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The Los Angeles Times (December 13,1992 ) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

Short Answer

Expert verified
a. The probability \(P(4)\) gives the likelihood that exactly 4 out of 6 passengers will rest or sleep on a specific route. b. The probability \(P(6)\) gives the likelihood that all 6 passengers will rest or sleep on a specific route. c. The probability \(P(x \geq 4)\) gives the likelihood that 4 or more passengers will rest or sleep on a specific route. Detailed solutions can be found above.

Step by step solution

01

Identify Known Variables

Firstly, the parameters of the problem need to be identified: number of attempts \(n=6\) and probability of success \(\pi=0.8\).
02

Calculate Probability for 4 Passengers

Apply the binomial probability formula, where \(x\) is the number of successes (e.g., passengers who rest or sleep), which in this case is 4: \[P(4) = C(6,4) \cdot (0.8)^4 \cdot (0.2)^2\] where \(C(6,4)\) represents the combinations of 6 items taken 4 at a time.
03

Calculate Probability for 6 Passengers

Use the binomial probability formula again, but this time for \(x=6\): \[P(6) = C(6,6) \cdot (0.8)^6 \cdot (0.2)^0\]
04

Calculate Cumulative Probability

Now the probability that 4 or more passengers will rest or sleep has to be calculated. This is equivalent to the sum of the probabilities of 4, 5, and 6 successes: \[P(x \geq 4) = P(4) + P(5) + P(6)\] where \(P(5) = C(6,5) \cdot (0.8)^5 \cdot (0.2)^1\). By calculating these values an approximate solution can be found.

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Key Concepts

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

Probability Theory
Understanding probability theory is essential for tackling problems involving likelihoods, like those in binomial distributions. Probability provides us with a way of quantifying the chance of an event happening. If you throw a fair six-sided die, the probability of any given number turning up is \( \frac{1}{6} \). The same logic applies when calculating the probability of a certain number of airline passengers choosing to sleep.
Probability theory involves:
  • Defining possible outcomes.
  • Assigning probabilities to these outcomes.
  • Calculating the likelihood of events using various probability rules and theorems.
In the context of our problem, the probability formula for a binomial distribution helps in determining how likely it is for a certain number of people to choose to sleep. This is governed by known parameters like the total number of passengers and the probability that a single passenger will sleep. This approach makes use of probabilities calculated from the uniform responses of surveyed passengers, translated into predictions for different scenarios.
Random Variables
In probability, a random variable is a way to map outcomes from a chance experiment to numerical values. In simpler terms, it's a function that gives us a number to represent the outcome of a random process.
For example, when tossing a coin, you might let "heads" represent 0 and "tails" represent 1. Random variables can take on several values, and the likelihood of each value arising can typically be represented by a probability distribution.
In our specific airline example, the random variable is "\(x\)", denoting the number of passengers out of six who choose to rest or sleep. Because the number of people choosing to sleep is random—though it has known probabilities—the outcomes are modeled using the concept of a random variable.
Here, \(x\) follows a binomial distribution, where the number of trials is 6 and the probability of "success", defined as a passenger sleeping, is 0.8. The binomial distribution is particularly useful when each trial is independent, and there is a fixed probability for each trial, both of which apply to our flight scenario.
Combinatorics
Combinatorics is a branch of mathematics focused on counting and arranging. It plays a vital role in probability theory, especially in calculating probabilities in binomial distributions where combinations come into play.
To find out probabilities like \(P(4)\), the binomial coefficient \( C(6,4) \) is used. This coefficient tells us how many ways we can choose 4 passengers out of 6 to rest or sleep, without considering the order. It can be calculated using the formula:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where \( n! \) denotes factorial of \( n \). In the airplane exercise, this calculation allows us to calculate how many ways passengers can arrange themselves into sleepers and non-sleepers.
Thus, combinatorics provides a systematic way of counting possible configurations and is deeply linked to probability distributions.

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Most popular questions from this chapter

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs [1980]: 307-325.) Let \(x\) denote the moisture loss for a randomly selected package. a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

A local television station sells \(15-\mathrm{sec}, 30-\mathrm{sec}\), and 60 -sec advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{lrrr} x & 15 & 30 & 60 \\ p(x) & .1 & .3 & .6 \end{array} $$ a. Find the average length for commercials appearing on this station. b. If a 15 -sec spot sells for $$\$ 500$$, a 30 -sec spot for $$\$ 800$$, and a 60 -sec spot for $$\$ 1000$$, find the average amount paid for commercials appearing on this station. (Hint: Consider a new variable, \(y=\) cost, and then find the probability distribution and mean value of \(y .\) )

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(x\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

Suppose that \(90 \%\) of all registered California voters favor banning the release of information from exit polls in presidential elections until after the polls in California close. A random sample of 25 California voters is to be selected. a. What is the probability that more than 20 voters favor the ban? b. What is the probability that at least 20 voters favor the ban? c. What are the mean value and standard deviation of the number of voters who favor the ban? d. If fewer than 20 voters in the sample favor the ban, is this at odds with the assertion that (at least) \(90 \%\) of the populace favors the ban? (Hint: Consider \(P(x<20)\) when \(\pi=.9 .)\)

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

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