/*! 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 95 In Exercises \(\mathrm{P} .95\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. Count the number of sixes in 10 dice rolls.

Short Answer

Expert verified
The given process describes a binomial random variable. The number of trials \(n\) is 10 and the probability of success \(p\) is 1/6.

Step by step solution

01

Identify the Type of Variable

Examine the question and the conditions for a binomial random variable. In this case, the process of rolling a die 10 times and counting the number of sixes meets the conditions of a binomial random variable.
02

Find the trial count (n)

The number of trials, \(n\), refers to how many times the experiment is repeated. In this scenario, the die is rolled 10 times, so \(n = 10\).
03

Find the probability of success (p)

The probability of success, \(p\), is the likelihood of the desired outcome. A six has a 1-in-6 chance of being rolled on a die, this makes the probability of success \(p = 1/6\).

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.

Probability of Success
Understanding the probability of success is fundamental in calculating the binomial random variable. It represents the likelihood of achieving the desired outcome in a single trial. In the context of dice rolls, if the desired outcome is rolling a six, we consider the structure of a regular die. A standard die has six faces, each with an equal chance of landing face up. Thus, the probability of rolling a six is one out of six possible outcomes, or
\[\begin{equation} p = \frac{1}{6}. \end{equation}\]
The concept rests on the assumption that each roll of the die is independent, meaning the result of one roll does not affect the outcome of the next roll. It's essential to grasp that the probability of success remains constant throughout all trials in a binomial experiment. This consistent probability is what underpins the calculations for binomial distributions and sets the stage for determining the likelihood of various scenarios involving multiple trials.
Trial Count
The trial count, denoted by 'n,' is another pillar in the study of binomial random variables. This numeric value represents the total number of independent trials conducted in the experiment. In the scenario of rolling a die, each roll is considered a separate trial. When evaluating whether an activity qualifies as a binomial random variable, not only should the individual trials be independent, but the number of trials should also be fixed beforehand.

In the given exercise, the die is rolled 10 times, which sets our trial count as
\[\begin{equation} n = 10. \end{equation}\]
Having a predetermined trial count is crucial; it enables the calculation of the distribution of outcomes and probabilities that govern the statistical analysis of binomial experiments. This fixed trial count allows for a structured approach to predict various outcomes based on the defined probability of success.
Binomial Distribution Conditions

Defining Characteristics

Binomial distribution conditions are specific requirements that a process must meet to be considered a binomial random variable. The conditions include a fixed number of trials, binary outcomes (success or failure), identical probability of success for each trial, and independence between trials. In our dice-rolling example, these conditions are satisfied as follows:
  • There is a fixed number of trials: 10 dice rolls.
  • Each roll results in a success (rolling a six) or a failure (rolling any other number).
  • The probability of rolling a six is constant at \[\begin{equation} p = \frac{1}{6} \end{equation}\] across all trials.
  • Rolling a die multiple times does not influence the result of subsequent rolls, implying that the trials are independent.

Variability of Outcomes

Even with all conditions met, the variability of the outcomes in a binomial distribution is expected. In other words, while the overall pattern of outcomes over many repetitions will match the binomial distribution, individual series of 10 dice rolls may yield different numbers of successes. Students often struggle with understanding why they don't always get the 'expected' number of sixes (which would be the mean of the distribution), but it's important to comprehend that probability describes the likelihood over the long term and not precise predictions for each experiment. By grasping these conditions and the concept of variability, students can better interpret binomial distributions and utilize them in practical scenarios, enhancing their statistical literacy.

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

Mean and Standard Deviation of Class Year In Exercise P.117, we discuss the random variable counting the number of seniors in a sample of four undergraduate students at a university, given that the proportion of undergraduate students who are seniors is \(0.25 .\) Find the mean and standard deviation of this random variable.

Find endpoint(s) on the given normal density curve with the given property. (a) The area to the right of the endpoint on a \(N(50,4)\) curve is about 0.01 (b) The area to the left of the endpoint on a \(N(2,0.05)\) curve is about 0.70 . (c) The symmetric middle area on a \(N(100,20)\) curve is about 0.95 .

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 \(90 ?\)

The Standard and Poor 500 (S\&P 500 ) is a weighted average of the stocks for 500 large companies in the United States. It is commonly used as a measure of the overall performance of the US stock market. Between January 1,2009 and January \(1,2012,\) the S\&P 500 increased for 423 of the 756 days that the stock market was open. We will investigate whether changes to the S\&P 500 are independent from day to day. This is important, because if changes are not independent, we should be able to use the performance on the current day to help predict performance on the next day. (a) What is the probability that the S\&P 500 increased on a randomly selected market day between January 1,2009 and January \(1,2012 ?\) (b) If we assume that daily changes to the \(S \& P\) 500 are independent, what is the probability that the S\&P 500 increases for two consecutive days? What is the probability that the S\&P 500 increases on a day, given that it increased the day before? (c) Between January 1, 2009 and January 1,2012 the S\&P 500 increased on two consecutive market days 234 times out of a possible \(755 .\) Based on this information, what is the probability that the S\&P 500 increases for two consecutive days? What is the probability that the S\&P 500 increases on a day, given that it increased the day before? d) Compare your answers to part (b) and part (c). Do you think that this analysis proves that daily changes to the S\&P 500 are not independent?

During the \(2015-16\) NBA season, Stephen Curry of the Golden State Warriors had a free throw shooting percentage of 0.908 . Assume that the probability Stephen Curry makes any given free throw is fixed at 0.908 , and that free throws are independent. (a) If Stephen Curry shoots two free throws, what is the probability that he makes both of them? (b) If Stephen Curry shoots two free throws, what is the probability that he misses both of them? (c) If Stephen Curry shoots two free throws, what is the probability that he makes exactly one of them?

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.