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Suppose we have a binomial experiment with 50 trials, and the probability of success on a single trial is 0.02. Is it appropriate to use the Poisson distribution to approximate the probability of two successes? Explain.

Short Answer

Expert verified
Yes, it is appropriate to use the Poisson distribution since \(np = 1\) satisfies the conditions for approximation.

Step by step solution

01

Identify Parameters

We begin by identifying the parameters of the binomial distribution. The number of trials is denoted as \(n = 50\), and the probability of success on each trial is \(p = 0.02\).
02

Compute the Expected Number of Successes

To determine whether the Poisson approximation is appropriate, calculate the expected value (mean) of successes using the formula \(np\). Thus, \(np = 50 \times 0.02 = 1.0\).
03

Check Poisson Approximation Conditions

The Poisson distribution can be used as an approximation to the binomial distribution when \(n\) is large, \(p\) is small, and \(np\) is less than or equal to 5. Here, \(np = 1\), which satisfies these conditions.
04

Determine Appropriateness

Since the expected number of successes \(np\) is equal to 1 and satisfies the conditions for Poisson approximation, it is indeed appropriate to use the Poisson distribution to approximate the probability of observing two successes in this scenario.

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Key Concepts

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

Binomial Experiment
A binomial experiment is a statistical experiment that consists of a fixed number of repeated trials, each of which can result in just two possible outcomes: success or failure. These experiments are interesting because they help us understand probabilities in scenarios where there is a repetitive action, like flipping a coin or performing a quality check. Here are the core characteristics of a binomial experiment:
  • The number of trials, denoted as \(n\), is fixed beforehand. In our exercise, \(n = 50\) trials are considered.
  • Each trial is independent, meaning the outcome of one trial does not affect the others.
  • The probability of success, \(p\), is constant across all trials. For our case, \(p = 0.02\).
Understanding these characteristics is crucial since they allow mathematicians and statisticians to use the binomial distribution to calculate probabilities of certain outcomes, like the probability of achieving exactly two successes in our example.
Probability of Success
The probability of success in a binomial experiment refers to the likelihood that a single trial results in a success. In the context of our exercise, each trial has a success probability, \(p\), of 0.02. This represents a very low chance of success on any given trial.
However, the overall probability of getting a certain number of successes over many trials can be significantly different from the single-trial success probability. This is where binomial and Poisson distributions come in handy. Let's remember what is important about the probability of success:
  • It remains constant for each trial in the experiment.
  • It is independent of the outcomes of other trials.
  • It affects the distribution type used to approximate probabilities.
In cases like ours, where trials are numerous and success probability is low, using a Poisson approximation can be more efficient than calculating directly from the binomial distribution. It simplifies complex calculations while maintaining accuracy.
Poisson Approximation Conditions
The Poisson approximation is a helpful tool used to estimate binomial probabilities when certain conditions are met. Specifically, it offers a simpler way to compute probabilities when dealing with a large number of trials and a small probability of success. Let's dive into the conditions necessary to use a Poisson approximation:
  • Large Number of Trials (\(n\)): The trials should be numerous. Although there's no strict rule, having \(n\) in the tens or higher is generally needed.
  • Small Probability of Success (\(p\)): The success probability should be low, typically \(p \leq 0.1\). In our exercise, \(p = 0.02\), which is small enough.
  • Small Expected Number of Successes (\(np\)): The mean, which is calculated as \(np\), should be relatively small, typically \( pn \leq 5 \). Our exercise yielded \(np = 1\), which fits this criterion.
Meeting these conditions ensures that the Poisson distribution is a reasonable approximation for our binomial experiment. Using it simplifies the calculations while providing fairly accurate probability estimates, especially helpful when direct computation using the binomial distribution is cumbersome.

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Most popular questions from this chapter

Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about \(57 \%\) of all people who take the state bar exam pass (Source: The Book of Odds by Shook and Shook, Signet). Let \(n=1,2,3, \ldots\) represent the number of times a person takes the bar exam until the first pass. (a) Write out a formula for the probability distribution of the random variable \(n .\) (b) What is the probability that Bob first passes the bar exam on the second \(\operatorname{try}(n=2) ?\) (c) What is the probability that Bob needs three attempts to pass the bar exam? (d) What is the probability that Bob needs more than three attempts to pass the bar exam? (e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

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Suppose you are a hospital manager and have been told that there is no need to worry that respirator monitoring equipment might fail because the probability any one monitor will fail is only \(0.01 .\) The hospital has 20 such monitors and they work independently. Should you be more concerned about the probability that exactly one of the 20 monitors fails, or that at least one fails? Explain.

Old Friends Information Service is a California company that is in the business of finding addresses of long-lost friends. Old Friends claims to have a \(70 \%\) success rate (Source: The Wall Street Journal). Suppose that you have the names of six friends for whom you have no addresses and decide to use Old Friends to track them. (a) Make a histogram showing the probability of \(r=0\) to 6 friends for whom an address will be found. (b) Find the mean and standard deviation of this probability distribution. What is the expected number of friends for whom addresses will be found? (c) Quota Problem How many names would you have to submit to be \(97 \%\) sure that at least two addresses will be found?

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