91Ó°ÊÓ

A manufacturer of flashlight batteries wishes to control the quality of its product by rejecting any lot in which the proportion of batteries having unacceptable voltage appears to be too high. To this end, out of each large lot ( 10,000 batteries), 25 will be selected and tested. If at least 5 of these generate an unacceptable voltage, the entire lot will be rejected. What is the probability that a lot will be rejected if a. \(5 \%\) of the batteries in the lot have unacceptable voltages? b. \(10 \%\) of the batteries in the lot have unacceptable voltages? c. \(20 \%\) of the batteries in the lot have unacceptable voltages? d. What would happen to the probabilities in parts (a) - (c) if the critical rejection number were increased from 5 to 6 ?

Step by step solution

01

Understand the Problem

We need to find the probability that out of 25 batteries sampled from a large lot, at least 5 of them have unacceptable voltage. This is equivalent to finding the probability that 5 or more batteries in the sample are defective, given certain probabilities of defects in the entire lot.

02

Define the Random Variable

Let the random variable \( X \) represent the number of batteries with unacceptable voltage in a sample of 25 batteries. Here, \( X \) follows a binomial distribution with parameters \( n = 25 \) (the number of trials or experiments) and \( p \) (the probability of success i.e., a battery having unacceptable voltage).

03

Compute Probability for Part (a)

For part (a) where \( p = 0.05 \):Use the formula for binomial probability: \[ P(X \, \geq \, 5) = 1 - P(X < 5) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right) \]Compute each term using the binomial probability formula:\[ P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \text{ for } k = 0, 1, 2, 3, 4 \]

04

Compute Probability for Part (b)

For part (b) where \( p = 0.10 \):Similarly, calculate:\[ P(X \, \geq \, 5) = 1 - P(X < 5) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right) \]Compute this using the binomial probability formula.

05

Compute Probability for Part (c)

For part (c) where \( p = 0.20 \):Calculate:\[ P(X \, \geq \, 5) = 1 - P(X < 5) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right) \]Use the binomial formula to compute each probability.

06

Investigation When Critical Rejection Number is 6

For all parts (a) to (c), calculate the probability when at least 6 batteries are defective, use:\[ P(X \, \geq \, 6) = 1 - P(X < 6) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \right) \]Re-calculate each part using the modified criteria.

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Key Concepts

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

Binomial Distribution

The concept of a binomial distribution is crucial in understanding how probabilities are calculated in quality control scenarios, like in the flashlight battery example. In a binomial distribution, you have a fixed number of trials, each with two possible outcomes: success or failure.

• For the flashlight batteries, a trial corresponds to testing one battery.
• A 'success' in this context could be a battery having unacceptable voltage, while a 'failure' is a normal voltage.

The parameters of a binomial distribution are:

• \( n \): the total number of trials.
• \( p \): the probability of success on an individual trial.

In this case, you test 25 batteries; hence \( n = 25 \). The probability \( p \) is the defect rate, for example, \( p = 0.05 \) if 5% of the batteries are defective.

Probability Calculation

Calculating the probability of a certain number of successes in a binomial distribution involves using the formula: \[ P(X = k) = \binom{n}{k} p^k(1-p)^{n-k} \]where:

• \( \binom{n}{k} \) is the binomial coefficient, indicating the number of ways to choose \( k \) successes out of \( n \) trials.
• \( p^k \) is the probability of getting exactly \( k \) successes.
• \( (1-p)^{n-k} \) is the probability of the remaining \( (n-k) \) trials resulting in failures.

To find the probability that at least 5 batteries have unacceptable voltage (i.e., \( P(X \geq 5) \)), we need to sum probabilities from \( P(X=0) \) to \( P(X=4) \) and subtract the total from 1:\[ P(X \geq 5) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right) \]This involves calculating each term individually. When the defect rate changes, so does the \( p \), affecting each computation.

Critical Rejection Number

The critical rejection number is the threshold number of defective items that triggers rejection of a batch. The manufacturer sets this number to ensure quality assurance by deciding the highest atomically acceptable number of defects. For the flashlight batteries example:

• If 5 or more batteries out of 25 show high voltage, the entire batch is rejected.

This number heavily influences the probability calculations. If it is raised to 6, you need to compute:\[ P(X \ge 6) = 1 - \left( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \right) \]Increasing the critical rejection number generally reduces the probability of the batch being rejected, as more defective items are allowed before a rejection occurs. This adjustment is critical for balancing product quality with manufacturing yield.