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Magazine Chapter 4: Problem 98
The manufacturing of semiconductor chips produces \(2 \%\) defective chips. Assume that the chips are independent and that a lot contains 1000 chips. Approximate the following probabilities: (a) More than 25 chips are defective. (b) Between 20 and 30 chips are defective.
Short Answer
Expert verified
(a) Approximately 0.1075; (b) Approximately 0.44.
Step by step solution
01
Understanding the Problem
We are given that the probability of a chip being defective is 0.02, or 2%, and we have a total of 1000 chips. We need to find probabilities related to the number of defective chips in a lot.
02
Identify the Distribution
The number of defective chips follows a binomial distribution since we have a fixed number of trials (1000 chips), independent trials, and a constant probability of success (a chip being defective). Thus, it is described by \( X \sim B(n = 1000, p = 0.02) \).
03
Approximate with Normal Distribution
Since \( n \) is large and \( p \) is not extremely small or close to 1, we can use the normal approximation to the binomial distribution. The mean (\( \mu \)) and the variance (\( \sigma^2 \)) are given by: \( \mu = np = 1000 \times 0.02 = 20 \), and \( \sigma^2 = np(1-p) = 1000 \times 0.02 \times 0.98 = 19.6 \). The standard deviation is \( \sigma = \sqrt{19.6} \approx 4.43 \).
04
Calculate Probability for Part (a)
We want the probability of more than 25 defective chips, i.e., \( P(X > 25) \). We use the continuity correction: \( P(X > 25) \approx P(Y > 25.5) \), where \( Y \) is normally distributed with mean 20 and standard deviation 4.43. Calculate the \( z \)-score: \( z = \frac{25.5 - 20}{4.43} \approx 1.24 \). Then, use the standard normal distribution table to find \( P(Z > 1.24) \approx 0.1075 \).
05
Calculate Probability for Part (b)
We need the probability of having between 20 and 30 defective chips, i.e., \( P(20 < X < 30) \). We use continuity correction: \( P(20.5 < Y < 29.5) \). Calculate the \( z \)-scores for 20.5 and 29.5: \( z_{20.5} = \frac{20.5 - 20}{4.43} \approx 0.11 \) and \( z_{29.5} = \frac{29.5 - 20}{4.43} \approx 2.14 \). Using the standard normal distribution table, find \( P(0.11 < Z < 2.14) = P(Z < 2.14) - P(Z < 0.11) \approx 0.9838 - 0.5438 = 0.44 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial distribution
The binomial distribution is a fundamental probability distribution in statistics used to model the number of successes in a series of independent experiments, where each experiment has two possible outcomes. In the context of manufacturing semiconductors, a defective chip is one such outcome.
Characteristics of a binomial distribution include:
Characteristics of a binomial distribution include:
- A fixed number of trials, which in our exercise equals the 1000 chips produced.
- Each trial is independent, meaning the defect status of one chip does not affect another.
- A constant probability of success, here "success" refers to a chip being defective, which is 2% or 0.02.
Normal approximation
When dealing with a large number of trials, like the 1000 chip tests, the binomial distribution can be approximated by a normal distribution. This approximation is helpful in simplifying calculations and making computations more efficient.
To use a normal approximation:
To use a normal approximation:
- The number of trials, \( n \), should be large.
- The probability \( p \) shouldn't be too close to 0 or 1.
Manufacturing defects
Manufacturing defects often follow predictable statistical patterns, especially in large production processes like semiconductor chip manufacturing. By defining a defective chip as the event of interest, we can calculate probabilities of various defect rates occurring.
In practical terms:
In practical terms:
- Defective chips can arise from material flaws, faulty machinery, or operational errors.
- Understanding the frequency and likelihood of defects helps in improving manufacturing quality and efficiency.
Continuity correction
The continuity correction is vital when transitioning from discrete to continuous probability distributions, as it helps improve the accuracy of probability estimations. In this particular context, it adjusts a discrete binomial probability to approximate a continuous normal probability.
This adjustment involves:
This adjustment involves:
- Adding or subtracting 0.5 from the discrete variable when approximating.
- It often provides a better approximation because it accounts for the fact that a binomial distribution is discrete, whereas a normal distribution is continuous.
