Problem 231 - Applied Statistics and Probability for Engineers

Question

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 is 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 1 or more washers are thicker than the target is at least \(0.90 ?\)

Step-by-Step Solution

Answer

You need to select a minimum of 5 washers for both parts (a) and (b).

Step 1: Understanding the Problem

We need to determine the minimum number of washers, \( n \), to select such that the probability that none of them exceed the target is less than \(0.10\). In the second part, we need the probability that at least one exceeds the target to be at least \(0.90\). Since 60% of washers exceed the target, we'll set this up using binomial probabilities.

Step 2: Probability Formula Introduction

The probability that a single washer does not exceed the target thickness is \(1 - 0.60 = 0.40\). Therefore, the probability that all \(n\) washers do not exceed is \(0.40^n\). We need to solve \(0.40^n < 0.10\) for part (a), and by complementary probability, \(1 - 0.40^n \geq 0.90\) for part (b).

Step 3: Solving Part (a)

Set up the inequality \(0.40^n < 0.10\) and solve for \(n\). Taking the logarithm of both sides, we get \(\log(0.40^n) < \log(0.10)\). Simplifying, \(n \log(0.40) < \log(0.10)\). Divide both sides by \(\log(0.40)\), remembering that the logarithm of a fraction is negative, changing the inequality direction: \(n > \frac{\log(0.10)}{\log(0.40)}\). Calculate to find \(n \approx 4.78\), we round up to get \(n = 5\).

Step 4: Solving Part (b)

We want \(1 - 0.40^n \geq 0.90\), equivalent to \(0.40^n \leq 0.10\). Reuse the results from part (a), which showed \(n > 4.78\), confirming that \(n = 5\) satisfies \(0.40^n \leq 0.10\). Therefore, the minimum number of washers is also \(n = 5\).

Key Concepts

Binomial Probability

The concept of binomial probability is crucial when dealing with situations that have two possible outcomes. For instance, in the washer exercise, a washer can either exceed the target thickness or not. Such problems are well-suited to the binomial probability model.

• A binomial experiment consists of a fixed number of independent trials.
• Each trial results in a success or a failure.
• The probability of success is the same for each trial.

In the washer example:- Each washer selection is a trial.- There are two outcomes: either the washer exceeds the target (success) or it does not (failure).- The probability of a washer exceeding the target, or success, is 0.60.To determine the probability that a certain number of washers exceed the target thickness, the binomial probability formula is used:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:- \( n \) is the total number of trials or washers picked.- \( k \) is the number of successes (washers exceeding the target).- \( p \) is the probability of success on a single trial.- \( \binom{n}{k} \) is the binomial coefficient.

Logarithmic Inequalities

Logarithmic inequalities are a helpful mathematical tool used to solve scenarios where you need to determine a range of values that allows a certain inequality to hold. In the exercise, they are particularly useful in transforming exponential expressions such as \(0.40^n < 0.10\). The logarithmic properties play an essential role:

• The logarithm of a number decreases as the base raised to an exponent.
• Logarithms convert products into sums, powers into products, and divisions into differences.

For the given exercise steps:- We transformed the inequality \(0.40^n < 0.10\) by applying logarithms to both sides.- Using the property \( \log(a^b) = b \log(a) \), the left side becomes \( n \log(0.40) \).- This allows us to isolate \( n \) by finding \( n > \frac{\log(0.10)}{\log(0.40)} \).This transformation turns complex multiplicative relationships into simpler linear expressions which can be easily evaluated to find the minimum number of trials needed in probabilistic scenarios.

Sampling with Replacement

Sampling with replacement means that each sample or selection is made independently, allowing the sample pool to return to its initial state after each pick. This method ensures that the probability of selecting any item does not change from one trial to the next. In the context of the washer example: - Sampling with replacement assumes each washer is returned to the batch after checking if it exceeds the target. - This ensures that probability remains constant for each washer selection, aligning with the requirements of a binomial distribution. Some key points about sampling with replacement include:

• Each selection is independent.
• The probability remains constant across trials or selections.
• It's ideal for large populations or infinite sample spaces.

With sampling with replacement, we maintain the assumptions necessary for the binomial probability calculations, ensuring that the statistical models used yield accurate and reliable results.