/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 In Exercises \(\mathrm{P} .95\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 99

In Exercises \(\mathrm{P} .95\) to \(\mathrm{P} .99,\) determine whether the process describes a binomial random variable. If it is binomial, give values for \(n\) and \(p .\) If it is not binomial, state why not. Worldwide, the proportion of babies who are boys is about 0.51 . We randomly sample 100 babies born and count the number of boys.

Short Answer

Expert verified
This process describes a binomial random variable. The number of trials \(n\) is 100 and the probability of success \(p\) is 0.51.

Step by step solution

01

Determine the fixed number of trials

In the given scenario, 100 babies are randomly sampled. So, the number of trials is fixed and defined as \(n=100\).
02

Determine the success and failure

The scenario deals with determining the gender of babies which has only two possible outcomes - either the baby is a boy (success) or not (failure). Hence, the second condition of a binomial process is also satisfied.
03

Verify the independence of trials

Since the births of individuals are independent events, selecting one baby does not affect the probability of the gender of the next baby. Thus, the trials are independent.
04

Check uniform probability of success

The problem states that the global proportion of male babies is approximately 0.51. This probability remains the same for every birth, satisfying the fourth condition of a binomial process. Hence, the probability of success is defined as \(p=0.51\).
05

Final evaluation

Since all four conditions for a binomial process are met, we can conclude that the described scenario represents a binomial random variable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is a branch of mathematics that deals with the analysis and interpretation of random events. It involves predicting the likelihood of various outcomes based on known parameters. In the case of the exercise, probability theory is used to determine whether a sequence of events鈥攆or example, the birth of boys鈥攊s governed by a set of consistent probabilities.

When assessing a problem through the lens of probability theory, four primary conditions must be examined:
  • Fixed Number of Trials: This involves a specified number of attempts or occurrences, such as sampling 100 babies in our exercise.
  • Binary Outcomes: Each trial should have two possible outcomes. Here, a baby can either be a boy or a girl.
  • Independent Trials: The outcome of one trial should not influence the outcome of another, such as one baby's gender not affecting another's.
  • Constant Probability: The probability of each outcome remains unchanged throughout all trials. In this case, the probability of a baby being a boy is consistently 0.51.
Together, these conditions verify that the process described can be analyzed using a binomial distribution, a key tool within probability theory.
Random Variables
Random Variables are a fundamental concept within probability theory used to represent uncertain numerical outcomes. In the context of our exercise, the random variable is the number of boys born out of 100 babies. This variable can take any value from 0 to 100, depending on the results of the sampling process.

There are two primary types of random variables:
  • Discrete Random Variables: These have countable outcomes, such as the quantity of boys born out of 100 in our scenario.
  • Continuous Random Variables: These can take any value within a range, useful for measuring things like height or temperature.
Random variables stand at the heart of statistical modeling, allowing researchers to capture and assess the chance for different outcomes in a given experimental setting.
Statistical Analysis
Statistical analysis involves collecting, reviewing, and documenting data to identify trends or test hypotheses. In this context, it means using the principles of the binomial distribution to predict the distribution of gender in a random sample of newborns.

The binomial distribution is a key tool for modeling binary outcome scenarios. It helps in estimating probabilities and making predictions about populations or processes. Here are some characteristics of the binomial distribution:
  • The mean, or expected value, of a binomial distribution is given by the formula: \( \mu = n \times p \), where \( n \) is the number of trials and \( p \) is the probability of success.
  • The variance, measuring spread of the distribution, is \( \sigma^2 = n \times p \times (1-p) \).
  • The distribution is symmetrical when \( p = 0.5 \) and becomes skewed as \( p \) moves away from 0.5.
Using these tools, statisticians can make informed predictions about real-world processes, such as estimating the expected number of boys in a newborn sampling of 100, based on known probabilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color blindness is much more common in men than in women (this is because the genes corresponding to the red and green receptors are located on the X-chromosome). Approximately \(7 \%\) of American men and \(0.4 \%\) of American women are red-green color-blind. \(^{5}\) (a) If an American male is selected at random, what is the probability that he is red-green color-blind? (b) If an American female is selected at random, what is the probability that she is NOT redgreen color-blind? (c) If one man and one woman are selected at random, what is the probability that neither are redgreen color-blind? (d) If one man and one woman are selected at random, what is the probability that at least one of them is red-green color-blind?

As in Exercise \(\mathrm{P} .35,\) we have a bag of peanut \(\mathrm{M} \& \mathrm{M}\) 's with \(80 \mathrm{M} \& \mathrm{Ms}\) in it, and there are 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is yellow? (b) If we select one at random, what is the probability that it is not brown? (c) If we select one at random, what is the probability that it is blue or green? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are red? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is yellow and the second one is blue?

Curving Grades on an Exam A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean \(\mu=75\) and standard deviation \(\sigma=10 .\) Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 62 and the standard deviation was 18\. To be fair, he decides to "curve" the scores to match the desired \(N(75,10)\) distribution. To do this, he standardizes the actual scores to \(z\) -scores using the \(N(62,18)\) distribution and then "unstandardizes" those \(z\) -scores to shift to \(N(75,10)\). What is the new grade assigned for a student whose original score was 47 ? How about a student who originally scores a \(90 ?\)

Use the fact that we have independent events \(\mathrm{A}\) and \(\mathrm{B}\) with \(P(A)=0.7\) and \(P(B)=0.6\). Find \(P(A\) or \(B)\).

In a bag of peanut \(M\) \& M's, there are \(80 \mathrm{M} \& \mathrm{Ms}\), with 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is red? (b) If we select one at random, what is the probability that it is not blue? (c) If we select one at random, what is the probability that it is red or orange? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are blue? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is green?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.