/*! 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 72 In Exercises 69 to \(72,\) expla... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 69 to \(72,\) explain whether the given random variable has a binomial distribution. Lefties Exactly 10\(\%\) of the students in a school are left-handed. Select 15 students at random from the school and define \(W=\) the number who are left handed.

Short Answer

Expert verified
Yes, the random variable has a binomial distribution.

Step by step solution

01

Definition of Binomial Distribution

A binomial distribution arises when there are a fixed number of independent trials, each resulting in a success or failure, with a constant probability of success. Mathematically, a random variable follows a binomial distribution when it satisfies these criteria.
02

Determining the Number of Trials

In this problem, 15 students are selected at random from the school. The number of trials ( ) is thus 15, which is a fixed number.
03

Identifying Success and Failure

We define the random variable \(W\) as the number of left-handed students. In this scenario, a 'success' is a student being left-handed, and a failure is being right-handed.
04

Checking for Constant Probability of Success

The problem states that 10% of the students are left-handed, so the probability of success (\(p\)) in each trial is 0.10. This probability remains consistent for each student chosen.
05

Determining Independence of Trials

Assume that the trials are independent, meaning the outcome for one student (being left-handed or not) does not affect another student's outcome. This is plausible under the assumption that the sample of 15 students is a small fraction compared to the total student body.
06

Conclusion on the Distribution

Since the number of trials is fixed, there are only two possible outcomes (left-handed or not), the probability of success is constant at 0.10, and trials are assumed to be independent, the random variable \(W\) follows a binomial distribution.

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.

Random Variable
In the context of the binomial distribution, a random variable is an essential concept. A random variable is a number that represents the outcome of a random phenomenon. In this exercise, we define the random variable \(W\) as the number of left-handed students selected from a group of 15. This means \(W\) can take on any integer value from 0 to 15, depending on how many of the 15 students happen to be left-handed. Each possible outcome where a student is either left or right-handed contributes to determining the final value of \(W\). Understanding this helps clarify how random variables are used to describe probabilities of real-world scenarios in a quantifiable way. Remember, the focus with random variables in a binomial context is binary outcomes: success or failure.
Probability of Success
In a binomial distribution, the 'probability of success' refers to the likelihood of achieving the desired outcome in each trial. For our scenario of selecting students, a success is defined as a student being left-handed. According to the problem, 10% of students are left-handed, so the probability of success, denoted as \(p\), is 0.10 in each trial. This probability remains the same for every student picked. These consistent chances make calculating binomial probabilities straightforward, as you apply the same success rate across all trials of your experiment or observation. Thus, knowing \(p = 0.10\) allows us to predict and analyze the number of left-handed students among those selected.
Independent Trials
Independent trials are a critical requirement for a binomial distribution. It means each trial or selection of a student in this case does not affect the others. One student's handedness doesn't impact another's in any way, ensuring the integrity of each trial. This assumption holds especially when the sample size is relatively small compared to the total population, mitigating any potential dependence. If the outcomes were dependent, the scenario wouldn't fit a binomial distribution model. Assuming independence simplifies our calculations, allowing for a straightforward application of binomial formulas. By adhering to this principle, we accurately model scenarios using the binomial distribution method.
Success and Failure
In binomial distributions, outcomes are often categorized simply as 'success' or 'failure'. For this exercise, 'success' is defined as the selection of a left-handed student and 'failure' is the selection of a right-handed student. These designations are key because the binomial distribution framework relies on only two possible outcomes for each trial. This binary outcome results in a straightforward probability calculation for various scenarios. To identify whether a distribution is indeed binomial, ensure that your outcomes can be neatly divided into two categories for every trial, such as left-handed or not. This simplicity allows for using the probability models and calculations associated with binomial distributions.

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

How many close friends do you have? An opinion poll asks this question of an SRS of 1100 adults. Suppose that the number of close friends adults claim to have varies from person to person with mean \(\mu=9\) and standard deviation \(\sigma=2.5\) We will see later that in repeated random samples of size 1100 , the mean response \(\overline{x}\) will vary according to the Normal distribution with mean 9 and standard deviation 0.075 . What is \(P(8.9 \leq \overline{x} \leq 9.1)\) , the probability that the sample result \(\overline{x}\) estimates the population truth \(\mu=9\) to within \(\pm 0.1 ?\)

In Exercises 69 to \(72,\) explain whether the given random variable has a binomial distribution. Long or short? Put the names of all the students in your class in a hat. Mix them up, and draw four names without looking. Let \(Y=\) the number whose last names have more than six letters.

Too cool at the cabin? During the winter months, the temperatures at the Starneses' Colorado cabin can stay well below freezing \(\left(32^{\circ} \mathrm{F} \text { or } 0^{\circ} \mathrm{C}\right)\) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at \(50^{\circ} \mathrm{F}\) . She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years' worth of data, the temperature \(T\) in the cabin at midnight on a randomly selected night follows a Normal distribution with mean \(8.5^{\circ} \mathrm{C}\) and standard deviation \(2.25^{\circ} \mathrm{C} .\) (a) Let \(Y=\) the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that \(\mathrm{F}=(9 / 5) \mathrm{C}+32 ) .\) Find the mean and standard deviation of \(\mathrm{Y}\) . (b) Find the probability that the midnight temperature in the cabin is below \(40^{\circ} \mathrm{F}\) . Show your work.

Checking independence In which of the following games of chance would you be willing to assume independence of \(X\) and \(Y\) in making a probability model? Explain your answer in each case. (a) In blackjack, you are dealt two cards and examine the total points \(X\) on the cards (face cards count 10 points). You can choose to be dealt another card and compete based on the total points \(Y\) on all three cards. (b) In craps, the betting is based on successive rolls of two dice. \(X\) is the sum of the faces on the first roll, and \(Y\) is the sum of the faces on the next roll.

Scrabble In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowels, 56 consonants, and 2 blank tiles in the bag. Cait chooses her 7 tiles and is surprised to discover that all of them are vowels. Can we use a binomial distribution to approximate this probability? Justify your answer.

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.