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 7 booked passengers from New York to Los Angeles, exactly 6 show up? Round your answer to three decimal places.
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 7 booked passengers from New York to Los Angeles, exactly 6 show up? Round your answer to three decimal places.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Probability and Statistics: Airline Passenger Boarding
#### 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 7 booked passengers from New York to Los Angeles, exactly 6 show up?
**Instruction:**
Round your answer to three decimal places.
---
In this scenario, to find the required probability, we use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( P(X = k) \) is the probability of k successes in n trials,
- \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose k successes from n trials,
- \( p \) is the success probability on a single trial (0.73 in this case since 73% of passengers show up),
- \( n \) is the number of trials (7 in this case),
- \( k \) is the number of successes we are interested in (6 in this case).
**Calculation Steps:**
1. **Determine the binomial coefficient**: \[% \binom{7}{6} = \frac{7!}{6!(7-6)!} = 7 \]
2. **Substitute the values into the binomial formula**:
\[ P(X = 6) = 7 \cdot (0.73)^6 \cdot (1-0.73)^{7-6} \]
3. **Calculate the probabilities**:
- \( (0.73)^6 \approx 0.198 \)
- \( (1-0.73) \approx 0.27 \)
4. **Combine the results**:
\[ P(X = 6) = 7 \cdot 0.198 \cdot 0.27 \approx 0.374 \]
Hence, the probability that exactly 6 out of 7 booked passengers will show up is approximately 0.374 when rounded to three decimal places.
Note: The use of a calculator or statistical software is recommended to minimize arithmetic errors and ensure accuracy.
---
#### Important Concepts:
- **Binomial Distribution**: A binomial distribution is appropriate for](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01405fe2-be5e-45a6-bebc-d67f8dba8499%2F7096d621-525f-4ada-a6d7-19648752b3b0%2F4mhb7ra_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability and Statistics: Airline Passenger Boarding
#### 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 7 booked passengers from New York to Los Angeles, exactly 6 show up?
**Instruction:**
Round your answer to three decimal places.
---
In this scenario, to find the required probability, we use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( P(X = k) \) is the probability of k successes in n trials,
- \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose k successes from n trials,
- \( p \) is the success probability on a single trial (0.73 in this case since 73% of passengers show up),
- \( n \) is the number of trials (7 in this case),
- \( k \) is the number of successes we are interested in (6 in this case).
**Calculation Steps:**
1. **Determine the binomial coefficient**: \[% \binom{7}{6} = \frac{7!}{6!(7-6)!} = 7 \]
2. **Substitute the values into the binomial formula**:
\[ P(X = 6) = 7 \cdot (0.73)^6 \cdot (1-0.73)^{7-6} \]
3. **Calculate the probabilities**:
- \( (0.73)^6 \approx 0.198 \)
- \( (1-0.73) \approx 0.27 \)
4. **Combine the results**:
\[ P(X = 6) = 7 \cdot 0.198 \cdot 0.27 \approx 0.374 \]
Hence, the probability that exactly 6 out of 7 booked passengers will show up is approximately 0.374 when rounded to three decimal places.
Note: The use of a calculator or statistical software is recommended to minimize arithmetic errors and ensure accuracy.
---
#### Important Concepts:
- **Binomial Distribution**: A binomial distribution is appropriate for
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