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Elementary Statistics Practice Test 2 Solutions
Chapter 4: Probability
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Elementary Statistics Practice Test 2 Solutions
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1 Probability Please read sections 3.1 – 3.3 in your .docxaryan532920
1
Probability
Please read sections 3.1 – 3.3 in your textbook
Def: An experiment is a process by which observations are generated.
Def: A variable is a quantity that is observed in the experiment.
Def: The sample space (S) for an experiment is the set of all possible outcomes.
Def: An event E is a subset of a sample space. It provides the collection of outcomes
that correspond to some classification.
Example:
Note: A sample space does not have to be finite.
Example: Pick any positive integer. The sample space is countably infinite.
A discrete sample space is one with a finite number of elements, { }1,2,3,4,5,6 or one that
has a countably infinite number of elements { }1,3,5,7,... .
A continuous sample space consists of elements forming a continuum. { }x / 2 x 5< <
2
A Venn diagram is used to show relationships between events.
A intersection B = (A ∩ B) = A and B
The outcomes in (A intersection B) belong to set A as well as to set B.
A union B = (A U B) = A alone or B alone or both
Union Formula
For any events A, B, P (A or B) = P (A) + P (B) – P (A intersection B) i.e.
P (A U B) = P (A) + P (B) – P (A ∩ B)
3
cA complement not A A ' A A = = = =
A complement consists of all outcomes outside of A.
Note: P (not A) = 1 – P (A)
Def: Two events are mutually exclusive (disjoint, incompatible) if they do not intersect,
i.e. if they do not occur at the same time. They have no outcomes in common.
When A and B are mutually exclusive, (A ∩ B) = null set = Ø, and P (A and B) = 0.
Thus, when A and B are mutually exclusive, P (A or B) = P (A) + P (B)
(This is exactly the same statement as rule 3 below)
Axioms of Probability
Def: A probability function p is a rule for calculating the probability of an event. The
function p satisfies 3 conditions:
1) 0 ≤ P (A) ≤1, for all events A in the sample space S
2) P (Sample Space S) = 1
3) If A, B, C are mutually exclusive events in the sample space S, then
P(A B C) P(A) P(B) P(C)∪ ∪ = + +
4
The Classical Probability Concept: If there are n equally likely possibilities, of which one
must occur and s are regarded as successes, then the probability of success is s
n
.
Example:
Frequency interpretation of Probability: The probability of an event E is the proportion of
times the event occurs during a long run of repeated experiments.
Example:
Def: A set function assigns a non-negative value to a set.
Ex: N (A) is a set function whose value is the number of elements in A.
Def: An additive set function f is a function for which f (A U B) = f (A) + f (B) when A and
B are mutually exclusive.
N (A) is an additive set function.
Ex: Toss 2 fair dice. Let A be the event that the sum on the two dice is 5. Let B be the
event that the sum on ...
1. A law firm wants to determine the trend in its annual billings .docxmonicafrancis71118
1. A law firm wants to determine the trend in its annual billings so that it can better forecast revenues. It plots the data on its billings for the past 10 years and finds that the scatter plot appears to be linear. What formula should they use to determine the trend line?
σ = ∑√(x - μ)2 ÷ N
F = s12 ÷ s22
t = (x̄ - μx-bar) ÷ s/√n
Tt = b0 + b1t
3 points
QUESTION 2
1. A set of subjects, usually randomly sampled, selected to participate in a research study is called:
Population
Sample
Mode group
Partial selection
3 points
QUESTION 3
1. If a researcher accepts a null hypothesis when that hypothesis is actually true, she has committed:
a type I error
a type II error
no error
a causation
3 points
QUESTION 4
1. A binomial probability distribution is a discrete distribution (i.e., the x-variable is discrete).
True
False
3 points
QUESTION 5
1. The tdistribution is wider and flatter (i.e., has more variation) than the normal distribution.
True
False
3 points
QUESTION 6
1. A physician wants to estimate the average amount of time that patients spend in his waiting room. He asks his receptionist to record the waiting times for 28 of his patients and finds that the sample mean (x̄) is 37 minutes and the sample standard deviation (s) is 12 minutes. What formula would you use to construct the 95% confidence interval for the population mean of waiting times?
t = (x̄ - μx-bar) ÷ s/√n
µ = ∑ x ÷ N
x̄ - t(s ÷ √n) < µ < x̄ + t(s ÷ √n)
z = (x - µ) ÷ σ
3 points
QUESTION 7
1. When the alternative hypothesis states that the difference between two groups can only be in one direction, we call this a:
One-tailed test
Bi-directional test
Two-tailed test
Non-parametric test
3 points
QUESTION 8
1. For any probability distribution, the probability of any x-value occurring within any given range is equal to the area under the distribution and above that range.
True
False
3 points
QUESTION 9
1. The formula for ____________ is (Row total X Column total)/T
Observed frequencies
Degrees of freedom
Expected frequencies
Sampling error
3 points
QUESTION 10
1. State Senator Hanna Rowe has ordered an investigation of the large number of boating accidents that have occurred in the state in recent summers. Acting on her instructions, her aide, Geoff Spencer, has randomly selected 9 summer months within the last few years and has compiled data on the number of boating accidents that occurred during each of these months. The mean number of boating accidents to occur in these 9 months was 31 (x̄), and the standard deviation (s) in this sample was 9 boating accidents per month. Geoff was told to construct a 90% confidence interval for the true mean number of boating accidents per month. What formula should Geoff use?
x̄ - t(s ÷ √n) < µ < x̄ + t(s ÷ √n)
F = s12 ÷ s22
z = (x - µ) ÷ σ
x̄ - z(σ ÷ √n) < µ < x̄ + z(σ ÷ √n)
.
It is a consolidation of basic probability concepts worth understanding before attempting to apply probability concepts for predictions. The material is formed from different sources. ll the sources are acknowledged.
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.
This presentation was provided by Steph Pollock of The American Psychological Association’s Journals Program, and Damita Snow, of The American Society of Civil Engineers (ASCE), for the initial session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session One: 'Setting Expectations: a DEIA Primer,' was held June 6, 2024.
The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
Delivering Micro-Credentials in Technical and Vocational Education and TrainingAG2 Design
Explore how micro-credentials are transforming Technical and Vocational Education and Training (TVET) with this comprehensive slide deck. Discover what micro-credentials are, their importance in TVET, the advantages they offer, and the insights from industry experts. Additionally, learn about the top software applications available for creating and managing micro-credentials. This presentation also includes valuable resources and a discussion on the future of these specialised certifications.
For more detailed information on delivering micro-credentials in TVET, visit this https://tvettrainer.com/delivering-micro-credentials-in-tvet/
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
Thinking of getting a dog? Be aware that breeds like Pit Bulls, Rottweilers, and German Shepherds can be loyal and dangerous. Proper training and socialization are crucial to preventing aggressive behaviors. Ensure safety by understanding their needs and always supervising interactions. Stay safe, and enjoy your furry friends!
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
2. SEGUNDA EVALUACIÓN TERCER CORTE:
PROBABILIDADES
Se presentan cinco problemas prácticos de probabilidades
para desarrollar la teoría sustentada por formulas
matemáticas
El objetivo es demostrar de manera ordenada los pasos
para encontrar la probabilidad de que un suceso ocurra
Los problemas serán resueltos mediante la formula
elemental de las probabilidades, se aplicaran otros artificios
como la regla de la suma y teoría de combinaciones
3. PROBLEMA 1: UN ESTUDIANTE RESPONDE AL AZAR A 4
PREGUNTAS DE VERDADERO Y FALSO. PARA EL CASO
HALLAR LOS SIGUIENTES VALORES
A) Escriba el espacio muestral
B) Escriba el suceso responder “falso” a una sola pregunta
C) Escriba el suceso responder “verdadero” al menos a 3
preguntas
D) Asígnele probabilidades a los eventos anteriores
4. SOLUCIÓN:
Espacio muestral
Planteamiento:
En este caso el experimento aleatorio son las cuatro
preguntas, entonces se tiene los siguientes resultados
posibles para la combinación de verdaderos y falsos en
el espacio muestral
Se usa la letra “V” para representar verdadero y “F” para
representar falso en las respuestas obtenidas para las
cuatro preguntas.
5. SOLUCIÓN:
Espacio muestral
Combinaciones:
S = { (V,V,V,V), (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V),
(V,V,F,F), (V,F,V,F), (V,F,F,V), (F,V,V,F), (F,V,F,V), (F,F,V,V),
(V,F,F,F), (F,V,F,F), (F,F,V,F), (F,F,F,V), (F,F,F,F) }
Para el caso, se tiene un total de cuatro preguntas y sus
cuatro respuestas, entonces se crea una combinación posible
de 4x4, que tiene como resultado un total de 16 posibles
resultados.
6. SOLUCIÓN:
El suceso responder “falso” a una sola pregunta
Se utiliza la letra R para denotar al suceso posible partiendo
de la información obtenida a través del espacio muestral y las
posibles combinaciones de respuestas
R = { (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V) }
Este suceso puede ocurrir para cuatro posibles
combinaciones dadas por el espacio muestral que
representan las respuestas de las preguntas.
7. SOLUCIÓN:
El suceso responder “verdadero” al menos a 3
preguntas
R = { (V,V,V,V), (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V) }
En este suceso se considera a todas las combinaciones con
tres verdaderos y además se considera el caso donde las
cuatro preguntas con respuesta verdadero, para un total de
cinco posibilidades
8. SOLUCIÓN:
Asígnele probabilidades a los eventos anteriores
𝑃 𝑅 =
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑅
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑆
=
𝑛(𝑅)
𝑛(𝑆)
Se aplica la formula general para hallar la probabilidad de
que ocurra un evento
9. SOLUCIÓN:
Asígnele probabilidades a los eventos anteriores
Caso: Responder falso a una pregunta
R = { (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V) }
S = { (V,V,V,V), (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V), (V,V,F,F), (V,F,V,F),
(V,F,F,V), (F,V,V,F), (F,V,F,V), (F,F,V,V), (V,F,F,F), (F,V,F,F), (F,F,V,F), (F,F,F,V),
(F,F,F,F) }
𝑃 𝑅 =
𝑛(𝑅)
𝑛(𝑆)
=
4
16
= 0,25
𝑷 𝑹 = 𝟎, 𝟐𝟓𝒙𝟏𝟎𝟎 = 𝟐𝟓%
Existe un 25% de probabilidades que ocurra este suceso
10. SOLUCIÓN:
Asígnele probabilidades a los eventos anteriores
Caso: Responder “verdadero” al menos a 3 preguntas
R = { (V,V,V,V), (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V) }
S = { (V,V,V,V), (V,V,V,F), (V,V,F,V), (V,F,V,V), (F,V,V,V), (V,V,F,F), (V,F,V,F), (V,F,F,V),
(F,V,V,F), (F,V,F,V), (F,F,V,V), (V,F,F,F), (F,V,F,F), (F,F,V,F), (F,F,F,V), (F,F,F,F) }
𝑃 𝑅 =
𝑛(𝑅)
𝑛(𝑆)
=
5
16
= 0,3125
𝑷 𝑹 = 𝟎, 𝟑𝟏𝟐𝟓𝒙𝟏𝟎𝟎 = 𝟑𝟏, 𝟐𝟓%
Existe un 31,25% de probabilidades que ocurra este suceso.
11. PROBLEMA 2: LA PROBABILIDAD DE QUE UN SERVICIO DE
PRUEBAS PARA CONSUMIDORES CALIFIQUE UN NUEVO
DISPOSITIVO ANTICONTAMINANTE DE AUTOS COMO MUY
MALO, MALO, REGULAR, BUENO, MUY BUENO O EXCELENTE
ES DE 0,07; 0,12; 0,17; 0,32; 0,21; Y 0,11, RESPECTIVAMENTE.
CUÁLES SON LAS PROBABILIDADES DE QUE LO CALIFIQUEN
COMO
A) Muy malo, malo, regular o bueno
B) Bueno, muy bueno o excelente
Tome en cuenta que los eventos son mutuamente
excluyentes
12. SOLUCIÓN:
Planteamiento General
Se tiene seis tipos diferentes de sucesos, que serán
representados de la siguiente forma para su
simplificación:
Muy Malo (MM) = 0,07
Malo (M) = 0,12
Regular (R) = 0,17
Bueno (B) = 0,32
Muy bueno (MB) = 0,21
Excelente (E) = 0,11
13. SOLUCIÓN:
Planteamiento General
En el enunciado nos indican que para el caso los
eventos son mutuamente excluyentes
Se aplica la regla de la suma para cuando los eventos
mutuamente excluyentes:
𝑃 𝐴𝑈𝐵 = 𝑃 𝐴 + 𝑃(𝐵)
14. SOLUCIÓN:
Muy malo, malo, regular o bueno
Para el caso se necesita sumar el resultado de cuatro sucesos
y toma la letra T para definirlo, entonces:
𝑃 𝑇 = 𝑃 𝑀𝑀 + 𝑃 𝑀 + 𝑃 𝑅 + 𝑃 𝐵
𝑃 𝑇 = 0,07 + 0,12 + 0,17 + 0,32 = 0,68
𝑃 𝑇 = 0,68 𝑥 100 = 68%
Existen altas probabilidades que un consumidor califique el
servicio como bueno o un grado menor a este. Para el caso
ocurrirá el 68% de los casos este conjunto de calificaciones
15. SOLUCIÓN:
Bueno, muy bueno o excelente
Para el caso se necesita sumar el resultado de cuatro sucesos
y toma la letra T para definirlo, entonces:
𝑃 𝑇 = 𝑃 𝐵 + 𝑃 𝑀𝐵 + 𝑃(𝐸)
𝑃 𝑇 = 0,32 + 0,21 + 0,11 = 0,64
𝑃 𝑇 = 0,64 𝑥 100 = 64%
Es un 64% probable que la calificación del
consumidor se encuentre enmarcada en este
conjunto de respuestas
16. PROBLEMA 3: SI LAS PROBABILIDADES DE QUE UNA FAMILIA,
ALEATORIAMENTE ELEGIDA EN UNA ENCUESTA REALIZADA
EN UNA GRAN CIUDAD, POSEA UN TELEVISOR DE COLOR,
EN BLANCO Y NEGRO O AMBOS SON, RESPECTIVAMENTE,
0,87; 0,36 Y 0,29. ¿CUÁL ES LA PROBABILIDAD DE QUE UNA
FAMILIA EN ESA CIUDAD POSEA UN TIPO O AMBAS CLASES?
18. SOLUCIÓN:
Planteamiento General
Se tiene tres tipos diferentes de sucesos, que serán
representados de la siguiente forma para su
simplificación:
TV a color (A) = 0,87
Tv en blanco y negro (B) = 0,36
Ambos (𝐴 ∩ 𝐵) = 0,29
19. SOLUCIÓN:
Probabilidad de tener un tipo o ambas
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)
𝑃 𝐴 ∪ 𝐵 = 0,87 + 0,36 − 0,29
𝑃 𝐴 ∪ 𝐵 = 0,94
𝑃 𝐴 ∪ 𝐵 = 0,94 𝑥 100 = 94%
Existen un 94% de probabilidades que una familia en esa ciudad posea un
tipo o ambas clases de televisor.
20. PROBLEMA 4: EL SUPERVISOR DE UN GRUPO DE 20
OBREROS PIDE LA OPINIÓN DE DOS DE ELLOS
SELECCIONADOS AL AZAR, SOBRE LAS NUEVAS
DISPOSICIONES DE SEGURIDAD EN LA CONSTRUCCIÓN. SI
12 ESTÁN A FAVOR DE LAS NUEVAS DISPOSICIONES Y LOS
OCHOS RESTANTES EN CONTRA ¿CUÁL ES LA PROBABILIDAD
DE QUE AMBOS TRABAJADORES ELEGIDOS POR EL
SUPERVISOR ESTÉN EN CONTRA DE LAS NUEVAS
DISPOSICIONES?
21. SOLUCIÓN:
Planteamiento General
Se aplica la formula general de las probabilidades
𝑃 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 𝑦 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 =
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑅
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑆
22. SOLUCIÓN:
Planteamiento General
Se necesita conocer el espacio muestral, se conoce
que de los 20 trabajadores 12 están a favor y 8 están
en contra, el número de combinaciones es amplia,
entonces se recurre al uso de la fórmula para hallar
el valor de S y el valor de R
𝐶𝑘
𝑛
=
𝑛!
𝑛 − 𝑘 ! 𝑘!
23. SOLUCIÓN:
Espacio muestral
Donde n corresponde al número total de trabajadores y k
como los agrupa el supervisor al azar, entonces se obtiene se
procede a buscar las posibles combinaciones
𝐶2
20
=
20!
20 − 2 ! 2!
=
20𝑥19𝑥18!
18! 𝑥2𝑥1
=
20𝑥19
2
= 190
Estoy quiere decir que el espacio muestral S, contiene un total
de 190 combinaciones que el supervisor puede seleccionar al
azar para solicitar la opinión.
24. SOLUCIÓN:
Suceso esperado dos opiniones en contra
Ahora cuando mediante combinaciones también se puede
conocer la posibilidad de que ambos trabajadores estén en
contra de las nuevas disipaciones. Para el caso el valor de n
será de todos los que están en contra y k como los agrupa el
supervisor al azar, entonces:
𝐶2
8
=
8!
8 − 2 ! 2!
=
8𝑥7𝑥6!
6! 𝑥2𝑥1
=
8𝑥7
2
= 28
Existen un total de 28 sucesos posibles R, donde ambos estén
en contra al momento de ser seleccionados al azar.
25. SOLUCIÓN:
Probabilidad de que ambas opiniones sean en contra
𝑃 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 𝑦 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 =
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑅
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑆
𝑃 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 𝑦 𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎 =
28
190
= 0,1473
𝑷 𝑹 = 𝟎, 𝟑𝟏𝟐𝟓𝒙𝟏𝟎𝟎 = 𝟏𝟒, 𝟕𝟑%
Las probabilidades 14,73%, es poco probable que ambos trabajadores
elegidos por el supervisor estén en contra de las nuevas disposiciones
26. PROBLEMA 5: EN UN EXPERIMENTO PARA ESTUDIAR LA RELACIÓN
ENTRE LA HIPERTENSIÓN Y EL HÁBITO DE FUMAR, SE REUNIERON
LOS SIGUIENTES DATOS EN 180 INDIVIDUOS
Si se selecciona aleatoriamente a uno de estos individuos,
encuentre la probabilidad de que la persona
A) Experimente hipertensión
B) Sea un no fumador
C) Experimente hipertensión, dado que es un fumador empedernido
D) Sea un no fumador, dado que no ha presentado problemas de
hipertensión
No fumadores Fumadores
moderados
Fumadores
empedernidos
Total
Hipertenso 21 36 30 87
No hipertenso 48 26 19 93
Total 69 62 49 180
27. SOLUCIÓN:
Planteamiento general
𝑃 𝑅 =
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑅
𝑛𝑢𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒𝑠 𝑑𝑒 𝑆
=
𝑛(𝑅)
𝑛(𝑆)
Se aplica la formula general para hallar la probabilidad de que ocurra un
evento
La condición es seleccionar aleatoriamente a uno de estos individuos,
entonces el espacio muestral será igual a todos los individuos
considerados en el estudio S =180
28. SOLUCIÓN:
Experimente hipertensión
El total de los hipertensos según la tabla es 87, entonces
𝑃 ℎ𝑖𝑝𝑒𝑟𝑡𝑒𝑛𝑠𝑜 =
87
180
= 0,4833
𝑃 ℎ𝑖𝑝𝑒𝑟𝑡𝑒𝑛𝑠𝑜 = 0,4833 𝑥 100 = 48,33%
La probabilidad de seleccionar al azar a uno de los individuos y sea
hipertenso es de 48,33%
29. SOLUCIÓN:
Experimente hipertensión
El total de los que experimente hipertensión, dado que es un
fumador empedernido según los datos específicos es de 30,
entonces
𝑃 hipertensión, fumador empedernido =
30
180
= 0,1667
𝑃(hipertensión, fumador empedernido) = 0,1667 𝑥 100 = 16,67%
La probabilidad de seleccionar al azar a uno de los individuos y experimente
hipertensión, dado que es un fumador empedernido es de 16,67%
30. SOLUCIÓN:
Experimente hipertensión
El total de los no fumador, dado que no ha presentado problemas de
hipertensión según los datos específicos es de 48, entonces
𝑃 no fumador, no problemas de hipertensión =
48
180
= 0,2667
𝑃(no fumador, no problemas de hipertensión) = 0,2667 𝑥 100 = 26,67%
La probabilidad de seleccionar al azar a uno de los individuos y sea no
fumador, dado que no ha presentado problemas de hipertensión es de
26,67%