/*! 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 67 The space shuttle flight control... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The space shuttle flight control system called Primary Avionics Software Set (PASS) uses four independent computers working in parallel. At each critical step, the computers "vote" to determine the appropriate step. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is \(0.0001 .\) Let \(X\) denote the number of computers that vote for a left roll when a right roll is appropriate. What are the mean and variance of \(X ?\)

Short Answer

Expert verified
Mean: 0.0004, Variance: 0.00039996.

Step by step solution

01

Define the Distribution

The problem involves determining the number of components that fail in a certain way (vote for a left roll instead of a right roll). Given that there are 4 independent computers each with a probability of error \( p = 0.0001 \), the distribution of \( X \), the number of computers voting incorrectly, follows a binomial distribution with parameters \( n = 4 \) and \( p = 0.0001 \).
02

Calculate the Mean

The mean of a binomial distribution \( X \) with \( n \) trials and probability of success \( p \) is calculated using the formula: \[ \mu = np \]. Substituting the given values, \( \mu = 4 \times 0.0001 = 0.0004 \).
03

Calculate the Variance

The variance of a binomial distribution \( X \) is given by \( \sigma^2 = np(1-p) \). Using the given values, compute the variance: \( \sigma^2 = 4 \times 0.0001 \times (1 - 0.0001) \). This calculation simplifies to \( \sigma^2 \approx 4 \times 0.0001 \times 0.9999 = 0.00039996 \).

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.

Understanding Probability in Binomial Distribution
Probability in the context of a binomial distribution refers to the chance that a certain event will occur during a series of trials. Each trial can be a success or a failure, and the trials are independent. In our scenario, a single trial consists of a computer making an error, like voting for a left roll when it should have voted for a right roll.
This probability is consistent across all trials, as each computer has the same fixed chance to err, which is given as 0.0001.
- **Independent Trials:** Each computer operates independently, meaning the error of one doesn't affect the others. - **Success and Failure:** A success could be the computer voting correctly, and a failure the incorrect vote occurring.
Given these characteristics, a probability of success (or correctness) and failure (or error) must sum up to 1. So, if the error probability is 0.0001, the success probability is 1 - 0.0001 = 0.9999. This fixed probability is crucial, as it allows us to model this scenario with a binomial distribution.
Calculating the Mean in a Binomial Distribution
The mean of a binomial distribution provides the expected number of successes or particular events within a series of trials. For a binomial distribution characterized by two parameters: the number of trials () and the probability of success (), the mean is calculated using the formula: \[ \mu = np \]
Here, represents the number of computers (4 in this case), and is the probability of error (0.0001). So, the mean or expected number of computers that will vote incorrectly is:
\[ \mu = 4 \times 0.0001 = 0.0004 \]
Despite its name, the mean doesn't necessarily imply that this number will occur every time; instead, it indicates the average result you'd expect over numerous repetitions of this voting process.
This calculation shows that on average, you would anticipate 0.0004 computers to vote incorrectly when subjected to the same conditions repeatedly.
Calculating Variance in a Binomial Distribution
Variance measures the extent to which the outcomes from a binomial distribution will vary from the mean. The higher the variance, the more spread out the values are in a distribution. This is crucial in predicting the reliability of systems like the space shuttle computers.In a binomial distribution, the variance is calculated as follows:\[ \sigma^2 = np(1-p) \]
Where represents the number of trials, and is the probability of the event happening (the error). For our case, = 4 and = 0.0001, so:\[ \sigma^2 = 4 \times 0.0001 \times (1 - 0.0001) = 0.00039996 \]
This variance tells us about the variability in the number of computers that may vote incorrectly. A smaller variance indicates that the number of errors is consistently close to the mean, pointing to predictability in the voting behavior. Understanding variance supports the reliability analysis of systems because it provides insights into how much the data spreads around the expected value or mean.

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

Derive the formula for the mean and standard deviation of a discrete uniform random variable over the range of integers \(a, a+1, \ldots b\).

When a computer disk manufacturer tests a disk, it writes to the disk and then tests it using a certifier. The certifier counts the number of missing pulses or errors. The number of errors on a test area on a disk has a Poisson distribution with \(\lambda=0.2\). (a) What is the expected number of errors per test area? (b) What percentage of test areas have two or fewer errors?

Suppose that \(X\) has a hypergeometric distribution with \(N=100, n=4,\) and \(K=20 .\) Determine the following: (a) \(P(X=1)\) (b) \(P(X=6)\) (c) \(P(X=4)\) (d) Mean and variance of \(X\)

Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is one star per 16 cubic light-years. (a) What is the probability of two or more stars in 16 cubic light-years? (b) How many cubic light-years of space must be studied so that the probability of one or more stars exceeds \(0.95 ?\)

Saguaro cacti are large cacti indigenous to the southwestern United States and Mexico. Assume that the number of saguaro cacti in a region follows a Poisson distribution with a mean of 280 per square kilometer. Determine the following: (a) Mean number of cacti per 10,000 square meters. (b) Probability of no cacti in 10,000 square meters. (c) Area of a region such that the probability of at least two cacti in the region is \(0.9 .\)

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.