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Chapter 3: Problem 182

A company performs inspection on shipments from suppliers in order to defect nonconforming products. Assume a lot contains 1000 items and \(1 \%\) are nonconforming. What sample size is needed so that the probability of choosing at least one nonconforming item in the sample is at least \(0.90 ?\) Assume the binomial approximation to the hypergeometric distribution is adequate.

Short Answer

Expert verified
The sample size needed is 230.

Step by step solution

01

Understand the Problem

We want to find out the number of items needed in a sample such that there's at least a 90% chance of finding one nonconforming item when 1% of the 1000-item lot are nonconforming.
02

Define the Probability

The probability of picking a nonconforming item is 0.01 per item. We want the probability of choosing at least one nonconforming item to be at least 0.90.
03

Set Up the Complementary Probability

The complementary probability (not picking any nonconforming items in the sample) should be at most 0.10. This is expressed as \( P(X = 0) \leq 0.10 \), where \( X \) is the number of nonconforming items in the sample.
04

Use the Binomial Probability Formula

Using the binomial formula, we calculate \( P(X = 0) = (1 - 0.01)^n \leq 0.10 \), where \( n \) is the sample size.
05

Solve the Inequality

Solve the inequality \( (0.99)^n \leq 0.10 \) to find the minimum sample size. Taking the natural logarithm of both sides gives: \( n \cdot \ln(0.99) \leq \ln(0.10) \).
06

Calculate the Required Sample Size

Rearranging the inequality, we get \( n \geq \frac{\ln(0.10)}{\ln(0.99)} \). Calculate the values: \( \ln(0.10) \approx -2.302 \) and \( \ln(0.99) \approx -0.01005 \). Thus, \( n \geq \frac{-2.302}{-0.01005} \approx 229.15 \).
07

Determine the Minimum Whole Number

Since we can't sample a fraction of an item, we round up to the nearest whole number. Therefore, the minimum sample size required is 230.

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Key Concepts

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

Sample Size Determination
To ensure that a sample has at a minimum a 90% probability of including at least one nonconforming item from a lot, determining the correct sample size is essential. This process relies on understanding the balance between the proportion of nonconforming items in the lot and the desired confidence level.
In this case, the lot has a size of 1000 items with a nonconforming rate of 1%. The goal is to select a sample size such that the probability of picking at least one nonconforming item is at least 90%. To find this sample size, we use statistical techniques such as inequalities and logarithms.
By setting up an inequality based on the binomial formula, we solve for the sample size that meets our probability requirement. Rounding up to the nearest whole number ensures practicality since partial items cannot be sampled. For this scenario, calculating leads to a sample size of 230 items.
Nonconforming Items
Nonconformity in items refers to products that do not meet specified quality standards. When inspecting shipments, identifying nonconforming items is crucial for maintaining product quality and customer satisfaction.
In the given exercise, out of a 1000-item lot, 1% of the items are expected to be nonconforming. These defective items can affect the overall reputation and performance of the company, highlighting the importance of rigorous quality control.
The presence of nonconforming items necessitates sampling and testing procedures. The detection of even a small number of these items can trigger corrective actions to address the underlying issues causing the defects.
Probability Calculation
In statistical quality control, calculating probabilities helps in making informed decisions about sample sizes. The probability that no nonconforming items are chosen, denoted as \( P(X = 0) \), is a complementary probability in this context.
The binomial distribution offers a model to estimate these probabilities. The probability of not choosing any flawed items from a sample must be less than or equal to 0.10 (10%). Using the formula \((0.99)^n \leq 0.10\), where 0.99 represents the probability of picking a conforming item, we set up an equation to solve for \( n \), the number of items to sample.
To solve \((0.99)^n \leq 0.10\), logarithms help by transforming the multiplicative inequality into an additive one, making it easier to calculate the required sample size.

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