/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 A manufacturer of semiconductor ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturer of semiconductor devices takes a random sample of 100 chips and tests them, classifying each chip as defective or nondefective. Let \(X_{i}=0\) if the chip is nondefective and \(X_{i}=1\) if the chip is defective. The sample fraction defective is $$\hat{P}=\frac{X_{1}+X_{2}+\cdots+X_{100}}{100}$$ What is the sampling distribution of the random variable \(\hat{P} ?\)

Short Answer

Expert verified
The sampling distribution of \(\hat{P}\) is approximately normal: \(N\left(p, \frac{p(1-p)}{100}\right)\).

Step by step solution

01

Understand the problem

We need to find the sampling distribution of the sample fraction defective, denoted as \(\hat{P}\). \(\hat{P}\) is the proportion of defective chips in a sample size of 100.
02

Define variables and model

Each chip is tested to be either nondefective \((0)\) or defective \((1)\). Hence, we have a binomial model where each \(X_i\) has a probability \(p\) of being defective.
03

Calculate expected value and variance

For \(\hat{P}\), the expected value \(E(\hat{P}) = p\) and the variance \(Var(\hat{P}) = \frac{p(1-p)}{n}\) where \(n = 100\).
04

Apply Central Limit Theorem

For large \(n\), the distribution of \(\hat{P}\) approximates a normal distribution by the Central Limit Theorem, even if each \(X_i\) is binary.
05

Conclusion about the sampling distribution

Therefore, the sampling distribution of \(\hat{P}\) is approximately \(N\left(p, \frac{p(1-p)}{100}\right)\), given that \(n=100\) is sufficiently large for this normal approximation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Binomial Model
When dealing with a situation where each outcome is binary, such as defective or nondefective chips, the **Binomial Model** is an ideal framework. In this scenario, each trial is independent, and there are only two possible outcomes — either successful (defective) or not (nondefective). This binary nature fits perfectly into the binomial distribution.
* The probability of success on a single trial is denoted by \( p \).
* For our case, if a chip is defective, it is considered a success.
  • Each chip test is an independent event.
  • The number of trials here is 100, which makes our calculations robust.
Understanding this model is crucial for determining the expected behavior when testing a large batch of semiconductors.
The total number of defects (successes) can be expressed as a sum of random variables, \( X_1 + X_2 + \cdots + X_{100} \), where each \( X_i \) follows a Bernoulli distribution. This sum defines our experiment in the binomial context.
Central Limit Theorem
The **Central Limit Theorem (CLT)** is a fundamental principle in statistics that assures us of stability and predictability in our distribution outputs, even if the underlying data originate from a non-normal distribution. The CLT states that, given a sufficiently large sample size, the sampling distribution of the sample mean (or proportion, in our case) will approximate a normal distribution.
* For our exercise with 100 chips, each outcome is binary, following a binomial framework.
* However, because \( n = 100 \) is large enough, the distribution of the sample proportion, \( \hat{P} \), becomes approximately normal.
  • This normal approximation allows us to formally describe the distribution with specific mean and variance values.
  • The beauty of CLT lies in providing a snapshot of predictability and reliability in data.
Thus, even as each \( X_i \) is binary and seemingly erratic, CLT provides a smooth, bell-shaped curve for the sampling distribution of \( \hat{P} \). This makes statistical inference feasible, aiding in real-world applications like our semiconductor testing.
Expected Value and Variance
Understanding **Expected Value and Variance** helps us grasp the average behavior and fluctuation potential of our random variables. For the sample proportion, \( \hat{P} \), three main considerations are:
1. **Expected Value (Mean):** This is the average outcome we set to anticipate. For \( \hat{P} \), it simplifies to \( E(\hat{P}) = p \). So, on average, the proportion of defective chips in our tests aligns with the actual defect probability, \( p \).
2. **Variance:** This indicates how much variation or spread we expect around the mean. For our sample proportion, it is \( Var(\hat{P}) = \frac{p(1-p)}{n} \), where \( n \) is 100 in this context.
  • Variance shrinks as \( n \) increases, offering more precise results with larger samples.
  • With a smaller variance, the statistic (here, \( \hat{P} \)) remains closer to its expected value.
The expected value gives us reassurance on what to anticipate, while variance informs about potential deviations, enhancing our overall prediction framework for the manufacturing quality check.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of \(n=49\) customers is observed. Find the probability that the average time waiting in line for these customers is (a) Less than 10 minutes (b) Between 5 and 10 minutes (c) Less than 6 minutes

The Rayleigh distribution has probability density function $$f(x)=\frac{x}{\theta} e^{-x^{2} / 2 \theta}, \quad x>0, \quad 0<\theta<\infty$$ (a) It can be shown that \(E\left(X^{2}\right)=2 \theta\). Use this information to construct an unbiased estimator for \(\theta\). (b) Find the maximum likelihood estimator of \(\theta\). Compare your answer to part (a). (c) Use the invariance property of the maximum likelihood estimator to find the maximum likelihood estimator of the median of the Raleigh distribution.

Two different plasma etchers in a semiconductor factory have the same mean etch rate \(\mu .\) However, machine 1 is newer than machine 2 and consequently has smaller variability in etch rate. We know that the variance of etch rate for machine 1 is \(\sigma_{1}^{2}\) and for machine 2 it is \(\sigma_{2}^{2}=a \sigma_{1}^{2}\). Suppose that we have \(n_{1}\) independent observations on etch rate from machine 1 and \(n_{2}\) independent observations on etch rate from machine 2 (a) Show that \(\hat{\mu}=\alpha \bar{X}_{1}+(1-\alpha) \bar{X}_{2}\) is an unbiased estimator of \(\mu\) for any value of \(\alpha\) between 0 and 1 . (b) Find the standard error of the point estimate of \(\mu\) in part (a). (c) What value of \(\alpha\) would minimize the standard error of the point estimate of \(\mu ?\) (d) Suppose that \(a=4\) and \(n_{1}=2 n_{2}\). What value of \(\alpha\) would you select to minimize the standard error of the point estimate of \(\mu\) ? How "bad" would it be to arbitrarily choose \(\alpha=0.5\) in this case?

A random sample of size \(n=16\) is taken from a normal population with \(\mu=40\) and \(\sigma^{2}=5 .\) Find the probability that the sample mean is less than or equal to \(37 .\)

When the population has a normal distribution, the estimator $$\begin{aligned}\hat{\sigma}=\operatorname{median}\left(\left|X_{1}-\bar{X}\right|,\left|X_{2}\bar{X}\right|\right.\\\&\left.\cdots,\left|X_{n}-\bar{X}\right|\right) / 0.6745\end{aligned}$$ is sometimes used to estimate the population standard deviation. This estimator is more robust to outliers than the usual sample standard deviation and usually does not differ much from \(S\) when there are no unusual observations. (a) Calculate \(\hat{\sigma}\) and \(S\) for the data \(10,12,9,14,18,15,\) and 16 . (b) Replace the first observation in the sample (10) with 50 and recalculate both \(S\) and \(\hat{\sigma}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.