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Binomial Distribution: Control Charts This problem will be referred to in the study of control charts (Section 6.1). In the binomial probability distribution, let the number of trials be \(n=3\), and let the probability of success be \(p=0.0228\). Use a calculator to compute (a) the probability of two successes. (b) the probability of three successes. (c) the probability of two or three successes.

Short Answer

Expert verified
(a) 0.00153, (b) 0.000012, (c) 0.001542.

Step by step solution

01

Understanding the Binomial Distribution

A binomial distribution models the number of successes in a fixed number of trials. Here, each trial results in success with a probability \( p = 0.0228 \) and there are \( n = 3 \) trials. We are tasked with finding the probability of achieving two successes and three successes.
02

Formula for Binomial Probability

The formula for the probability of getting exactly \( k \) successes in \( n \) independent Bernoulli trials is given by \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient \( \frac{n!}{k!(n-k)!} \).
03

Probability of Two Successes \( (a) \)

Substitute \( n = 3 \), \( k = 2 \), and \( p = 0.0228 \) in the formula: \[ P(X = 2) = \binom{3}{2} (0.0228)^2 (1 - 0.0228)^{3-2} \] \[ P(X = 2) = 3 \times 0.0228^2 \times 0.9772 = 0.00153 \].
04

Probability of Three Successes \( (b) \)

Substitute \( n = 3 \), \( k = 3 \), and \( p = 0.0228 \) in the formula: \[ P(X = 3) = \binom{3}{3} (0.0228)^3 (1 - 0.0228)^{3-3} \] \[ P(X = 3) = 1 \times 0.0228^3 \times 1 = 0.000012 \].
05

Probability of Two or Three Successes \( (c) \)

Add the probabilities from Steps 3 and 4 to find the probability of two or three successes: \[ P(X = 2 \text{ or } X = 3) = P(X = 2) + P(X = 3) = 0.00153 + 0.000012 = 0.001542 \]

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Key Concepts

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

Control Charts
Control charts are essential tools in statistical quality control. They are used to monitor processes to ensure they remain stable over time. By analyzing data from a process using control charts, we can determine if variations are within an acceptable range or if something unusual is happening. The concept is based on collecting data points over time and plotting them to see how they deviate from a central value, usually the average. Each point on a control chart represents a sample taken from the process, and typically, control limits define the boundaries within which the process should operate without issue. For processes that generate binomial data—like yes/no or pass/fail outcomes—using control charts involves analyzing the proportion of successes or failures over time. This analysis helps identify shifts or trends in the process performance which might require further investigation or corrective actions.
Control charts provide a visual way to quickly assess whether a process is in-control and stable, or if it requires intervention.
Probability of Success
In the context of binomial distribution, the probability of success refers to the likelihood that a single trial will result in a successful outcome. In our exercise, this probability is represented by the variable \( p \), which in this case is 0.0228. This means that each trial has a 2.28% chance of success.
Probability of success is crucial because it is used to determine the probabilities of different numbers of successes in multiple trials. This value is constant throughout all trials and is assumed to be the same every time a trial is conducted.
  • Fixed Probability: In a binomial distribution, the probability of success (\( p \)) must remain fixed within the series of trials.
  • Consistency: This probability represents the underlying parameter driving the random variable defined by the binomial distribution.
Understanding the probability of success helps us accurately calculate the likelihood of observing a specified number of successes across a series of trials in binomial experiments.
Bernoulli Trials
Bernoulli trials are the building blocks of the binomial distribution. Each trial represents a single event with two possible outcomes: success and failure. The characteristic feature of Bernoulli trials is that they are independent and identically distributed, meaning the outcome of one trial does not affect the others, and each trial has the same probability of success.
In the given problem, we perform three Bernoulli trials, each with a probability of success \( p = 0.0228 \). These trials are independent, as the result of one does not influence the others.
  • Independence: Each trial does not affect the others, ensuring the integrity of statistical analysis.
  • Probability Consistency: The reliability of statistical outcomes depends on the unchanging probability of success across trials.
Understanding Bernoulli trials is essential for effectively applying the binomial distribution formula, as they provide the framework for modeling the number of successes in a series of independent experiments.
Binomial Coefficient
The binomial coefficient is a crucial part of the binomial probability formula. It represents the number of ways to choose \( k \) successes from \( n \) independent trials and is denoted as \( \binom{n}{k} \). The mathematical calculation is expressed as \( \frac{n!}{k!(n-k)!} \), where \( ! \) stands for factorial, the product of all positive integers up to a given number.
In our example, to find the probability of exactly two successes in three trials, we calculate the binomial coefficient \( \binom{3}{2} = 3 \). Similarly, for three successes, \( \binom{3}{3} = 1 \).
  • Combination Role: The coefficient helps determine the number of combinations without considering the order of successes.
  • Foundation for Probability: It provides the essential count of success configurations needed to apply the formula accurately.
Mastering the binomial coefficient computation is vital for solving binomial distribution problems, as it dictates the number of success combinations possible out of a set number of trials.

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Most popular questions from this chapter

Criminal Justice: Parole USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 . (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) Quota Problem How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Airlines: Lost Bags USA Today reported that for all airlines, the number of lost bags was May: \(6.02\) per 1000 passengers December: \(12.78\) per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\). What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

Fishing: Lake Trout At Fontaine Lake Camp on Lake Athabasca in northern Canada, history shows that about \(30 \%\) of the guests catch lake trout over 20 pounds on a 4 -day fishing trip (Source: Athabasca Fishing Lodges, Saskatoon, Canada). Let \(n\) be a random variable that represents the first trip to Fontaine Lake Camp on which a guest catches a lake trout over 20 pounds. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) Find the probability that a guest catches a lake trout weighing at least 20 pounds for the first time on trip number \(3 .\) (c) Find the probability that it takes more than three trips for a guest to catch a lake trout weighing at least 20 pounds. (d) What is the expected number of fishing trips that must be taken to catch the first lake trout over 20 pounds? Hint: Use \(\mu\) for the geometric distribution and round.

Critical Tbinking Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about \(6.9\). Use the criterion that it is unusual to have data values more than \(2.5\) standard deviations above the mean or \(2.5\) standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b) Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c) Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.

Basic Computation: Expected Value For a fundraiser, 1000 raffle tickets are sold and the winner is chosen at random. There is only one prize, \(\$ 500\) in cash. You buy one ticket. (a) What is the probability you will win the prize of \(\$ 500\) ? (b) Your expected earnings can be found by multiplying the value of the prize by the probability you will win the prize. What are your expected earnings? (c) Interpretation If a ticket costs \(\$ 2\), what is the difference between your "costs" and "expected earnings"? How much are you effectively contributing to the fundraiser?

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