to a general
hospital follow the Poisson Law withhospital follow the Poisson Law with λλ = 12 cases per year.= 12 cases per year.
Find probability that in the next yearFind probability that in the next year exactly threeexactly three cases will becases will be
admitted to the hospital.admitted to the hospital.
f (x) = ef (x) = e--λλ
λλ xx
/ x/ x !!
P (X =3) = eP (X =3) = e--λλ
λλ xx
/ x/ x !!
== 2.71832.7183 ––1212
1212 33
/ 3/ 3 !!
= .00177= .00177
24. Example 6: Average number of certain aquatic organism in aExample 6: Average number of certain aquatic organism in a
water sample taken from a pond is found to be 2. If the numberwater sample taken from a pond is found to be 2. If the number
follow the Poisson Law, find the probability that next sample willfollow the Poisson Law, find the probability that next sample will
containcontain one or fewerone or fewer organism.organism.
P (XP (X ≤ 1≤ 1) = P (X) = P (X =0=0) + P (X =) + P (X = 11))
= e= e--λλ
λλ xx
/ x/ x ! +! + ee--λλ
λλ xx
/ x/ x !!
== 2.71832.7183 ––22
22 00
/ 0/ 0 ! +! + 2.71832.7183 ––22
22 11
/ 1/ 1 !!
= .1353 + .27067= .1353 + .27067
= .406= .406
We can get answer easily by consulting Table CWe can get answer easily by consulting Table C
P (XP (X ≤ 1≤ 1) =) = .406.406
25. IV. Continuous Probability DistributionsIV. Continuous Probability Distributions
HistogramHistogram
Frequency polygonFrequency polygon
IfIf nn approaches infinity,approaches infinity,
- width of class interval approaches zero.- width of class interval approaches zero.
- frequency polygon approaches a- frequency polygon approaches a smooth curvesmooth curve representingrepresenting
graphically the distribution of a continuous random variable (eg.graphically the distribution of a continuous random variable (eg.
age)age)
Total area under the curve is 1Total area under the curve is 1
Relative frequency of values between two points onRelative frequency of values between two points on xx axis isaxis is
equal to total area bounded by the curve, theequal to total area bounded by the curve, the xx axis, and twoaxis, and two
respective perpendicular lines.respective perpendicular lines.
26. Finding area under a smooth curve:Finding area under a smooth curve:
To find area, no cells in case of a smooth curveTo find area, no cells in case of a smooth curve →→ alternatealternate
methodmethod by integral calculusby integral calculus
A density function is a formula to represent the distribution ofA density function is a formula to represent the distribution of
continuous random variablecontinuous random variable
Definition:Definition: A nonnegative functionA nonnegative function f (x)f (x) is called ais called a distributiondistribution
(sometimes called a probability density function)(sometimes called a probability density function) of theof the
continuous random variablecontinuous random variable XX if theif the total areatotal area bounded by itsbounded by its
curve and the axis is equal tocurve and the axis is equal to 11 and if theand if the subareasubarea under theunder the
curve bounded by the curve, thecurve bounded by the curve, the xx axis, and perpendicularsaxis, and perpendiculars
erected at any two pointserected at any two points aa andand bb give the probability thatgive the probability that XX isis
between the pointsbetween the points aa andand bb..
27. IV. The Normal DistributionIV. The Normal Distribution
The Gaussian distribution [Carl Friedrich Gauss (1777-1855)]The Gaussian distribution [Carl Friedrich Gauss (1777-1855)]
The normal density is:The normal density is:
f (x) = (1/f (x) = (1/√√ 22ππσσ ) e) e-(x--(x- µµ)2/2)2/2σσ22
( -( -∞∞< x << x < ∞)∞)
Characteristics of the normal distributionCharacteristics of the normal distribution
1. It is symmetrical about the mean,1. It is symmetrical about the mean, µ . The curve on either side ofµ . The curve on either side of
µ is mirror image of the other sideµ is mirror image of the other side
2. The mean, the median and mode are all equal.2. The mean, the median and mode are all equal.
3. The total area under the curve above the x axis is3. The total area under the curve above the x axis is one squareone square
unitunit. Normal distribution is probability distribution. 50% of the. Normal distribution is probability distribution. 50% of the
area is to the right of a perpendicular erected at the mean andarea is to the right of a perpendicular erected at the mean and
50% is to the left.50% is to the left.
4. 1 SD from the mean in both directions, the area is 68%.4. 1 SD from the mean in both directions, the area is 68%.
For 2 SD and 3 SD, areas are 95% and 99.7% respectivelyFor 2 SD and 3 SD, areas are 95% and 99.7% respectively
28. 5. The normal distribution is determined by5. The normal distribution is determined by µ andµ and σσ
(Different values of(Different values of µµ cause the distribution graph shift along the xcause the distribution graph shift along the x
axis and different values ofaxis and different values of σσ cause the degree of flatness orcause the degree of flatness or
peakedness of distribution graphs)peakedness of distribution graphs)
The standard normal distribution :The standard normal distribution :
Also called ‘Also called ‘unit normal distributionunit normal distribution’ because its mean is zero and’ because its mean is zero and
SD is 1SD is 1
Table D is used to calculate the probabilityTable D is used to calculate the probability
Example 7: Given the standard normal distribution, find the areaExample 7: Given the standard normal distribution, find the area
under the curve, above the z-axis between z= -∞ and z = 2under the curve, above the z-axis between z= -∞ and z = 2
Answer 7: by consulting Table D the area is 0.9772Answer 7: by consulting Table D the area is 0.9772
29. Example 8: What is the probability that a z picked at randomExample 8: What is the probability that a z picked at random
from the population of z’s will have a value between -2.55 andfrom the population of z’s will have a value between -2.55 and
+2.55 ?+2.55 ?
Answer 8 : P (Answer 8 : P (-2.55 < z < +2.55)-2.55 < z < +2.55)
= 0.9946 – 0.005= 0.9946 – 0.005
= 0.9892= 0.9892
Example 9: What proportion of z value are between -2.47 andExample 9: What proportion of z value are between -2.47 and
1.53?1.53?
Answer 9 : P (Answer 9 : P (-2.47-2.47 ≤≤ zz ≤≤ 1.53)1.53)
= 0.9370 – 0.0068= 0.9370 – 0.0068
= 0.9302= 0.9302
30. Example 10: Given the standard normal distribution,Example 10: Given the standard normal distribution,
findfind P (z ≥2.71)P (z ≥2.71)
Answer 10 : P (z ≥2.71)Answer 10 : P (z ≥2.71)
= 1 –= 1 – P (z ≤ 2.71)P (z ≤ 2.71)
= 1 - 0.9966= 1 - 0.9966
= 0.0034= 0.0034
31. Normal Distribution ApplicationsNormal Distribution Applications
- To makeTo make probabilityprobability statement by using normal distributionstatement by using normal distribution
- Example 11: Amount of time children (8-15 yr) spent in theExample 11: Amount of time children (8-15 yr) spent in the
upright position in 24 hr followed a normal distribution with aupright position in 24 hr followed a normal distribution with a
mean of 5.4 hr and SD of 1.3 hr. Find the prob. that a childmean of 5.4 hr and SD of 1.3 hr. Find the prob. that a child
selected at random spends less than 3 hr in upright position inselected at random spends less than 3 hr in upright position in
24 hr period.24 hr period.
- Answer 11:Answer 11:
z = x-z = x- µµ // σσ
= 3 – 5.4 / 1.3= 3 – 5.4 / 1.3
= - 1.85= - 1.85
By consulting Table DBy consulting Table D
P (x < 3) = P ( z < - 1.85) = 0.0322P (x < 3) = P ( z < - 1.85) = 0.0322
32. - Example 12: In a study of a common breath metabolite,Example 12: In a study of a common breath metabolite,
ammonia concentration (ppb) in a subject followed a normalammonia concentration (ppb) in a subject followed a normal
distribution over 30 days with a mean of 491ppb and SD 119distribution over 30 days with a mean of 491ppb and SD 119
ppb. What is the prob. that on a random day, the subject’sppb. What is the prob. that on a random day, the subject’s
ammonia conc. is between 292 and 649 ppb?ammonia conc. is between 292 and 649 ppb?
- Answer 12:Answer 12:
z = x-z = x- µµ // σσ
P ( 292 ≤ x ≤ 649 ) = P (P ( 292 ≤ x ≤ 649 ) = P ( x-x- µµ // σσ ≤ z ≤≤ z ≤ x-x- µµ // σσ))
= P (= P ( 292 - 491 / 119292 - 491 / 119 ≤ z ≤ 649≤ z ≤ 649 - 491 / 119- 491 / 119))
= P (= P ( - 1.67- 1.67 ≤ z ≤≤ z ≤ 1.331.33))
= 0.9082 – 0.0475= 0.9082 – 0.0475
= 0.8607= 0.8607
33. Example 13: In a population of 10,000 of the children inExample 13: In a population of 10,000 of the children in
example 11, how many would you expect to be upright moreexample 11, how many would you expect to be upright more
than 8.5 hr? (mean, 5.4 hr and SD, 1.3 hr)than 8.5 hr? (mean, 5.4 hr and SD, 1.3 hr)
- Answer 13:Answer 13:
z = x-z = x- µµ // σσ
P ( x ≥ 8.5 ) = 1 - P ( x ≤ 8.5 )P ( x ≥ 8.5 ) = 1 - P ( x ≤ 8.5 )
= 1 - P ( z == 1 - P ( z = x-x- µµ // σσ ))
= 1 - P ( z =8.5= 1 - P ( z =8.5 – 5.4 / 1.3– 5.4 / 1.3 ))
= 1 - P ( z = 2.38 )= 1 - P ( z = 2.38 )
= 1 – 0. 9913= 1 – 0. 9913
= 0. 0087= 0. 0087
For population , we expect 10,000 (0.0087) = 87 children to spendFor population , we expect 10,000 (0.0087) = 87 children to spend
more than 8.5 hr upright.more than 8.5 hr upright.