Problem 22 - Basic Statistics for Business & Economics

Question

A manufacturer of window frames knows from long experience that \(5 \%\) of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames: a. None will need adjustment? b. At least one will need adjustment? c. More than two will need adjustment?

Step-by-Step Solution

Answer

a. 0.3585 b. 0.6415 c. 0.0771

Step 1: Define the Problem

We are dealing with a binomial distribution problem because each window frame either has a defect (success) or does not have a defect (failure). The probability of a window frame having a defect (requiring adjustment) is given as \( p = 0.05 \), and the probability of no defect is \( 1-p = 0.95 \). The sample size is \( n = 20 \). We are tasked with finding the probabilities for different scenarios of defects.

Step 2: Calculate Probability for Scenario (a)

For scenario (a), where none of the window frames need adjustment, we are looking for \( P(X = 0) \), where \( X \) is the random variable representing the number of defective frames in the sample. The binomial probability formula is:\[P(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k}\]Substituting the given values for \( k = 0 \), \( n = 20 \), \( p = 0.05 \):\[P(X = 0) = \binom{20}{0} \, (0.05)^0 \, (0.95)^{20}\]Simplifying:\[P(X = 0) = (0.95)^{20} = 0.3585\]

Step 3: Calculate Probability for Scenario (b)

For scenario (b), at least one frame needs adjustment, which is the complement of none needing adjustment. Thus, we have:\[P(X \geq 1) = 1 - P(X = 0)\]Using the result from Step 2:\[P(X \geq 1) = 1 - 0.3585 = 0.6415\]

Step 4: Calculate Probability for Scenario (c)

For scenario (c), more than two frames need adjustment, we need to find \( P(X > 2) \), which is:\[P(X > 2) = 1 - P(X \leq 2)\]Calculate \( P(X \leq 2) \) by summing the probabilities for \( X = 0 \), \( X = 1 \), and \( X = 2 \):\[\P(X = 1) = \binom{20}{1} \, (0.05)^1 \, (0.95)^{19} = 20 \, \times \, 0.05 \, \times \, (0.95)^{19} = 0.3774\\]\[\P(X = 2) = \binom{20}{2} \, (0.05)^2 \, (0.95)^{18} = 190 \, \times \, (0.05)^2 \, \times \, (0.95)^{18} = 0.1871 \]Sum these probabilities:\[P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\]\[P(X \leq 2) = 0.3585 + 0.3774 + 0.1871 = 0.9229\]Thus:\[P(X > 2) = 1 - 0.9229 = 0.0771\]

Key Concepts

Probability

Probability is a fundamental concept in statistics and mathematics that measures how likely an event is to occur. Understanding probability helps us predict and quantify uncertainty in various real-life situations.

Let's break down the basics:

• Probability values range from 0 to 1. A probability of 0 means the event will not occur, while a probability of 1 means the event will definitely occur.
• The probability of an event is often expressed as a fraction or a decimal. For example, a probability of 0.5 means there's a 50% chance the event will occur.
• Probabilities can also be expressed as percentages. Thus, a 0.05 probability can be read as a 5% chance.

In the context of the exercise, the probability that a single window frame is defective is 0.05, meaning there is a 5% chance. This basic understanding of probability is crucial, as it sets the foundation for working with more complex statistical models like the binomial distribution.

Defective Items

A defective item refers to a product that does not meet quality standards, requiring adjustments or repairs before it can be used or sold. Identifying defective items is essential for maintaining quality and customer satisfaction. In manufacturing processes, it's important to estimate the probability of items being defective to handle quality control effectively.

For example, in the case of window frames, identifying the rate at which frames are defective (5% in the problem) helps the manufacturer determine how often defects occur. Knowing the likelihood of defects enables the company to plan for repairs and minimize production losses. Managing defective items is crucial in ensuring overall product quality and operational efficiency.

Sample Size

Sample size refers to the number of observations or items selected from a population for analysis. It plays a crucial role in determining the reliability and accuracy of statistical conclusions. In probability and statistics, the sample size affects the certainty of the results.

Here's why sample size matters:

• Representativeness: A larger sample size is more likely to represent the entire population accurately. A small sample might not capture the diversity and variances within a population effectively.
• Statistical Confidence: With a larger sample, statisticians can be more confident that their results are not due to random chance.
• Reduced Margin of Error: Larger samples usually result in smaller margins of error, leading to more precise estimates.

In the window frame exercise, the sample size is 20. This means we're examining 20 window frames to determine how likely defects are to occur. Understanding the concept of sample size helps in appreciating the applicability and implications of statistical analysis.

Binomial Probability Formula

The binomial probability formula is a statistical method used to calculate the probability of observing a specific number of successes in a fixed number of independent experiments. Each experiment should have only two possible outcomes: success or failure.

The binomial probability formula is given by:\[P(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k}\]where:

• \(n\) is the total number of trials.
• \(k\) is the number of successful trials (e.g., number of defective items).
• \(p\) is the probability of a success on an individual trial.
• \(1-p\) is the probability of failure.
• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

In the given exercise, this formula helps in determining the probability of finding none, at least one, or more than two defective window frames in the sample of 20. By substituting the values corresponding to each scenario, we can find the specific probabilities, providing a powerful tool for quality control and forecasting in manufacturing.