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Suppose that a lot of washers is large enough that it can be assumed that the sampling is done with replacement. Assume that \(60 \%\) of the washers exceed the target thickness. (a) What is the minimum number of washers that need to be selected so that the probability that none are thicker than the target is less than \(0.10 ?\) (b) What is the minimum number of washers that need to be selected so that the probability that one or more washers are thicker than the target is at least \(0.90 ?\)

Short Answer

Expert verified
Minimum 3 washers for both parts.

Step by step solution

01

Understanding the Problem

We are given that 60% of washers exceed the target thickness. We need to find the minimum number of washers to ensure a probability of less than 0.10 for none exceeding the target (for part a) and at least 0.90 for one or more exceeding the target (for part b).
02

Establishing Probability Formula

The probability that a single washer is not thicker than the target is 40% or 0.40. If you select "n" washers, the probability that none are thicker than the target is \((0.40)^n\).
03

Setting Up Equation for Part (a)

We need \((0.40)^n < 0.10\). This inequality will help us find the minimum number of washers, "n".
04

Solving the Inequality for Part (a)

Take the logarithm of both sides to solve for n: \[\log((0.40)^n) < \log(0.10) \]This simplifies to:\[n \cdot \log(0.40) < \log(0.10)\]Divide by \(\log(0.40)\):\[n > \frac{\log(0.10)}{\log(0.40)}\]Calculate using a calculator: \(n > 2.72\). Therefore, the minimum integer n is 3.
05

Setting Up and Solving for Part (b)

The probability that one or more washers exceed the target is \(1 - (0.40)^n\). We want this probability to be at least 0.90, thus:\[1 - (0.40)^n \geq 0.90\]Rearrange to:\[(0.40)^n \leq 0.10\] This is the same condition as in case (a), thus n is 3.

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Key Concepts

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

Sampling with Replacement
Sampling with replacement is a common approach in probability theory where each item selected from the population is put back before the next one is drawn. This ensures that each draw is independent and the probabilities do not change. Imagine having a big jar of marbles: when you pick one, note its type, and then return it to the jar before picking another. This is how sampling with replacement works.
This method is crucial when you have a large set, like the washers in our exercise. Here, the assumption that the number of washers is "large enough" suggests that each pick does not alter the overall makeup of the group. Therefore, each washer has the same probability of being picked every time.
  • In this exercise, 60% of washers exceed the target thickness, staying constant due to replacement.
  • This means each draw is independent, allowing us to apply certain probability computations easily, like multiplying probabilities of individual events.
Understanding this principle is essential for accurately calculating probabilities and is often used when assessing defects or quality in manufacturing scenarios.
Inequality Solving
Inequality solving is a mathematical process used to find the values of variables that satisfy a given inequality. In the exercise, we solve inequalities to find out how many samples are needed to reach a specific probability.
For Part (a) of the exercise, we faced the inequality \((0.40)^n < 0.10\).This inequality helps us understand the requirements for the number of washers needed so that none exceed the thickness.
  • This involves using logarithms to isolate the variable, "n". By applying \\log\((\(0.40)^n\) < \log\(0.10\)\), we can simplify and solve it.
  • An inequality indicates a condition where values must consistently meet a threshold, which can often apply in quality control or statistical confidence levels.
By solving these inequalities, we determine that the minimum number of washers required is 3. This process demonstrates how inequalities help in making decisions based on statistical probability thresholds.
Logarithms
Logarithms are a mathematical tool used to solve equations and inequalities involving exponential relationships. They are especially valuable when working with probabilities, as seen in the exercise.
Here, we used logarithms to solve inequities:
  • When we had the inequality \\((0.40)^n < 0.10\), directly solving for \(n\) using logarithms simplifies the equation to \\[n \cdot \log(0.40) < \log(0.10)\].
  • By dividing both sides by \(\log(0.40)\), we arrive at the expression \(n > \frac{\log(0.10)}{\log(0.40)}\).
  • This use of logarithms allows us to solve for \(n\) efficiently and determine the minimum number of washers.
Logarithms transform multiplicative relationships into additive ones, which are often easier to handle, making them a powerful tool in probability and many other areas of mathematics.

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Most popular questions from this chapter

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