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6.81 FlightView surveyed 2600 North American airline passengers and reported that approximately \(80 \%\) said that they carry a smartphone when they travel. Suppose that the actual percentage is \(80 \% .\) Consider randomly selecting six passengers and define the random variable \(x\) to be the number of the six selected passengers who travel with a smartphone. The probability distribution of \(x\) is the binomial distribution 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 travel with a smartphone. c. Calculate \(P(x \geq 4)\).

Short Answer

Expert verified
a. \(P(X = 4) \approx 0.24576\): The probability that exactly 4 out of 6 selected passengers travel with a smartphone is approximately 0.24576. b. \(P(X = 6) \approx 0.262144\): The probability that all 6 selected passengers travel with a smartphone is approximately 0.262144. c. \(P(X \geq 4) \approx 0.90112\): The probability that at least 4 of the 6 selected passengers travel with a smartphone is approximately 0.90112.

Step by step solution

01

Calculate p(4)

To find the probability P(X = 4), plug in the values for n, x, and p into the binomial probability formula: \[P(X = 4) = \binom{6}{4}(0.8)^4 (1 - 0.8)^{6-4}\] Use the combination formula for \(\binom{n}{x}\): \[\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = 15\] Now calculate the probability: \[P(X = 4) = 15(0.8)^4(0.2)^2 = 15(0.4096)(0.04) \approx 0.24576\] The probability that exactly 4 out of 6 selected passengers travel with a smartphone is approximately 0.24576.
02

Calculate p(6)

To find the probability P(X = 6), plug in the values for n, x, and p into the binomial probability formula: \[P(X = 6) = \binom{6}{6}(0.8)^6(1 - 0.8)^{6-6}\] Calculate the combination for \(\binom{n}{x}\): \[\binom{6}{6} = \frac{6!}{6!(6-6)!} = \frac{6!}{6!0!} = 1\] Now calculate the probability: \[P(X = 6) = 1(0.8)^6(0.2)^0 = 1(0.262144)(1) \approx 0.262144\] The probability that all 6 selected passengers travel with a smartphone is approximately 0.262144.
03

Calculate P(x ≥ 4)

To find the probability P(X ≥ 4), we will calculate the probabilities for x = 4, 5, and 6, and then add them together: \[P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)\] We have already found the values for P(X = 4) and P(X = 6) in Steps 1 and 2. Next, calculate P(X = 5): \[P(X = 5) = \binom{6}{5}(0.8)^5(1 - 0.8)^{6-5}\] Calculate the combination for \(\binom{n}{x}\): \[\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = 6\] Now calculate the probability: \[P(X = 5) = 6(0.8)^5(0.2)^1 = 6(0.32768)(0.2) \approx 0.393216\] Now add the probabilities together to find P(X ≥ 4): \[P(X \geq 4) = 0.24576 + 0.393216 + 0.262144 \approx 0.90112\] The probability that at least 4 of the 6 selected passengers travel with a smartphone is approximately 0.90112.

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Key Concepts

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

Binomial Probability Formula
Understanding the binomial probability formula is like having a roadmap to navigate the likelihood of certain outcomes within a set number of trials. It is applicable when the trials are independent, the number of trials is fixed, and each trial results in a 'success' or 'failure'.

In our exercise, where passengers may or may not carry a smartphone, the formula is expressed as: \[P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}\]Here, \(n\) represents the number of trials (or passengers), \(x\) is the number of successful outcomes we're interested in, \(p\) is the probability of success on an individual trial, and \((1-p)\) is the probability of failure. Additionally, the term \(\binom{n}{x}\) is the number of ways to choose \(x\) successes out of \(n\) trials, which leads us to the concept of combinations.
Probability Distribution
A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. Essentially, it's a comprehensive list of all outcomes and their associated probabilities.

The binomial distribution is a specific type of probability distribution with only two possible outcomes - often termed as 'success' and 'failure'. In our exercise, the binomial distribution is employed to understand the spread of probabilities across different numbers of smartphone-carrying passengers. Thus, it helps in predicting how likely it is to encounter a certain number of passengers with smartphones among the six randomly selected.
Combinations in Probability
Imagine selecting a team from a group of players. The number of ways you can do this is referred to as a combination. In probability, combinations allow us to calculate the likelihood of various outcomes when order doesn't matter.

The formula to find the number of combinations is given by: \[\binom{n}{x} = \frac{n!}{x!(n-x)!}\]This is a critical component of the binomial probability formula, determining how many ways \(x\) successes can occur in \(n\) trials. Our exercise involves computing combinations to evaluate the probabilities of 4, 5, or 6 passengers carrying a smartphone out of 6.
Probability Interpretation
Interpreting probability is the art of making sense of the numbers. It tells us how to understand these values in a real-world context.

In the exercise, when we calculate a probability such as \(p(4)\), we're essentially asking, 'What is the chance that exactly four out of six passengers carry a smartphone?' The answer, about 24.576%, gives us insight into the likelihood of this specific scenario. Similarly, when we explore probabilities for at least a certain number of successes, such as \(P(x \geq 4)\), we're aggregating multiple probabilities to understand a broader context, like 'What's the chance that at least four passengers are carrying a smartphone?' In this case, it's roughly 90.112%, indicating a high likelihood for this event to occur. Thus, interpreting these probabilities helps in decision-making and predicting outcomes in uncertain scenarios.

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Most popular questions from this chapter

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