/*! 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 30 With and without replacement. In... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

With and without replacement. In the following situations assume that half of the specified population is male and the other half is female. (a) Suppose you're sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you're sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your findings from parts (a) and (b), explain whether or not this assumption is reasonable.

Short Answer

Expert verified
(a) 0.25 with; 2/9 without. (b) 0.25 for both. (c) Independence is reasonable in large populations.

Step by step solution

01

With Replacement - Small Population

In a room with 10 people, the probability of choosing a female is 0.5 (since half are female). When sampling with replacement, the probability remains the same for each draw. Therefore, the probability of drawing two females in a row = \(0.5 \times 0.5 = 0.25\).
02

Without Replacement - Small Population

For a small population with 10 people (5 females), after one female is chosen, only 9 people remain with 4 females. Thus, the probability of drawing two females without replacement is \(\frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}\).
03

With Replacement - Large Population

In a stadium with 10,000 people, the probability of choosing a female remains 0.5 for each draw with replacement. Thus, the probability of drawing two females in a row = \(0.5 \times 0.5 = 0.25\).
04

Without Replacement - Large Population

In the large 10,000-person stadium with 5,000 females, after choosing one female, 9,999 people remain. The probability is \(\frac{5000}{10000} \times \frac{4999}{9999} \approx 0.25\).
05

Evaluating Independence Assumption

In small populations, such as with 10 people, the probabilities differ notably between with and without replacement. However, in large populations like the stadium, the probabilities show minimal difference, which supports treating the events as independent.

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 91Ó°ÊÓ!

Key Concepts

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

Sampling methods
When it comes to understanding probability, it's crucial to grasp the concept of sampling methods. Sampling is the process of selecting a subset of individuals from a population to estimate characteristics or behaviors. There are two primary types of sampling methods:
  • With replacement: In this method, every time an individual is selected, they are placed back into the population before the next draw. This means they could be selected again, and the population size remains constant for each draw.
  • Without replacement: Once an individual is selected, they are not placed back into the population. This reduces the population size and changes the selection probabilities for subsequent draws.
For the exercise in the example, we looked at both methods to determine the probability of selecting females from populations of differing sizes. Properly understanding which method you're using is vital, as it affects the outcomes of probability calculations.
Independent events
An independent event is one where the outcome of one event does not affect the outcome of another. In probability, this means knowing one event's result gives no information about another event. Sampling with replacement often allows events to remain independent.
In our exercise scenario with a stadium of 10,000 individuals, even when sampling without replacement, the probability is approximately maintained (0.25 for drawing two females). This is because the adjustment in the probability is negligible due to the large number—making it reasonable to assume independence.
However, in smaller populations, like a room of only 10 people, without replacement leads to dependent events. The probability changes significantly because each draw affects the next, which we saw when the probability of selecting two females altered noticeably to \(\frac{2}{9}\). Whether events are independent or dependent can drastically change your probability computations.
Probability theory
At the heart of this exercise is probability theory, a fundamental concept that helps us understand the likelihood of events. Probability ranges between 0 and 1, where 0 means an event is impossible, and 1 indicates certainty. It tells us how likely something is to happen and is crucial in fields like statistics, mathematics, and data science.
  • Formula: For calculating the probability of multiple events, especially when they are independent, we use the multiplication rule. The probability of sequential independent events happening is the product of their individual probabilities.
  • Example: In our exercise, when sampling females with replacement, each draw remains at a probability of 0.5. Thus, multiplying to find the chance of two consecutive female draws: \(0.5 \times 0.5 = 0.25\).
Understanding these rules helps in calculating probabilities correctly, as we clearly see in the adjustments needed when sampling without replacement.
Population size
Population size can drastically impact probability calculations and assumptions in sampling exercises. It refers to the number of individuals within a given group from which you are sampling. In probability, large populations and small populations may lead to different treatment and strategy results.
In large populations, like the stadium example with 10,000 people, the effects of sampling without replacement are minimal because removing a single person from such a large group hardly changes the probability for subsequent selections. This allows for treating each draw as if it were independent even when it isn’t entirely true.
On the contrary, in a smaller population, like the one with 10 people, removing one significantly alters the remaining group's composition. Thus, the probability calculations in the case of "without replacement" are quite different in smaller groups, as evident with the notable dip from 0.25 to \(\frac{2}{9}\) for selecting two females in sequence. Recognizing how population size interacts with your probability model is critical for making accurate assessments.

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

LA weather, Part I. 1 ' The average daily high temperature in June in LA is \(77^{\circ} \mathrm{F}\) with a standard deviation of \(5^{\circ} \mathrm{F}\), Suppose that the temperatures in June closely follow a normal distribution. (a) What is the probability of observing an \(83^{\circ} \mathrm{F}\) temperature or higher in LA during a randomly chosen day in June? (b) How cool are the coldest 10\% of the days (days with lowest average high temperature) during June in LA?

GRE scores, Part III. \(151, \sigma=7\) ) for the verbal part of the exam and \(N(\mu=153, \sigma=7.67)\) for the quantitative part. Suppose performance on these two sections is independent. Use this information to compute each of the following: (a) The probability of a combined (verbal + quantitative) score above 320 . (b) The score of a student who scored better than \(90 \%\) of the test takers overall.

Nearsighted children. \({ }^{90} \mathrm{~B}^{2}\) It is believed that nearsightedness affects about \(8 \%\) of all children. We are interested in finding the probability that fewer than 12 out of 200 randomly sampled children will be nearsighted. (a) Estimate this probability using the normal approximation to the binomial distribution. (b) Estimate this probability using the distribution of the sample proportion. (c) How do your answers from parts (a) and (b) compare?

Bernoulli, the mean. Use the probability rules from Section 3.5 to derive the mean of a Bernoulli random variable, i.e. a random variable \(X\) that takes value 1 with probability \(p\) and value 0 with probability \(1-p .\) That is, compute the expected value of a generic Bernoulli random variable.

Area under the curve, Part I. What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph. (a) \(Z<-1.35\) (b) \(Z>1.48\) (c) \(-0.42\)

See all solutions

Recommended explanations on Math Textbooks

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.