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American Airlines claims that the average number of people who pay for in- flight moves, when the plane is fully loaded, is 42 with a standard deviation of 8. A sample of 36 fully loaded planes is taken. What is the probability that fewer than 38 people paid for the in- flight moves?
  • a)
    0.0013
  • b)
    0.125
  • c)
    0.012
  • d)
    0.2311
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
American Airlines claims that the average number of people who pay for...
Understanding the Problem
American Airlines claims that the average number of people paying for in-flight movies is 42, with a standard deviation of 8. We want to find the probability that fewer than 38 people pay when a sample of 36 planes is taken.
Key Parameters
- Population Mean (μ): 42
- Population Standard Deviation (σ): 8
- Sample Size (n): 36
Calculating the Sample Mean and Standard Deviation
- Sample Mean (x̄): 38
- Sample Standard Deviation (σx̄): σ / √n = 8 / √36 = 8 / 6 = 1.33
Finding the Z-Score
To calculate the probability, we first need to find the Z-score for the sample mean of 38:
- Z = (x̄ - μ) / σx̄
- Z = (38 - 42) / 1.33 = -3.01
Using the Z-Score
Now, we will look up the Z-score of -3.01 in the standard normal distribution table or use a calculator.
- The Z-score of -3.01 corresponds to a probability of approximately 0.0013.
Conclusion
Thus, the probability that fewer than 38 people paid for in-flight movies is:
- P(X < 38)="" ≈="" />
The correct answer is option 'A'. This low probability indicates that it is quite rare for fewer than 38 passengers to purchase in-flight movies on a fully loaded plane.
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American Airlines claims that the average number of people who pay for in- flight moves, when the plane is fully loaded, is 42 with a standard deviation of 8. A sample of 36 fully loaded planes is taken. What is the probability that fewer than 38 people paid for the in- flight moves?a)0.0013b)0.125c)0.012d)0.2311Correct answer is option 'A'. Can you explain this answer?
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