91Ó°ÊÓ

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the cdf of \(X\). First determine \(F(0)=P(X \leq 0), F(1)\), and \(F(2)\), and then obtain \(F(x)\) for all other \(x\).

Step by step solution

01

List All Possible Outcomes

Consider a lot of five boards labeled as 1, 2, 3, 4, and 5. We need to choose any two boards for inspection. This is a combination problem where we choose 2 out of 5, calculated as \( \binom{5}{2} = 10 \) combinations. The possible combinations or outcomes can be listed as: \((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\).

02

Probability Distribution of X

Define \( X \) as the number of defective boards selected. Given that only boards 1 and 2 are defective, calculate the probability distribution for \( X \): - \( X = 0 \): No defective boards selected. Outcomes: \((3, 4), (3, 5), (4, 5)\). Probability: \( P(X = 0) = \frac{3}{10} \).- \( X = 1 \): One defective board selected. Outcomes: \((1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\). Probability: \( P(X = 1) = \frac{6}{10} \).- \( X = 2 \): Both defective boards selected. Outcome: \((1, 2)\). Probability: \( P(X = 2) = \frac{1}{10} \).

03

Calculate Cumulative Distribution Function (CDF)

To obtain the cumulative distribution function \( F(x) \):- \( F(0) = P(X \leq 0) = P(X = 0) = \frac{3}{10} \).- \( F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{3}{10} + \frac{6}{10} = \frac{9}{10} \).- \( F(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{3}{10} + \frac{6}{10} + \frac{1}{10} = 1 \).Thus, \( F(x) \) is defined as: - For \( x < 0 \), \( F(x) = 0 \). - For \( 0 \leq x < 1 \), \( F(x) = \frac{3}{10} \). - For \( 1 \leq x < 2 \), \( F(x) = \frac{9}{10} \). - For \( x \geq 2 \), \( F(x) = 1 \).

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Key Concepts

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

Quality Control

Quality control is an essential process in manufacturing that ensures products meet certain standards of quality. It's like a checkpoint that catches any defects before the product reaches the customer. In the exercise, quality control is illustrated through the inspection of computer boards, where a sample from a batch is selected for testing. This process helps identify defective products among a large group.

• **Purpose:** The main goal is to minimize defects and improve product reliability.
• **Method:** By selecting two boards out of five, the manufacturer inspects them for defects.
• **Outcome:** Two defective boards (1 and 2) in the batch are discovered during inspections.

This helps in making informed decisions about accepting or rejecting a batch, reducing the likelihood of defective products reaching the customers.

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function, or CDF, of a random variable provides valuable insight into the probability that the variable takes a value less than or equal to a given level. In other words, it tells us about the probability accumulated up to a specific point.
Imagine stacking probabilities like building blocks; the CDF shows how high your stack has gotten at any point. In our context:

• **Understanding the CDF:** For calculating the probabilities, the CDF helps us see the likelihood of selecting a certain number or fewer defective boards.
• **Example Calculations:** For example, to find \( F(0) \), we calculate \( P(X \leq 0) \), which means no defective boards, gives us a probability of \( \frac{3}{10} \).
• **Total Probability:** As we progress to \( F(1) \), we add up all probabilities up to having one defective board, equaling \( \frac{9}{10} \), and complete with \( F(2) = 1 \), signifying all possible outcomes.

It's a comprehensive way to visualize and handle probabilities, ensuring we understand the entire distribution of outcomes.

Combinatorics

Combinatorics is a branch of mathematics centered around counting, arranging, and finding patterns. It plays a crucial role in solving problems where the goal is to determine the number of ways something can happen.
In our exercise, combinatorics is used for listing all possible outcomes of selecting boards. This is achieved using a principle known as combinations. A combination focuses on selecting items from a set such that the order of selection doesn't matter.

• **Combination Formula:** To find how many ways we can choose 2 boards from 5 (without regard to order), we use the formula \( \binom{5}{2} = 10 \).
• **Practical Application:** Listing out these combinations helps in detailing potential inspection outcomes like \((1,2), (1,3), (4,5)\), etc.
• **Relevance:** It's essential for calculating probabilities related to quality control and other situations where selection matters.

This mathematical tool is indispensable for understanding how different outcomes can arise from a set of possibilities.