From experience, an airline knows that only 73% of the passengers booked on a flight from New York to Los Angeles actually board their flight. If this percentage is correct, what is the probability that, in a random sample of 4 booked passengers from New York to Los Angeles, exactly 2 show up? Round your answer to three decimal places. 0 S ? X
From experience, an airline knows that only 73% of the passengers booked on a flight from New York to Los Angeles actually board their flight. If this percentage is correct, what is the probability that, in a random sample of 4 booked passengers from New York to Los Angeles, exactly 2 show up? Round your answer to three decimal places. 0 S ? X
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 10CYU
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![### Probability of Passengers Showing Up for Their Flight
**Problem Statement:**
From experience, an airline knows that only 73% of the passengers booked on a flight from New York to Los Angeles actually board their flight. If this percentage is correct, what is the probability that, in a random sample of 4 booked passengers from New York to Los Angeles, exactly 2 show up?
**Instructions:**
Round your answer to three decimal places.
**Input Section:**
- A text box is provided to input your answer.
- There are also three buttons:
- "X": Likely to clear or reset your input.
- "↻": Likely to refresh or redo your previous action.
- "?": Likely to provide hints or help related to the question.
Consider this scenario to apply the binomial probability formula, where:
- \( n \) = 4 (number of trials, i.e., passengers)
- \( k \) = 2 (number of successes, i.e., passengers actually showing up)
- \( p \) = 0.73 (probability of success on a single trial)
- \( q \) = 1 - p = 0.27 (probability of failure on a single trial)
Use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
**Example Calculation:**
To find the probability of exactly 2 out of 4 passengers showing up:
\[ P(X = 2) = \binom{4}{2} (0.73)^2 (0.27)^2 \]
\[ P(X = 2) = \frac{4!}{2!(4-2)!} (0.5329) (0.0729) \]
\[ P(X = 2) = \frac{24}{4 \cdot 2} (0.5329) (0.0729) \]
\[ P(X = 2) = 6 (0.5329) (0.0729) \]
\[ P(X = 2) = 6 (0.0388) \]
\[ P(X = 2) = 0.2328 \]
After rounding to three decimal places, the probability is approximately 0.233.
Enter the rounded result in the provided text box, and use the buttons](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d7590ab-1e8b-4f01-9940-34be16d61a30%2F7db81ed7-159b-4ef7-8524-60800aaec777%2Fvchz6xg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability of Passengers Showing Up for Their Flight
**Problem Statement:**
From experience, an airline knows that only 73% of the passengers booked on a flight from New York to Los Angeles actually board their flight. If this percentage is correct, what is the probability that, in a random sample of 4 booked passengers from New York to Los Angeles, exactly 2 show up?
**Instructions:**
Round your answer to three decimal places.
**Input Section:**
- A text box is provided to input your answer.
- There are also three buttons:
- "X": Likely to clear or reset your input.
- "↻": Likely to refresh or redo your previous action.
- "?": Likely to provide hints or help related to the question.
Consider this scenario to apply the binomial probability formula, where:
- \( n \) = 4 (number of trials, i.e., passengers)
- \( k \) = 2 (number of successes, i.e., passengers actually showing up)
- \( p \) = 0.73 (probability of success on a single trial)
- \( q \) = 1 - p = 0.27 (probability of failure on a single trial)
Use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
**Example Calculation:**
To find the probability of exactly 2 out of 4 passengers showing up:
\[ P(X = 2) = \binom{4}{2} (0.73)^2 (0.27)^2 \]
\[ P(X = 2) = \frac{4!}{2!(4-2)!} (0.5329) (0.0729) \]
\[ P(X = 2) = \frac{24}{4 \cdot 2} (0.5329) (0.0729) \]
\[ P(X = 2) = 6 (0.5329) (0.0729) \]
\[ P(X = 2) = 6 (0.0388) \]
\[ P(X = 2) = 0.2328 \]
After rounding to three decimal places, the probability is approximately 0.233.
Enter the rounded result in the provided text box, and use the buttons
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