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When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

Short Answer

Expert verified
a. Probability is approximately 0.00013. b. Probability is approximately 0.104.

Step by step solution

01

Define the Problem

We want to find the probability that at least 10% of the 250 boards are defective. Additionally, we also need the probability of exactly 10 defectives. The defect rate is 5%, and the total number of boards is 250.
02

Understand the Approximation

For large samples, the binomial distribution can be approximated using a normal distribution if certain conditions are met. These conditions are: the sample size is large enough, i.e., both np and n(1-p) are greater than 5. Here, n = 250 and p = 0.05, so np = 12.5 and n(1-p) = 237.5.
03

Normal Approximation to Binomial

Since both conditions from Step 2 are satisfied, we can use the normal distribution to approximate the binomial distribution. Calculate the mean and the standard deviation for the distribution of defectives: Mean \(\mu = np = 12.5\) Standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{12.5 \times 0.95} \approx 3.44\).
04

Find Probability for Part (a)

To find the probability that at least 10% of the 250 boards are defective, compute the probability that there are at least 25 defectives using the normal distribution. Convert the problem to the z-score: \(P(X \geq 25)\). The z-score is \(z = \frac{25 - 12.5}{3.44} \approx 3.65\). Using a standard normal distribution table or calculator, \(P(Z \geq 3.65) \approx 0.00013\).
05

Calculate for Part (b) Using Binomial Formula

Here, we need the probability of exactly 10 defectives. Use the binomial probability formula: \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\), where \(k = 10\). Thus, \(P(X = 10) = \binom{250}{10}(0.05)^{10}(0.95)^{240}\) Calculate this to find the approximate probability.
06

Calculation for Part (b)

Using the binomial coefficient and probability calculations: \(\binom{250}{10} \approx 4.292 \times 10^{11}\), \((0.05)^{10} = 9.7656 \times 10^{-14}\), and \((0.95)^{240} \approx 0.00582\). Thus, \(P(X = 10) \approx 4.292 \times 10^{11} \times 9.7656 \times 10^{-14} \times 0.00582 \approx 0.104\).

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Key Concepts

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

Binomial Distribution
Let's start by understanding the binomial distribution, which is essential when dealing with problems involving two outcomes, such as defective versus non-defective. In our exercise, each circuit board can either be defective or not, making the binomial distribution a suitable choice. This distribution represents the probability of achieving a given number of "successes" (defective boards in this case) in a fixed number of independent trials, where each trial has the same probability of success.
  • The number of trials: 250 circuit boards
  • Probability of success (defective board): 5% or 0.05
  • Random variable: Number of defective boards in the batch
In such cases, the binomial distribution requires knowing the probability of success and the total number of trials to calculate probabilities for various outcomes, which we have done step-by-step for questions about these boards.
Probability Calculation
When working with probabilities in a binomial context, it often involves calculating the likelihood of certain outcomes. Let's break it down using our example. For the probability calculation, you'll typically use either the binomial formula directly or approximate it with a normal distribution if certain conditions apply. Here's how each works:
  • **Direct Binomial Calculation:** Useful for small sample sizes or exact outcomes. This involves using the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.
  • **Normal Approximation:** For large sample sizes, it simplifies calculations. By checking conditions \(np \text{ and } n(1-p) > 5\), this exercise shows us how to calculate approximate probabilities with the normal distribution. We do this by calculating the mean \(\mu = np\) and standard deviation \(\sigma = \sqrt{np(1-p)}\), and then converting to a z-score to find probabilities.
This approach makes it manageable to calculate the probability that at least a certain percentage of boards are defective.
Defective Rate Analysis
In manufacturing, defective rate analysis is crucial for maintaining quality control. This analysis helps determine how frequently defective items occur in a batch, which directly impacts production standards. In our example, performing such an analysis involves calculating probabilities to understand potential defects in a shipment. Here’s why this is essential:
  • **Quality Assurance:** Knowing that 5% of boards are defective can help set benchmarks and mitigation strategies. It isn’t just about detecting faults but also about improving processes to minimize them.
  • **Risk Management:** By predicting defects, companies can manage risks better, ensuring supply chain reliability and customer satisfaction.
  • **Cost Management:** Reducing defects can significantly lower waste costs, enhancing overall profitability.
Each probability we calculate provides a data-driven perspective on defect management, facilitating informed, strategic decisions in manufacturing settings.

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