/*! 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 81 The Los Angeles Times (December ... [FREE SOLUTION] | 91Ó°ÊÓ

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The Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

Short Answer

Expert verified
The probability that exactly four passengers prefer to rest or sleep is represented by p(4). The probability that all six passengers prefer to rest or sleep is given by p(6). The probability that at least four passengers prefer to rest or sleep is given by \( P(x \geq 4) \). To find the exact probabilities, calculations need to be done as per the steps given.

Step by step solution

01

Calculation of p(4)

The formula for binomial distribution is: \( P(x) = C(n,x) * p^x * (1 - p)^(n-x) \) where C(n,x) is the combination of n items taken x at a time. To find p(4): substitute n=6, x=4 and p=0.8 into the binomial formula: \( P(4) = C(6,4) * 0.8^4 * 0.2^2 \)
02

Calculation of p(6)

To find p(6): substitute n=6, x=6 and p=0.8 into the binomial formula: \( P(6) = C(6,6) * 0.8^6 * 0.2^0 \)
03

Calculation of \(P(x \geq 4)\)

The probability that x is greater than or equal to 4, \( P(x \geq 4) \) can be computed by summing up probabilities: P(4), P(5) and P(6). So, we already have P(4) and P(6) from Steps 1 and 2. We calculate \( P(5) = C(6,5) * 0.8^5 * 0.2 \) and find \( P(x \geq 4) = P(4) + P(5) + P(6) \)

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Key Concepts

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

Probability Calculation
Calculating probability in the context of binomial distribution is an essential process that helps us understand the likelihood of various outcomes. The binomial distribution applies to situations where there are only two possible outcomes, such as success or failure.

To calculate probabilities using the binomial distribution, you use the formula: \[ P(x) = C(n, x) \cdot p^x \cdot (1 - p)^{n-x} \]where:
  • \( C(n, x) \) is the number of combinations of \( n \) items taken \( x \) at a time.
  • \( p \) is the probability of success.
  • \( 1 - p \) is the probability of failure.
  • \( n \) is the number of trials.
  • \( x \) is the number of successful trials.
To find the probability of exactly 4 passengers liking to rest or sleep from 6, substitute into the formula to calculate \( P(4) \), and for all 6 passengers, calculate \( P(6) \). These calculations help predict outcomes based on known probabilities. Following through these steps ensures a clear understanding of how each scenario's likelihood is derived.
Random Variable
A random variable is a fundamental concept in probability and statistics, representing a variable whose possible values are numerical outcomes of a random phenomenon. Let's explore the concept through an example in this exercise.

Here, our random variable \( x \) represents the number of passengers among six who prefer to sleep or rest. In a binomial distribution, the random variable is defined by two parameters: the number of trials \( n \) and the probability \( p \) of success on each trial. For our exercise, \( n = 6 \) (six passengers), and \( p = 0.8 \) (80% probability of each liking to rest or sleep).

Random variables can take on different values based on the possible outcomes of the experiment. By using the random variable \( x \), we assess different scenarios like finding probabilities for exactly 4, exactly 6, or at least 4 passengers liking to rest. Grasping this concept enables one to manage and analyze uncertain situations effectively.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, and it plays a crucial role in calculating probabilities for a binomial distribution. Understanding how to compute combinations is key to solving the exercise at hand.

In the binomial distribution formula, \( C(n, x) \) represents the number of ways to choose \( x \) successes from \( n \) trials. This is known as "combinations," and it is calculated as:\[C(n, x) = \frac{n!}{x!\,(n-x)!} \]

In our example, we compute combinations for different scenarios:
  • For \( P(4) \), it's \( C(6,4) \), determining how many ways 4 out of 6 passengers can prefer resting.
  • For \( P(6) \), it's \( C(6,6) \), representing the total ways all 6 passengers can like to rest.
  • Similarly, calculate \( C(6,5) \) for 5 out of 6 passengers.
Combinatorics provides a foundational tool for determining probabilities in various scenarios, adding depth to the analysis of random events.

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Most popular questions from this chapter

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An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

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