A binomial random variable X is defined as the number of successes achieved in the n trials of a Bernoulli process. Describe an event in your life that fits the properties of a Bernoulli process, being sure to explain how each property is met by your event. Finally, state the number of trials and the number of successes for your event. Be specific.
Respond with thoughtful substantive responses to at least two classmates postings.
Save your time - order a paper!
Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlinesOrder Paper Now
I mean the most relateable is taking the mastery exercises for class. My personal goal is to get at least an 80% on each quiz. If there are 15 questions per quiz then I would have to get 12 out of 15 problems correct. If each question has 5 choices to choose from, if I would guess the correct answer to each question the equation would be as such:
n= 15 questions total p= probability of success = 1/5 = .2 q= probability of failure = 1-.2 = .8 r= 12 successful results
So long story short, if you want to get at least an 80% on your quiz don’t even try to guess.
According to Anderson, Sweeney & Williams (2016) the Bernoulli process consists of 4 properties (FICT):
1. Fixed amount of experiments
2. Independence of each experiment
3. Consistency of likelihoods
4. Two likely outcomes
The only instance I can easily relate this process to in my life is determining the probability of having a third child and that child being a girl. The fixed amount of experiments or trials is equal to 8 because there are only 8 (n) possible combinations of 3 sets of boys and girls. Each experiment or trial is independent because the probability of success will not change since the likelihood of boy or girl is constant, which goes for the final property as well.
Lastly, the number of trials for this instance would be 8 since there are only 8 gender pairings possible for a set of 3 children. Also, the probability of success would be 3/8 = .375 according to the following chart:
|1st Child||2nd Child||3rd Child||# of Girls|
|# of Girls(x)||P(x)|
If anyone would like to see the formulas for this let me know and I’ll be happy to attach the spreadsheet.
Anderson, D., Sweeney, D., & Williams, T. (2016). Essentials of Modern Business Statistics with Microsoft® Excel®, 6th Edition. Boston, MA: Cengage Learning.