/*! 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 24 Approximately \(10 \%\) of the g... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 24

Approximately \(10 \%\) of the glass bottles coming off a production line have serious flaws in the glass. If two bottles are randomly selected, find the mean and variance of the number of bottles that have serious flaws.

Short Answer

Expert verified
Mean: 0.20; Variance: 0.18.

Step by step solution

01

Define Random Variable

Let the random variable \( X \) represent the number of bottles with serious flaws. Since we are selecting 2 bottles, \( X \) can take values 0, 1, or 2.
02

Identify Distribution Type

The scenario follows a Binomial Distribution because there are a fixed number of trials (2 bottles), each with two possible outcomes (flawed or not flawed). The probability of a flaw in a bottle is \( p = 0.10 \).
03

Use Binomial Distribution Formulas

For a Binomial Distribution \( X \sim B(n, p) \), where \( n \) is the number of trials and \( p \) is the probability of success, the mean \( \mu \) and variance \( \sigma^2 \) are given by \( \mu = np \) and \( \sigma^2 = np(1-p) \).
04

Calculate Mean

Substitute \( n = 2 \) and \( p = 0.10 \) into the formula for the mean: \[ \mu = np = 2 \times 0.10 = 0.20. \]
05

Calculate Variance

Substitute \( n = 2 \) and \( p = 0.10 \) into the formula for the variance: \[ \sigma^2 = np(1-p) = 2 \times 0.10 \times (1 - 0.10) = 2 \times 0.10 \times 0.90 = 0.18. \]

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 91影视!

Key Concepts

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

Random Variable
In statistics, a random variable is a concept that helps us understand the different outcomes of a random experiment. Imagine picking bottles randomly from a production line. A random variable assigns a numerical value to each possible outcome.
For our problem, let's denote the random variable as \( X \). \( X \) represents the number of bottles with serious flaws among the selected ones. Since we are selecting 2 bottles, the possible values that \( X \) can assume are 0, 1, or 2.
This random variable is discrete because it takes specific integer values rather than a continuum of values. Understanding the possible values of a random variable is crucial for further calculations, such as mean and variance.
Mean and Variance
Mean and variance are key parameters of a probability distribution, providing significant insights into the distribution's characteristics.
**Mean:** Often referred to as the expected value, the mean gives an overall expected outcome if the experiment is repeated many times. For a binomial distribution, the mean is calculated using the formula \( \mu = np \), where \( n \) represents the number of trials, and \( p \) is the probability of success. In our glass bottle example, \( n = 2 \) and \( p = 0.10 \). Calculating the mean, we have: \[ \mu = 2 \times 0.10 = 0.20 \]
**Variance:** This measures the spread or variability of the distribution. It tells us the degree to which the random variable outcomes deviate from the mean. For a binomial distribution, variance is computed as \( \sigma^2 = np(1-p) \). Hence for our case: \[ \sigma^2 = 2 \times 0.10 \times (1-0.10) = 0.18 \]
Understanding the mean and variance helps interpret how many bottles are likely to have flaws and how much variability we can expect in the number of flawed bottles.
Probability of Flaws
The probability of flaws refers to the likelihood of encountering flawed bottles in the selection process. In our scenario, this probability is fixed at \( p = 0.10 \) for each bottle, which means each bottle independently has a 10% chance of having serious flaws. The fixed probability applies to each trial in a Binomial distribution.
A Binomial distribution is structured around a fixed number of independent trials (in this case, 2 bottles), with each trial having just two possible outcomes: success (flawed) or failure (not flawed). By understanding the probability of success \( p \) and its complement \( 1-p \), we can determine the probability distribution's mean and variance, and make predictions about the number of flawed bottles.
By evaluating these probabilities and knowing the inherent variability through variance, one can manage expectations around quality control during production.

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

In the 18th century, the Chevalier de Mere asked Blaise Pascal to compare the probabilities of two events. Below, you will compute the probability of the two events that, prior to contrary gambling experience, were thought by de Mere to be equally likely. a. What is the probability of obtaining at least one 6 in four rolls of a fair die? Answer \(t\) b. If a pair of fair dice is tossed 24 times, what is the probability of at least one double six?

If \(Y\) has a geometric distribution with success probability \(p,\) show that $$P(Y=\text { an odd integer })=\frac{p}{1-q^{2}}$$.

The number of people entering the intensive care unit at a hospital on any single day possesses a Poisson distribution with a mean equal to five persons per day. a. What is the probability that the number of people entering the intensive care unit on a particular day is equal to 2 ? Is less than or equal to 2 ? b. Is it likely that \(Y\) will exceed \(10 ?\) Explain.

Given that we have already tossed a balanced coin ten times and obtained zero heads, what is the probability that we must toss it at least two more times to obtain the first head?

This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let \(Y\) be a random variable such that $$p(-1)=\frac{1}{18}, \quad p(0)=\frac{16}{18}, \quad p(1)=\frac{1}{18}$$ a. Show that \(E(Y)=0\) and \(V(Y)=1 / 9\) b. Use the probability distribution of \(Y\) to calculate \(P(|Y-\mu| \geq 3 \sigma) .\) Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when \(k=3\) *.c. In part.(b) we guaranteed \(E(Y)=0\) by placing all probability mass on the values \(-1,0,\) and1, with \(p(-1)=p(1) .\) The variance was controlled by the probabilities assigned to \(p(-1)\) and \(p(1) .\) Using this same basic idea, construct a probability distribution for a random variable \(X\) that will yield \(P\left(\left|X-\mu_{X}\right| \geq 2 \sigma_{X}\right)=1 / 4\) * d. If any \(k>1\) is specified, how can a random variable \(W\) be constructed so that \(P\left(\left|W-\mu_{W}\right| \geq k \sigma_{W}\right)=1 / k^{2} ?\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.