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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) Two successes: use the binomial probability formula to get approximately 0.0015. (b) Three successes: approximately 0.0000118. (c) Two or three successes: approximately 0.0015 + 0.0000118.

Step by step solution

01

Understand the Binomial Probability Formula

The binomial probability formula is used to determine the probability of obtaining exactly \(k\) successes in \(n\) independent trials, where each trial has a success probability \(p\). It is given by \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(\binom{n}{k}\) is the binomial coefficient.
02

Calculate for Two Successes

To find the probability of two successes \((k=2)\), substitute \(n=3\), \(k=2\), and \(p=0.0228\) into the formula: \[P(X=2) = \binom{3}{2} (0.0228)^2 (1-0.0228)^{1} = 3 \times 0.00051984 \times 0.9772\]. Use a calculator to compute this value.
03

Calculate for Three Successes

For three successes \((k=3)\), substitute \(n=3\), \(k=3\), and \(p=0.0228\) into the formula: \[P(X=3) = \binom{3}{3} (0.0228)^3 (1-0.0228)^{0} = 1 \times 0.00001183 \times 1\]. Use a calculator to get the result.
04

Compute Probability of Two or Three Successes

The probability of having two or three successes is the sum of the probabilities found in steps 2 and 3: \[P(X=2 \text{ or } X=3) = P(X=2) + P(X=3)\]. Add the values computed from the previous steps using a calculator.

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Key Concepts

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

Control Charts and Their Role
Control charts play a vital role in monitoring processes and ensuring they remain within set parameters. They are statistical tools used in quality control to track data over time, identify patterns, and detect inconsistencies. In manufacturing or production, control charts help ascertain whether a process is stable and predictable.

There are different types of control charts, such as X-bar charts for monitoring the mean of a process, and R charts for monitoring its variability. By understanding the expected distribution of outcomes, control charts can help distinguish between common cause variation (inherent in the process) and special cause variation (due to external factors). This is where concepts of probability and distributions, like the binomial distribution, come into play.

When dealing with binomial distributions, control charts often help visualize whether the frequency of successes in trials lies within acceptable limits, allowing managers to make data-driven decisions.
Understanding the Binomial Probability Formula
The binomial probability formula is fundamental when dealing with scenarios that include multiple trials with two possible outcomes, success or failure. The formula calculates the probability of exactly a certain number of successes in a fixed number of independent trials.

Given as:
  • \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
This equation utilizes:
  • The binomial coefficient \(\binom{n}{k}\): It represents the different ways to choose \(k\) successes out of \(n\) trials.
  • \(p^k\): The probability of achieving \(k\) successes.
  • \((1-p)^{n-k}\): The probability of having the remaining \(n-k\) trials as failures.
By substituting the values into the formula, you can calculate the probability of specific outcomes in a variety of real-world situations, like quality control in manufacturing.
Probability of Success in Trials
In a binomial setting, the probability of success, denoted by \(p\), is a crucial parameter. It defines how likely it is for each individual trial to result in success. This probability remains constant across all trials, ensuring that each trial is independent and identically distributed.

For example, if a company has a 2.28% success rate for a specific product feature, then \(p = 0.0228\). In this scenario, every trial or test of the feature has the same probability of success, maintaining consistency in expectations over repeated trials.

The probability of success directly influences outcomes when using the binomial probability formula. A low probability of success might seem discouraging in isolation, but it provides a realistic picture of the scenario when evaluating multiple trials.
Explaining the Binomial Coefficient
The binomial coefficient, represented as \(\binom{n}{k}\), is integral to the binomial probability formula as it determines the number of ways to pick \(k\) successes from \(n\) trials. Often termed 'combinations', the binomial coefficient calculates how groups can be formed without considering the order of successes.

Mathematically, it is defined as:
  • \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
Where \(!\) denotes the factorial operation, multiplying a number by all positive integers less than it.

For instance, in a situation with 3 trials (\(n=3\)) and wanting to find the probability of exactly 2 successes (\(k=2\)), the binomial coefficient will tell us the number of such groupings possible. Utilizing this helps efficiently solve probability problems, a vital skill in fields like statistics and quality control.

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Most popular questions from this chapter

: Syringes The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose the batch contains \(1 \%\) defective syringes. (a) Make a histogram showing the probabilities of \(r=0,1,2,3,4,5,6,7\), and 8 defective syringes in a random sample of eight syringes. (b) Find \(\mu .\) What is the expected number of defective syringes the inspector will find? (c) What is the probability that the batch will be accepted? (d) Find \(\sigma\).

For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

The following is based on information taken from The Wolf in the Southwest: The Making of an Endangered Species, edited by David Brown (University of Arizona Press). Before 1918 , approximately \(55 \%\) of the wolves in the New Mexico and Arizona region were male, and \(45 \%\) were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately \(70 \%\) of wolves in the region are male, and \(30 \%\) are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (a) Before 1918 , in a random sample of 12 wolves spotted in the region, what is the probability that 6 or more were male? What is the probability that 6 on more were female? What is the probability that fewer than 4 were female? (b) Answer part (a) for the period from 1918 to the present.

According to the college registrar's office, \(40 \%\) of students enrolled in an introductory statistics class this semester are freshmen, \(25 \%\) are sophomores, \(15 \%\) are juniors, and \(20 \%\) are seniors. You want to determine the probability that in a random sample of five students enrolled in introductory statistics this semester, exactly two are freshmen. (a) Describe a trial. Can we model a trial as having only two outcomes? If so, what is success? What is failure? What is the probability of success? (b) We are sampling without replacement. If only 30 students are enrolled in introductory statistics this semester, is it appropriate to model 5 trials as independent, with the same probability of success on each trial? Explain. What other probability distribution would be more appropriate in this setting?

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in Western Civilization each term. Let \(n=1,2,3, \ldots\) represent the number of times a student takes Western Civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass Western Civilization? (e) What is the expected number of attempts at Western Civilization Susan must make to have her (first) pass? Hint Use \(\mu\) for the geometric distribution and round.

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